14.6 - Graphs of Velocity And Acceleration in SHM

In the previous section, we saw three functions:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{ A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$
In this section, we will see an analysis of their graphs.

For simplicity, we will assume that $\small{\phi = 0}$. So we are going to plot the following three functions:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{ A\cos(\omega t)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t)}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t)}
\\ \end{array}}$
It is shown in fig.14.29 below:

Fig.14.29

   ♦ x is plotted in red color
   ♦ v is plotted in yellow color
   ♦ a is plotted in green color
• The following assumptions are made:
   ♦ $\small{A}$ = 2.5 units
   ♦ $\small{T}$ = 6 s. So $\small{\omega = \frac{2 \pi}{T} = \frac{2 \pi}{6} = \frac{\pi}{3}}$

Based on the assumptions, we can calculate the three amplitudes:
1. Amplitude of displacement function = $\small{A}$ = 2.5 units. This is indicated by the two horizontal magenta dotted lines in the top most graph.  
2. Amplitude of velocity function = $\small{\omega A = \frac{2.5 \pi}{3}}$ = 2.62 units. This is indicated by the two horizontal magenta dotted lines in the middle graph.
3. Amplitude of acceleration function = $\small{\omega^2 A = \frac{2.5 \pi^2}{9}}$ = 2.74 units. This is indicated by the two horizontal magenta dotted lines in the bottom graph.

Let us see the connection between the graphs and the horizontal oscillating spring that we saw in fig.14.9 is section 14.1.
First we will consider displacement. It can be written in 4 steps:

1. The stop-watch is turned on at the instant when the mass is at the right extreme A. So at t=0, in red graph, abscissa is zero and ordinate is 2.5 (In our present case, We will avoid using the familiar words, x-coordinate and y-coordinate. This is because, along the horizontal axis, we are plotting t, not x. And along the vertical axis, we are plotting x, v and a, not y. See the definition for abscissa and ordinate here)

2. Let us solve the equation $\small{x(t) = A\cos(\omega t) = 0}$
• That is., we are solving $\small{2.5\cos\left(\frac{\pi t}{3}  \right)= 0}$
We get: t = 1.5, 4.5, 7.5, 10.5, . . .
• These are indicated by the intersection of the red graph with the vertical red and magenta dashed lines. Those intersection points lie on the horizontal axis.
• That means, for those intersection points, ordinate is zero. That means, displacement is zero. That means, those intersection points corresponds to the instants at which the mass is at the equilibrium position.

3. Let us solve the equation $\small{x(t) = A\cos(\omega t) = A}$
• That is., we are solving $\small{2.5\cos\left(\frac{\pi t}{3}  \right)= 2.5}$
We get: t = 0, 6, 12, . . .
• These are indicated by the intersection of the red graph with the vertical axis and the vertical white dashed lines. Those intersection points lie on the upper magenta horizontal dashed line.
• That means, for those intersection points, ordinate is 2.5. That means, displacement is 2.5. That means, those intersection points corresponds to the instants at which the mass is at maximum displacement on the +ve side. The mass is at the extreme right.

4. Let us solve the equation $\small{x(t) = A\cos(\omega t) = -A}$
• That is., we are solving $\small{2.5\cos\left(\frac{\pi t}{3}  \right)= -2.5}$
We get: t = 3, 9, 15, . . .
• These are indicated by the intersection of the red graph with the vertical green dashed lines. Those intersection points lie on the lower magenta horizontal dashed line.
• That means, for those intersection points, ordinate is −2.5. That means, displacement is −2.5. That means, those intersection points corresponds to the instants at which the mass is at maximum displacement on the −ve side. The mass is at extreme left.

---

Next we will consider velocity. It can be written in 3 steps:

1. Let us solve the equation $\small{v(t) = -\omega A\sin(\omega t) = 0}$
• That is., we are solving $\small{\frac{-2.5 \pi}{3}\sin\left(\frac{\pi t}{3}  \right)= 0}$
We get: t = 0, 3, 6, 9, 12, . . .
• These are indicated by the intersection of the yellow curve with:
   ♦ the vertical axis
   ♦ vertical green dashed lines
   ♦ vertical yellow dashed lines
Those intersection points lie on the horizontal axis.
• That means, for those intersection points, ordinate is zero. That means, velocity is zero. That means, those intersection points corresponds to the instants at which the mass is stationary.
• At those intersection points, we can deduce a connection between red and yellow curves.
   ♦ When t = 0, the mass is at +A. Velocity here is zero
   ♦ When t = 3, the mass is at −A. Velocity here is zero
   ♦ When t = 6, the mass is at +A. Velocity here is zero
   ♦ When t = 9, the mass is at −A. Velocity here is zero
   ♦ When t = 12, the mass is at +A. Velocity here is zero
• So at extreme points, velocity is zero.

2. Let us solve the equation $\small{v(t) = -\omega A\sin(\omega t) = \omega A}$
• That is., we are solving $\small{\frac{-2.5 \pi}{3}\sin\left(\frac{\pi t}{3}  \right)= \frac{2.5 \pi}{3}}$
We get: t = 4.5, 10.5, 16.5, . . .
• These are indicated by the intersection of the yellow curve with:
   ♦ vertical magenta dashed lines
Those intersection points lie on the upper magenta horizontal dashed line.
• That means, for those intersection points, ordinate is 2.62. That means, velocity is 2.62. That means, those intersection points corresponds to the instants at which the mass has the maximum +ve velocity.
• At those intersection points, we can deduce a connection between red and yellow curves.
   ♦ When t = 4.5, the mass is at the equilibrium position, and is traveling towards the right. Velocity here is maximum. Since the velocity is towards the right, it is a +ve velocity
   ♦ When t = 10.5, the mass is at the equilibrium position, and is traveling towards the right. Velocity here is maximum. Since the velocity is towards the right, it is a +ve velocity.
• So at equilibrium position, if the mass is traveling towards the right, then the velocity will be the maximum +ve velocity.

3. Let us solve the equation $\small{v(t) = -\omega A\sin(\omega t) = -\omega A}$
• That is., we are solving $\small{\frac{-2.5 \pi}{3}\sin\left(\frac{\pi t}{3}  \right)= \frac{-2.5 \pi}{3}}$
We get: t = 1.5, 7.5, 13.5, . . .
• These are indicated by the intersection of the yellow curve with:
   ♦ vertical red dashed lines
Those intersection points lie on the lower magenta horizontal dashed line.
• That means, for those intersection points, ordinate is −2.62. That means, velocity is −2.62. That means, those intersection points corresponds to the instants at which the mass has the maximum −ve velocity.
• At those intersection points, we can deduce a connection between red and yellow curves.
   ♦ When t = 1.5, the mass is at the equilibrium position, and is traveling towards the left. Velocity here is maximum. Since the velocity is towards the left, it is a −ve velocity
   ♦ When t = 7.5, the mass is at the equilibrium position, and is traveling towards the left. Velocity here is maximum. Since the velocity is towards the left, it is a −ve velocity.
• So at equilibrium position, if the mass is traveling towards the left, then the velocity will be the maximum −ve velocity.

---

Finally, we will consider acceleration. It can be written in 3 steps:

1. Let us solve the equation $\small{a(t) = -\omega^2 A\cos(\omega t) = 0}$
• That is., we are solving $\small{\frac{(-1)2.5^2 \pi}{9}\cos\left(\frac{\pi t}{3}  \right)= 0}$
We get: t = 1.5, 4.5, 7.5, 10.5, 13.5, . . .
• These are indicated by the intersection of the green curve with:
   ♦ vertical red dashed lines
   ♦ vertical magenta dashed lines
Those intersection points lie on the horizontal axis.
• That means, for those intersection points, ordinate is zero. That means, acceleration is zero. That means, those intersection points corresponds to the instants at which the acceleration is zero.
• At those intersection points, we can deduce a connection between red, yellow and green curves.
When t = 1.5,
   ♦ The mass is at equilibrium position
   ♦ Velocity here is maximum −ve
   ♦ Acceleration here is zero
When t = 4.5,
   ♦ The mass is at equilibrium position
   ♦ Velocity here is maximum +ve
   ♦ Acceleration here is zero
When t = 7.5,
   ♦ The mass is at equilibrium position
   ♦ Velocity here is maximum −ve
   ♦ Acceleration here is zero
When t = 10.5,
   ♦ The mass is at equilibrium position
   ♦ Velocity here is maximum +ve
   ♦ Acceleration here is zero
• So at equilibrium position, if the mass is traveling towards the right, then the velocity will be the maximum +ve velocity. And acceleration will be zero.

2. Let us solve the equation $\small{a(t) = -\omega^2 A\cos(\omega t) = \omega^2 A}$
• That is., we are solving $\small{\frac{(-1)2.5^2 \pi}{9}\cos\left(\frac{\pi t}{3}  \right)= \frac{2.5^2 \pi}{9}}$
We get: t = 3, 9, 15, . . .
• These are indicated by the intersection of the green curve with:
   ♦ vertical green dashed lines
• Those intersection points lie on the upper magenta horizontal dashed line.
• That means, for those intersection points, ordinate is 2.74. That means, acceleration is 2.74. That means, those intersection points corresponds to the instants at which the mass has the maximum +ve acceleration.
• At those intersection points, we can deduce a connection between red, yellow and green curves.
When t = 3,
   ♦ The mass is at −A
   ♦ Velocity here is zero
   ♦ Acceleration here is +ve maximum because, it is towards right
When t = 9,
   ♦ The mass is at −A
   ♦ Velocity here is zero
   ♦ Acceleration here is +ve maximum because, it is towards right
• So at the −ve extreme point, velocity is zero and acceleration is +ve maximum

3. Let us solve the equation $\small{a(t) = -\omega^2 A\cos(\omega t) = -\omega^2 A}$
• That is., we are solving $\small{\frac{(-1)2.5^2 \pi}{9}\cos\left(\frac{\pi t}{3}  \right)= \frac{(-1)2.5^2 \pi}{9}}$
We get: t = 0, 6, 12, . . .
• These are indicated by the intersection of the green curve with:
   ♦ vertical axis
   ♦ vertical white dashed lines
• Those intersection points lie on the lower magenta horizontal dashed line.
• That means, for those intersection points, ordinate is −2.74. That means, acceleration is −2.74. That means, those intersection points corresponds to the instants at which the mass has the maximum −ve acceleration.
• At those intersection points, we can deduce a connection between red, yellow and green curves.
When t = 0,
   ♦ The mass is at +A
   ♦ Velocity here is zero
   ♦ Acceleration here is −ve maximum because, it is towards left
When t = 6,
   ♦ The mass is at +A
   ♦ Velocity here is zero
   ♦ Acceleration here is −ve maximum because, it is towards left
When t = 12,
   ♦ The mass is at +A
   ♦ Velocity here is zero
   ♦ Acceleration here is −ve maximum because, it is towards left
• So at the +ve extreme point, velocity is zero and acceleration is −ve maximum

---

In general, we can write 3 points:
1. When the mass is at equilibrium position,
   ♦ velocity (+ve or −ve) is maximum
   ♦ acceleration is zero
This is indicated by the red and magenta vertical dashed lines
2. When the mass is at +A,
   ♦ velocity is  zero
   ♦ acceleration is maximum −ve
This is indicated by the vertical axis and white vertical dashed lines
3. When the mass is at −A,
   ♦ velocity is  zero
   ♦ acceleration is maximum +ve
This is indicated by the green vertical dashed lines

---

Now we will see a solved example

Solved example 14.5
A body oscillates with SHM according to the equation (in SI units) $\small{x = 5 \cos\left(2\pi t + \frac{\pi}{4} \right)}$.
At t = 1.5 s, calculate the (a) displacement (b) speed and (c) acceleration of the body.
Solution:
1. For any SHM, we can write three equations:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{ A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$

2. In our present case, we are given the first one as:
$\small{x = 5 \cos\left(2\pi t + \frac{\pi}{4} \right)}$

3. Comparing them, we can write:
A = 5 m, $\small{\omega = 2\pi}$ rad/s and $\small{\phi = \frac{\pi}{4}}$ rad

4. So for this problem, the three equations can be written as:
(a) Displacement:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {=}    &{ 5{\rm{(m)}}\cos\left[2\pi(\rm{rad/s}) t(\rm{s}) + \frac{\pi}{4}(\rm{rad}) \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {=}    &{5{\rm{(m)}}\cos\left[2\pi t(\rm{rad}) + \frac{\pi}{4}(\rm{rad}) \right]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {=}    &{5{\rm{(m)}}\cos\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ \end{array}}$
(b) velocity:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{v(t)}    & {=}    &{-2\pi(\rm{rad/s})5{\rm{(m)}}\sin\left[2\pi(\rm{rad/s}) t(\rm{s}) + \frac{\pi}{4}(\rm{rad}) \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {=}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ \end{array}}$

(c) acceleration:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a(t)}    & {=}    &{-[2\pi(\rm{rad/s})]^2~5{\rm{(m)}}\cos\left[2\pi(\rm{rad/s}) t(\rm{s}) + \frac{\pi}{4}(\rm{rad}) \right]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {=}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ \end{array}}$

4. So for this problem, the three equations are:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{5{\rm{(m)}}\cos\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[2\pi t + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ \end{array}}$

5. Now we can write the required quantities
(a) Displacement:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(1.5)}    & {~=~}    &{5{\rm{(m)}}\cos\big[\left[2\pi (1.5) + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{5{\rm{(m)}}\cos\big[\left[3\pi + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{5(m)\cos\big[\frac{13 \pi}{4} (\rm{rad})\big]}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{-3.535{\rm{(m)}}}
\\ \end{array}}$

• Recall that cosine of any angle is a mere number. It does not have any unit.

(b) velocity:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{v(1.5)}    & {~=~}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\left[2\pi(1.5) + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\left[3\pi + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-10\pi(\rm{rad/s}){\rm{(m)}}\sin\big[\frac{13 \pi}{4} (\rm{rad})\big]}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{22.214{\rm{(m/s)}}}
\\ \end{array}}$

• Recall that, sine of any angle is a mere number. It does not have any unit.

• Also recall that, radians is the ratio of two lengths. So it is a dimensionless quantity. Therefore, we do not write it as a unit in the final result. An angle measured in radians, is a mere number. Details here.

(c) acceleration:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a(1.5)}    & {~=~}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[2\pi (1.5) + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\left[3\pi + \frac{\pi}{4} \right](\rm{rad})\big]}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-20\pi^2(\rm{rad/s})^2{\rm{(m)}}\cos\big[\frac{13 \pi}{4} (\rm{rad})\big]}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{139.58{\rm{(m/s^2)}}}
\\ \end{array}}$

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In the next section, we will see the force law for SHM.

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14.5 - Velocity and Acceleration in Simple Harmonic Motion

In the previous section, we saw simple harmonic motion (S.H.M). In this section, we will see velocity and acceleration in S.H.M.

Velocity in simple harmonic motion

• In fig.14.25 below, the magenta sphere is in uniform circular motion.
   ♦ Radius of the circle is A
   ♦ Angular speed of the sphere is $\small{\omega}$

Fig.14.25

• We know that, the "projection of the sphere on the x-axis" will be in SHM between (−A,0) and (A,0). We want an expression which can be used to calculate the "velocity of this projection" at any instant t.
This can be done in 4 steps:

1. Let at any instant t, the position of the sphere be P.
We know that, at any instant, the tangential velocity of the sphere will be $\small{\omega\,A}$. This is indicated by the white tangential arrow at P, in the fig.14.25 above.

2. Now we drop two perpendicular green dashed lines, onto the x-axis.
• One through P and the other through the head of $\small{\omega\,A}$
• These two perpendicular lines will give the projection of $\small{\omega\,A}$ on the x-axis.

3. This projection is the velocity of the SHM at the instant t. It is denoted as $\small{v(t)}$. So our next aim is to find the expression for $\small{v(t)}$

4. Fig.14.26 below shows an enlarged view of the first quadrant.

Fig.14.26

• The white dotted line is parallel to the x-axis and it passes through P. So by the rule of alternate angles, angle between OP and the dotted line is $\small{(\omega t~+~\phi)}$
• The tangential velocity $\small{\omega\,A}$ is perpendicular to OP. So the angle between the tangential velocity and the dotted line will be $\small{\left[\frac{\pi}{2} - (\omega t + \phi) \right]}$
• So the horizontal component of the tangential velocity will be $\small{\omega\,A \cos\left[\frac{\pi}{2} - (\omega t + \phi) \right]\,=\,\omega\,A \sin (\omega t + \phi)}$
See identity 5 in the list of trigonometric identities.
• But in the final expression, we must include a −ve sign because, the horizontal component is acting towards the −ve side of the x-axis.
• Thus we get: $\small{v(t)\,=\,-\omega\,A \sin (\omega t + \phi)}$

---

Note:
In math classes, it is proved that, derivative of displacement, w.r.t time, will give velocity. This is true in the case of SHM also.

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{\frac{dx}{dt}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{v(t)}    & {~=~}    &{\frac{d\left[A\cos(\omega t\,+\,\phi) \right]}{dt}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ \end{array}}$

• This is the same expression that we obtained above

---

Acceleration in simple harmonic motion

• In fig.14.27 below, we consider the same magenta sphere again.
   ♦ Radius of the circle is A
   ♦ Angular speed of the sphere is $\small{\omega}$

Fig.14.27

• We know that, the "projection of the sphere on the x-axis" will be in SHM between (−A,0) and (A,0). We want an expression which can be used to calculate the "acceleration of this projection" at any instant t.
This can be done in steps:

1. Let at any instant t, the position of the sphere be P.
We know that, at any instant, the centripetal acceleration experienced by the sphere will be $\small{v^2\,A}$ or $\small{\omega^2\,A}$. This is indicated by the white radial arrow at P, in the fig.14.27 above.

2. Now we drop two perpendicular green dashed lines, onto the x-axis.
• One through P and the other through the head of $\small{\omega\,A}$
• These two perpendicular lines will give the projection of $\small{\omega^2\,A}$ on the x-axis.

3. This projection is the acceleration of the SHM at the instant t. It is denoted as $\small{a(t)}$. So our next aim is to find the expression for $\small{a(t)}$

4. Fig.14.28 below shows an enlarged view of the first quadrant.

Fig.14.28

• The white dotted line is parallel to the x-axis and it passes through the head of $\small{\omega^2\,A}$. So the angle between $\small{\omega^2\,A}$ and the dotted line is $\small{(\omega t~+~\phi)}$
• So the horizontal component of $\small{\omega^2\,A}$ will be $\small{\omega^2\,A \cos(\omega t + \phi)}$
• But in the final expression, we must include a −ve sign because, the horizontal component is acting towards the −ve side of the x-axis.
• Thus we get: $\small{a(t)\,=\,-\omega^2\,A \cos (\omega t + \phi)}$

---

Note:
In math classes, it is proved that, derivative of velocity, w.r.t time, will give acceleration. This is true in the case of SHM also.

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{a(t)}    & {~=~}    &{\frac{dv}{dt}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{a(t)}    & {~=~}    &{\frac{d\left[-A \omega\sin(\omega t\,+\,\phi) \right]}{dt}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$

• This is the same expression that we obtained above

---

Now we will see two important properties related to acceleration.
Property 1:
This can be explained in 4 steps:
1. We obtained: $\small{a(t)\,=\,-\omega^2\,A \cos (\omega t + \phi)}$
2. But in the previous section, we saw that: $\small{x(t)\,=\,A \cos (\omega t + \phi)}$
3. So we can write: $\small{a(t)\,=\,-\omega^2\,x(t)}$
4. Suppose that, $\small{\omega}$ is a constant. Then we can write:
acceleration is proportional to displacement.

Property 2:
This can be explained in steps:
1. In property 1, we saw that:
Acceleration = A constant  × displacement
2. But there is a −ve sign in the expression for acceleration. So we can write:
   ♦ If the displacement is −ve, acceleration will be +ve.
   ♦ If the displacement is +ve, acceleration will be −ve.

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Let us see how the two properties are applicable to the horizontal oscillating spring that we saw in fig.14.9 is section 14.1. It can be written as four cases.

Case 1
This can be written in 3 steps:
1. Consider the case when the mass is traveling from x=0 to x=A
In this case, the displacement will be +ve at any instant.
2. Since the displacement is +ve, acceleration will be −ve at any instant.
• That is, acceleration will be acting towards the −ve side of the x-axis
• That means, acceleration will be acting opposite to the direction of motion.
• That means, the mass will be experiencing deceleration.
• Indeed, due to this deceleration, the mass comes to a stop at x = A
3. Also, the deceleration experienced by the mass is not uniform.
• Greater the distance from the equilibrium position, greater will be the deceleration. That means, as the mass approaches x=A, it will be experiencing more and more deceleration.

Case 2
This can be written in 3 steps:
1. Consider the case when the mass is traveling from x=A to x=0
In this case, the displacement will be +ve at any instant.
2. Since the displacement is +ve, acceleration will be −ve at any instant.
• That is, acceleration will be acting towards the −ve side of the x-axis
• That means, acceleration will be acting in the direction of motion.
• Indeed, due to this acceleration, the mass will have the maximum possible velocity at x = 0
3. Also, the acceleration experienced by the mass is not uniform.
• Lesser the distance from the equilibrium position, lesser will be the acceleration. That means, as the mass approaches x = 0, it will be experiencing less and less acceleration.

Case 3
This is the case when the mass is traveling from x=0 to x=−A
The reader may write all the 3 steps in detail

Case 4
This is the case when the mass is traveling from x=−A to x=0
The reader may write all the 3 steps in detail

---

We have seen three functions:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(t)}    & {~=~}    &{ A\cos(\omega t\,+\,\phi)}
\\ {~\color{magenta}    2    }    &{}    &{v(t)}    & {~=~}    &{-\omega A\sin(\omega t\,+\,\phi)}
\\ {~\color{magenta}    3    }    &{}    &{a(t)}    & {~=~}    &{-\omega^2 A\cos(\omega t\,+\,\phi)}
\\ \end{array}}$
In the next section, we will see an analysis of their graphs

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Chapter 7.39 - More Solved examples on Rolling motion

In the previous section, we saw rolling along inclined planes. We saw a solved example also. In this section we will see some solved examples related to rolling motion in general.

Solved example 7.52
A disc rotating about its axis with angular speed $\mathbf\small{|\vec{\omega}_0|}$ is placed lightly (without any translational push) on a perfectly frictionless table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in fig. 7.148? Will the disc roll in the direction indicated ?

Fig.7.148

Solution:
1. Since there is no friction, the disc will not roll. It will only spin 
• This is because, there is no external force to cause linear motion 
• We have seen the details here.
2. The angular speed will remain the same $\mathbf\small{|\vec{\omega}_0|}$ because, there is no external torque
• When the disc spins with an angular velocity, every point on it will have a certain linear velocity
3. This linear velocity is given by the equation: $\mathbf\small{|\vec{v}|=|\vec{r}|\times|\vec{\omega}|}$
 Where
• $\mathbf\small{|\vec{v}|}$ is the linear velocity of the particle
• $\mathbf\small{|\vec{r}|}$ is the distance of the particle from the center
• $\mathbf\small{|\vec{\omega}|}$ is the angular velocity with which the object rotates about the axis passing through the center
4. So in our present case, we have:
$\mathbf\small{|\vec{v}_A|=R\times|\vec{\omega}_0|}$ (towards right)
$\mathbf\small{|\vec{v}_B|=R\times|\vec{\omega}_0|}$ (towards left)
$\mathbf\small{|\vec{v}_C|=0.5R\times|\vec{\omega}_0|}$ (towards right)

Solved example 7.53
Explain why friction is necessary to make the disc in fig.7.148 roll in the direction indicated
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling
begins
(b) What is the force of friction after perfect rolling begins
Solution:
• If there is no friction, the disc will not roll. It will only spin. 
• This is because, there is no external force to cause linear motion
• We have seen the details here
Part (a):
1. The angular velocity is shown in the clockwise direction
• So the disc will move towards the right
2. Initially there is no perfect rolling. There will be slipping
• The friction will be kinetic in nature
• So the direction of the frictional force will be opposite to the direction of motion. That is., towards the left
3. So before perfect rolling, the frictional torque will be acting in the clockwise direction
Part (b):
1. This is a self propagating disc
• Once the perfect rolling begins, the point of contact with the ground will be at rest
2. An external force of friction will come into play only if the disc tries to accelerate
• Since our disc is not accelerating, the frictional force will be zero

Solved example 7.54
A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Solution:
1. At the bottom of the inclined plane, the potential energy is zero
2. Calculation of kinetic energy at bottom:
(i) Rotational kinetic energy 
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(\frac{m\,R^2}{2}\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{4}}$ 
• Note that for a solid cylinder, $\mathbf\small{I=\frac{m\,R^2}{2}}$
(ii) Translational kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
(iii) So total kinetic energy = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{4}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{3m|\vec{v}_{CM}|^2}{4}}$
3. Let the cylinder roll up to a height of 'h' m
• Then the potential energy at that height = mgh
4. Equating the two energies, we get: $\mathbf\small{mgh=\frac{3m|\vec{v}_{CM}|^2}{4}}$
$\mathbf\small{\Rightarrow h=\frac{3|\vec{v}_{CM}|^2}{4g}}$
• Substituting the values, we get: $\mathbf\small{\Rightarrow h=\frac{3(5)^2}{4(9.8)}=1.913\,\rm{m}}$
5. This 'h = 1.913 m' is the vertical distance. It is shown in fig.7.149 below:

Fig.7.149

• The sloping distance 's' can be calculated as follows:
$\mathbf\small{\sin \theta = \frac{h}{s}}$
$\mathbf\small{\Rightarrow \sin 30 =0.5= \frac{1.913}{s}}$
• Thus we get s = ^1.913⁄_0.5 = 3.826 m
Part (b):
1. The cylinder rolls back down from the height 1.913 m
• Since there is conservation of energy, we can write:
The linear velocity with which the center of the cylinder reaches the bottom will be the same 5 ms^-1
• So, for this rolling back motion, the final velocity v is 5 ms^-1
2. We have to find the acceleration with which this rolling back occurs
• For a disc and also for a solid cylinder, we have:
Eq.7.40: $\mathbf\small{|\vec{a}|= \frac{2|\vec{g}|\sin \theta}{3}}$
(We derived this equation in the previous section)
Substituting the values, we get: $\mathbf\small{|\vec{a}|= \frac{2(9.8)\sin 30}{3}=3.27}$ ms^-2
3. The initial velocity of this back ward rolling is zero
• We can use the equation: $\mathbf\small{|\vec{v}|=|\vec{u}|+|\vec{a}|t}$
• Substituting the values, we get: 5 = 0 + 3.27t
⇒ t = 1.53 s

Solved example 7.55
In the fig.7.150 below, a spring is compressed through a distance of 25 cm using a sphere and then the sphere is let go.

Fig.7.150

(a) What will be the velocity of the center of mass of the sphere when it is at the bottom of the inclined plane ?
(b) What will be the maximum distance traveled by the sphere along the incline ?
Mass of the sphere = 1.5 kg
Radius of the sphere = 8 cm
Spring constant k = 300 N m^-1
Solution:
Part (a):
1. When the spring is compressed, 'spring potential energy' is stored up in it
• When the spring is released, the energy becomes available for the sphere
2. Applying law of conservation of energy, we can write: 
Potential energy stored in the spring
= Rotational kinetic energy at B + translational kinetic energy at B
= Potential energy at C
3. Potential energy stored in the spring
= $\mathbf\small{\frac{1}{2}kx^2=\frac{1}{2}(300)(0.25)^2=9.375}$ J
4. Rotational kinetic energy at B
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(\frac{2m\,R^2}{5}\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{5}}$ 
• Note that for a sphere, $\mathbf\small{I=\frac{2m\,R^2}{5}}$
6. Translational kinetic energy at B
= $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
7. So we get:
Rotational kinetic energy at B + translational kinetic energy at B
= $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{5}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{7m|\vec{v}_{CM}|^2}{10}}$
Substituting the values, we get:
$\mathbf\small{\frac{7m|\vec{v}_{CM}|^2}{10}=\frac{7(1.5)|\vec{v}_{CM}|^2}{10}=1.05|\vec{v}_{CM}|^2}$
8. Equating this to the result in (3), we get: $\mathbf\small{1.05|\vec{v}_{CM}|^2=9.375}$
$\mathbf\small{\Rightarrow|\vec{v}_{CM}|=2.99}$ ms^-1
Part (b):
1. From (2) in part (a), we have:
Potential energy stored in the spring = Potential energy at C
2. From (3) in part (a), we have:
Potential energy stored in the spring = 9.375 J
3. Potential energy at C
= mgh = (1.5)(9.8)(h) = 14.7 h J
4. Equating the results in (2) and (3), we get: 9.375 = 14.7 h
h = 0.638 m
5. So s = ^0.638⁄_{sin 30} = 1.276 m

Solved example 7.56
A string is wound around a solid cylinder. The free end of the string is attached firmly to the ceiling as shown in fig.7.151 below. The cylinder is let go from a height of h from the ground. What will be the velocity of the CM of the cylinder when it just reaches the ground ?

Fig.7.151

Solution:
1. Applying law of conservation of energy, we can write: 
Potential energy of the cylinder at height h
= Rotational kinetic energy of cylinder at ground + translational kinetic energy of cylinder at ground
2. Potential energy of the cylinder when it is at a height h = mgh
3. Rotational kinetic energy at ground
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(\frac{m\,R^2}{2}\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{4}}$ 
• Note that for a solid cylinder, $\mathbf\small{I=\frac{m\,R^2}{2}}$
4. Translational kinetic energy at B
= $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
5. So we get:
Rotational kinetic energy at B + translational kinetic energy at B
= $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{4}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{3m|\vec{v}_{CM}|^2}{4}}$
6. Equating the energies, we get: $\mathbf\small{mgh=\frac{3m|\vec{v}_{CM}|^2}{4}}$
$\mathbf\small{\Rightarrow|\vec{v}_{CM}|=\sqrt{\frac{4gh}{3}}}$

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We have completed a discussion on rotational motion. In the next chapter, we will see gravitation

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Chapter 7.38 - Rolling along in Inclined planes

In the previous section, we saw the maximum force that can be applied on a wheel which is not able to self propagate. In this section we will see the case of objects rolling along inclined planes

1. Consider the disc shown in fig.7.147 (a) below

Fig.7.147(a)

• It has a mass m and radius R
• It is rolling down an inclined plane. The angle of inclination is θ
■ We want the disc to have 'pure rolling'. That is., there must not be any slip
2. This disc is not able to self propagate because there is no engine attached
• But it's self weight will cause it to move
• This will be clear when we draw all the force vectors acting on the wheel. They are shown in fig.b
We have:
(i) The weight $\mathbf\small{m|\vec{g}|}$ acting downwards from the center C
    ♦ The component $\mathbf\small{m|\vec{g}|\cos \theta}$ acts parallel to the inclined plane
    ♦ The component $\mathbf\small{m|\vec{g}|\sin \theta}$ acts perpendicular to the inclined plane
(ii) The normal reaction $\mathbf\small{|\vec{N}|}$ acting perpendicular to the plane
    ♦ This acts at the point of contact
    ♦ The magnitude is equal to $\mathbf\small{m|\vec{g}|\cos \theta}$  
(This force is not shown in the fig)
(iii) The frictional force  $\mathbf\small{|\vec{f_s}|=\mu_s|\vec{N}|=\mu_sm|\vec{g}|\cos \theta}$
    ♦ This force acts parallel to the inclined plane 
• Note that, we use the 𝝁_s instead of 𝝁_k because, we want pure rolling
• Also recall that, when there is no self propagation, the frictional force will be opposite to the direction of motion
3. Our aim is to find the maximum force acting on the disc so that no slipping occurs
• We will separate the calculations into two sets:
Set 1 comprises of the calculations related to rotation
Set 2 comprises of the calculations related to translation
4. Set 1:
(i) If there is a rolling motion, there has to be a torque
• This torque is provided by the frictional force $\mathbf\small{|\vec{f_s}|}$
• Because, all other forces are acting through the center of the wheel
• So we have: $\mathbf\small{|\vec{\tau}|=|\vec{f_s}|R=\mu_sm|\vec{g}|R \cos \theta}$
• $\mathbf\small{\mu_sm|\vec{g}|R \cos \theta}$ is the maximum frictional force possible
• So we can write: $\mathbf\small{|\vec{\tau}_{max}|=\mu_sm|\vec{g}|R \cos \theta}$
(ii) But $\mathbf\small{|\vec{\tau}|=I\,|\vec{\alpha}|}$
• Where:
    ♦ $\mathbf\small{I}$ is the moment of inertia of the wheel
    ♦ $\mathbf\small{|\vec{\alpha}|}$ is the angular acceleration of the wheel
(iii) For pure rolling, we have Eq.7.36 that we derived in the previous section: $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$
• Rearranging this, we get: $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{a}|}{R}}$
(iv) For our present case, we must consider the maximum allowable angular acceleration
• So we must denote it as $\mathbf\small{|\vec{\alpha}_{max}|}$
• So we have: $\mathbf\small{|\vec{\alpha}_{max}|=\frac{|\vec{a}_{max}|}{R}}$
(v) Thus from (ii), we get: $\mathbf\small{|\vec{\tau}_{max}|=I\,\frac{|\vec{a}_{max}|}{R}}$
• We know the I of the disc: $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\tau}_{max}|=\left(\frac{mR^2}{2}\right)\,\frac{|\vec{a}_{max}|}{R}=\frac{mR|\vec{a}_{max}|}{2}}$  
(vi) So the result in (i) becomes: $\mathbf\small{\frac{mR|\vec{a}_{max}|}{2}=\mu_sm|\vec{g}|R \cos \theta}$
• From this we get: $\mathbf\small{|\vec{a}_{max}|= 2\mu_s|\vec{g}|\cos \theta}$
5. Set 2:
 We have: Net force = mass × acceleration
• In our present case:
Net force = $\mathbf\small{m|\vec{g}|\sin \theta-|\vec{f_s}|}$
= $\mathbf\small{m|\vec{g}|\sin \theta-\mu_s\,m|\vec{g}|\cos \theta}$
= $\mathbf\small{m|\vec{a}_{max}|}$
6. The two sets are complete
• We take the result from Set 1 (vi)
• And apply it in the result from Set 2
• We get:
Net force 
= $\mathbf\small{m|\vec{g}|\sin \theta-\mu_s\,m|\vec{g}|\cos \theta}$
= $\mathbf\small{m\times 2\mu_s|\vec{g}|\cos \theta}$
$\mathbf\small{\Rightarrow m|\vec{g}|\sin \theta=3\mu_s\,m|\vec{g}|\cos \theta}$   
7. The quantity $\mathbf\small{|\vec{F}_{max}|}$ is the force that causes the disc to roll
• It is denoted as $\mathbf\small{|\vec{F}_{max}|}$  
8. Thus we get:
Eq.7.38: 
For a disc rolling down an inclined plane:
$\mathbf\small{|\vec{F}_{max}|}$
= $\mathbf\small{m|\vec{g}|\sin \theta}$
= $\mathbf\small{m\times2\mu_s|\vec{g}|\cos \theta+m\times\mu_s|\vec{g}|\cos \theta=3\mu_sm|\vec{g}|\cos \theta}$
= $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$
• Note that, this equation is valid only for a disc and a solid cylinder. For other types of objects, suitable equations must be derived using the appropriate values of I 
9. We see that $\mathbf\small{m|\vec{g}|\sin \theta}$ is the force that causes the disc to move
• This $\mathbf\small{m|\vec{g}|\sin \theta}$ must not exceed a 'certain value'
• This 'certain value' is: $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$
10. So we can write:
$\mathbf\small{m|\vec{g}|\sin \theta}$ should always be less than or equal to $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$
• That is: $\mathbf\small{m|\vec{g}|\sin \theta\;\;\leq \;\;3\mu_sm|\vec{g}|\cos \theta}$  
• Where
    ♦ $\mathbf\small{m|\vec{g}|\sin \theta}$ is the actual force acting
    ♦ $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$ is the maximum allowable force
11. If the left side ever becomes greater than the right side, we will have to make adjustments
• We cannot change mass because, it occurs on both sides
• We see that 𝝁_s occurs on the right side only
• So, if the left side increases beyond the limit, we can increase the right side by increasing 𝝁_s.
An example:
• m = 5 kg, θ = 35^o and 𝝁_s = 0.2
Then:
(i) Actual force acting = $\mathbf\small{m|\vec{g}|\sin \theta}$ = 28.1 N
(ii) Maximum allowable force = $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$ = 24.1 N
(iii) We have to increase (ii). For that, we will increase the 𝝁_s. This can be achieved by increasing the roughness of surface
(iv) But what must be the new value of 𝝁_s ?
• To find that, we consider the inequality:
$\mathbf\small{m|\vec{g}|\sin \theta\;\;\leq \;\;3\mu_sm|\vec{g}|\cos \theta}$
• If we multiply or divide both sides of an inequality by a same quantity, the 'inequality' will not change
(v) Here, we will divide both sides by $\mathbf\small{m|\vec{g}|\cos \theta}$. We get:
$\mathbf\small{\tan \theta\;\;\leq \;\;3\mu_s}$
• Dividing both sides by 3, we get:
$\mathbf\small{\frac{\tan \theta}{3}\;\;\leq \;\;\mu_s}$
(vi) That means 𝝁_s must be just greater than $\mathbf\small{\frac{\tan \theta}{3}}$
• If this condition is satisfied, the allowable force will be just sufficient
12. So for our present case, 𝝁_s must be just greater than (^{tan 35}⁄₃) = 0.233
• Let us provide a 𝝁_s of 0.25   
13. Let us check with the new data:
• m = 5 kg, θ = 35^o and 𝝁_s = 0.25
Then:
(i) Actual force acting = $\mathbf\small{m|\vec{g}|\sin \theta}$ = 28.1 N
(ii) Maximum allowable force = $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$ = 30.1 N
Note: In actual situations, various safety factors must be multiplied to the newly obtained 𝝁_s. Only then it will be considered safe. We will learn about them in higher classes 
14. If the 𝝁_s cannot be changed, then the only option is to change θ 
(i) To find the new value of θ, we consider the inequality:
$\mathbf\small{m|\vec{g}|\sin \theta\;\;\leq \;\;3\mu_sm|\vec{g}|\cos \theta}$
• If we multiply or divide both sides of an inequality by a same quantity, the 'inequality' will not change
(ii) Here, we will divide both sides by $\mathbf\small{m|\vec{g}|\cos \theta}$. We get:
$\mathbf\small{\tan \theta\;\;\leq \;\;3\mu_s}$
Thus we get:
7.39: $\mathbf\small{\theta\;\;\leq \;\;\tan^{-1}(3\mu_s)}$
• Note that, this relation is valid only for a disc and a solid cylinder. For other types of objects, suitable relations must be derived using the appropriate values of I
(iii) That means, θ should be brought down so that, it is just less than $\mathbf\small{tan^{-1}(3\mu_s)}$
(iv) If this condition is satisfied, the allowable force will be just sufficient
15. So for our present case, θ must be just less than [tan^-1 (3 × 0.2)] = 30.96^o
• Let us provide a θ of 30
16. Let us check with the new data:
• m = 5 kg, θ = 30^o and 𝝁_s = 0.2
Then:
(i) Actual force acting = $\mathbf\small{m|\vec{g}|\sin \theta}$ = 24.5 N
(ii) Maximum allowable force = $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$ = 25.5 N
Note: In actual situations, various safety factors must be multiplied to the newly obtained θ. Only then it will be considered safe. We will learn about them in higher classes

---

Let us write a summary of the above discussion. The summary can be written in 3 steps:

1. We have a disc rolling down an inclined plane
• From the given data, we are  able to find the following items:
(i) The maximum force that will be causing the motion:
$\mathbf\small{|\vec{F}_{max}|}$ = $\mathbf\small{m|\vec{g}|\sin \theta}$
(ii) The maximum allowable force:
$\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$
• Using the above two items, we can determine whether the disc will slip or not 
2. The condition at which the two items become equal is the 'limiting condition'
• At that condition, 'maximum allowable net force' will be acting on the disc
• This is given by: Net force = $\mathbf\small{m|\vec{g}|\sin \theta-|\vec{f_s}|}$
= $\mathbf\small{m|\vec{g}|\sin \theta-\mu_s\,m|\vec{g}|\cos \theta}$ 
3. That means, at that condition, the disc will be experiencing $\mathbf\small{|\vec{a}_{max}|}$ the 'maximum allowable linear acceleration' 
• And we know the method to find that maximum allowable linear acceleration:
$\mathbf\small{|\vec{a}_{max}|= 2\mu_s|\vec{g}|\cos \theta}$

---

1. But what if the disc is rolling well within the limit?

This will occur under the following conditions:
• 𝝁_s is greater than the limiting value
or
• θ is less than the limiting value   
2. In such a situation, the acceleration $\mathbf\small{|\vec{a}|}$  experienced by the disc will be less than $\mathbf\small{|\vec{a}_{max}|}$
3. In such a situation, the force ($\mathbf\small{m|\vec{g}|\sin \theta}$) which causes motion of the disc will be well within the safe limit
4. We want to know the $\mathbf\small{|\vec{a}|}$ in such a situation
5. This is a special situation
The reason for the situation becoming 'special' can be written in 6 steps:
(i) We usually calculate the static frictional force using 𝝁_s 
(ii) But such a calculation gives the 'maximum static frictional force which is possible'
(iii) When the disc is rolling well with in the safe limit, this maximum static frictional force will not be acting on it
(iv) Because, we know that, static friction is a self adjusting force
    ♦ It increases when the applied force increases
    ♦ It decreases when the applied force decreases
■ Here we experience the difference between rolling motion and sliding motion
• The sliding motion begins only when the full available frictional force is exceeded
• The rolling motion does not wait for 'the full available frictional force to be exceeded'
(v) In our present case, the applied force $\mathbf\small{m|\vec{g}|\sin \theta}$ is so less that, the full static frictional force is not mobilized
• So we do not know exactly how much static friction is acting  
• So we cannot use 𝝁_s. 
6. However, we can derive a formula, which does not involve the coefficient
We will separate the calculations into two sets:
• Set 1 comprises of the calculations related to rotation
• Set 2 comprises of the calculations related to translation
7. Set 1:
(i) $\mathbf\small{|\vec{\tau}|=I|\vec{\alpha}|=\frac{mR^2}{2}|\vec{\alpha}|=\frac{mR^2}{2}\times\frac{|\vec{a}|}{R}=\frac{mR|\vec{a}|}{2}}$
(ii) But $\mathbf\small{|\vec{\tau}|=\mu m|\vec{g}|\cos \theta \times R}$
(iii) Equating the results, we get: $\mathbf\small{|\vec{\tau}|=\mu\,m|\vec{g}|\cos \theta \times R=\frac{mR|\vec{a}|}{2}}$
$\mathbf\small{\Rightarrow \mu|\vec{g}|\cos \theta=\frac{|\vec{a}|}{2}}$
8. Set 2:
Net force = $\mathbf\small{m|\vec{g}|\sin \theta-\mu\;m|\vec{g}|\cos \theta=m|\vec{a}|}$
9. The two sets are complete
• Now we take the result from Set 1 (iii) 
• And apply it in the result from Set 2
We get:
Net force = $\mathbf\small{m|\vec{g}|\sin \theta-m\times\frac{|\vec{a}|}{2}=m|\vec{a}|}$
$\mathbf\small{\Rightarrow m|\vec{g}|\sin \theta=\frac{3m|\vec{a}|}{2}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{2|\vec{g}|\sin \theta}{3}}$
10. We can write:
When a disc is rolling down an incline with out slipping, the linear acceleration is given by:
Eq.7.40: $\mathbf\small{|\vec{a}|=\frac{2|\vec{g}|\sin \theta}{3}}$
• Note that, this equation is valid only for a disc and a solid cylinder. For other types of objects, suitable equations must be derived using the appropriate values of I

Solved example 7.51
A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30^o. The coefficient of static friction  𝝁_s = 0.25.
(a) How much is the force of friction acting on the cylinder ?
(b) What is the work done against friction during rolling ?
(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid,
and not roll perfectly ?
Solution:
• Both solid cylinder and disc have the same I. So the equations that we derived above are applicable for this problem
• First we need to know whether the maximum frictional force is mobilized or not
• For that we do the following 4 steps:
1. We will use the relation in Eq.7.38: For a disc rolling down an inclined plane:
$\mathbf\small{|\vec{F}_{max}|}$ = $\mathbf\small{m|\vec{g}|\sin \theta}$
• This maximum force must be less than or equal to $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta}$ 
2. Substituting the values we get:
$\mathbf\small{m|\vec{g}|\sin \theta=49\,\rm{N}}$
3. Similarly, we get: $\mathbf\small{3\mu_sm|\vec{g}|\cos \theta=63.65\,\rm{N}}$
4. Result in (2) is less than the result in (3)
• So it is safe. The cylinder will not slip. Also there is no need to mobilize the maximum available frictional force
• Now we can find the answers to the given questions
Part (a):
1. To find the actual acceleration, we use
Eq.7.40: $\mathbf\small{|\vec{a}|=\frac{2|\vec{g}|\sin \theta}{3}}$
• Substituting the values, we get: $\mathbf\small{|\vec{a}|=3.27\,\rm{m\,s^{-2}}}$
2. We have:
Net force = Force acting downwards along the slope - Frictional force
⇒ Net force = [$\mathbf\small{m|\vec{g}|\sin \theta}$ - Frictional force]
⇒ Net force = [49 - Frictional force]
3. But net force = Mass × acceleration = (10 × 3.27) = 32.7 N
4. So the result in (2) becomes:
32.7 = [49 - Frictional force]
⇒ Frictional force = 49 - 32.7 = 16.3 N
Part (b):
1. We have: Work done = Force × Displacement   
2. The frictional force acts on every particle on the periphery of the cylinder
3.  Consider the instant at which a particle experiences the frictional force
• At that instant, the displacement of that particle is zero
• At the next instant, the frictional force no longer acts on that particle
    ♦ Because, the succeeding particle takes up it's position
4. So displacement is always zero
5. So no work is done against the frictional force
Part (c):
• We will use the relation in 7.39: $\mathbf\small{\theta\;\;\leq \;\;\tan^{-1}(3\mu_s)}$
• Substituting the values we get: θ ≤ tan^-1 (3 × 0.25)
⇒ θ ≤ tan^-1 (0.75)
⇒ θ ≤ 36.87
• So if θ is greater than 36.87^o, the cylinder will begin to skid

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We have completed a discussion on rolling motion. In the next section, we will see a few more solved examples

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