In the previous section, we saw work done by a torque. In this section, we will see power
1. We have: $\mathbf\small{dW=|\vec{\tau}|\;d\theta}$
• This is the work done in the time interval during which, the body turns through dθ
• Let this time interval be dt
• So dW joules of work is done in a time interval of dt seconds
2. Thus we get:
• Power (P) produced by the torque
= Work done by the torque in 1 second
= $\mathbf\small{\frac{dW}{dt}=\frac{|\vec{\tau}|\;d\theta}{dt}=|\vec{\tau}|\left(\frac{d\theta}{dt}\right)}$
3. Consider the term $\mathbf\small{\left(\frac{d\theta}{dt}\right)}$
• Angular displacement is being divided by time. It will give angular velocity ω
• So the result in (2) becomes:
Eq.7.28: $\mathbf\small{P=|\vec{\tau}|\omega}$
1. We have: $\mathbf\small{dW=|\vec{\tau}|\;d\theta}$
• This is the work done in the time interval during which, the body turns through dθ
• Let this time interval be dt
• So dW joules of work is done in a time interval of dt seconds
2. Thus we get:
• Power (P) produced by the torque
= Work done by the torque in 1 second
= $\mathbf\small{\frac{dW}{dt}=\frac{|\vec{\tau}|\;d\theta}{dt}=|\vec{\tau}|\left(\frac{d\theta}{dt}\right)}$
3. Consider the term $\mathbf\small{\left(\frac{d\theta}{dt}\right)}$
• Angular displacement is being divided by time. It will give angular velocity ω
• So the result in (2) becomes:
Eq.7.28: $\mathbf\small{P=|\vec{\tau}|\omega}$
Now we will derive an interesting result related to rotational motion. But first, we will derive it for linear motion
1. A rigid body (of mass m) in linear motion, has an initial velocity of v1 ms-1
• A force F acts on it for a time interval of Δt seconds
♦ Direction of F is same as the direction of motion
• As a result, the velocity of the body increases to v2
• During the Δt seconds, the body undergoes a displacement of Δs
2. Work done by the force = Force × displacement = $\mathbf\small{F\times \Delta s}$
3. Applying work-energy theorem, we can write:
• If no work is lost against friction or air resistance, all the work done by the force will be utilized to increase the kinetic energy of the body (Details here)
• Also note that, in this case, the body is rigid. So there is no internal motion of particles. All the external work will indeed be utilized for increasing the kinetic energy of the body
4. Now, increase in kinetic energy = $\mathbf\small{0.5m\,v_2^2-0.5m\,v_1^2=0.5m(v_2^2-v_1^2)}$
• Equating this to the result in (2), we get: $\mathbf\small{0.5m(v_2^2-v_1^2)=F\times \Delta s}$
5. Dividing both sides by Δt, we get: $\mathbf\small{\frac{0.5m(v_2^2-v_1^2)}{\Delta t}=\frac{F\times \Delta s}{\Delta t}=F\times\frac{\Delta s}{\Delta t}}$
6. $\mathbf\small{\frac{\Delta s}{\Delta t}}$ normally gives velocity
• But in our present case, the velocity is not uniform
♦ This is because of the action of the force F
♦ Because of the F, the body will be moving with an acceleration
• So $\mathbf\small{\frac{\Delta s}{\Delta t}}$ will give us the average velocity
• That means: $\mathbf\small{\frac{\Delta s}{\Delta t}=\frac{v_1+v_2}{2}}$
7. So the result in (5) becomes: $\mathbf\small{\frac{0.5m(v_2^2-v_1^2)}{\Delta t}=F\frac{(v_1+v_2)}{2}}$
$\mathbf\small{\Rightarrow \frac{0.5m(v_2+v_1)(v_2-v_1)}{\Delta t}=F\frac{(v_1+v_2)}{2}=0.5F(v_1+v_2)}$
$\mathbf\small{\Rightarrow \frac{0.5m(v_2-v_1)}{\Delta t}=0.5F}$
$\mathbf\small{\Rightarrow \frac{m(v_2-v_1)}{\Delta t}=F}$
8. But $\mathbf\small{\frac{(v_2-v_1)}{\Delta t}}$ is the acceleration a
• So the result in (7) becomes: $\mathbf\small{m\,a=F}$
• This is Newton's second law of motion
1. A rigid body (of mass m) in linear motion, has an initial velocity of v1 ms-1
• A force F acts on it for a time interval of Δt seconds
♦ Direction of F is same as the direction of motion
• As a result, the velocity of the body increases to v2
• During the Δt seconds, the body undergoes a displacement of Δs
2. Work done by the force = Force × displacement = $\mathbf\small{F\times \Delta s}$
3. Applying work-energy theorem, we can write:
• If no work is lost against friction or air resistance, all the work done by the force will be utilized to increase the kinetic energy of the body (Details here)
• Also note that, in this case, the body is rigid. So there is no internal motion of particles. All the external work will indeed be utilized for increasing the kinetic energy of the body
4. Now, increase in kinetic energy = $\mathbf\small{0.5m\,v_2^2-0.5m\,v_1^2=0.5m(v_2^2-v_1^2)}$
• Equating this to the result in (2), we get: $\mathbf\small{0.5m(v_2^2-v_1^2)=F\times \Delta s}$
5. Dividing both sides by Δt, we get: $\mathbf\small{\frac{0.5m(v_2^2-v_1^2)}{\Delta t}=\frac{F\times \Delta s}{\Delta t}=F\times\frac{\Delta s}{\Delta t}}$
6. $\mathbf\small{\frac{\Delta s}{\Delta t}}$ normally gives velocity
• But in our present case, the velocity is not uniform
♦ This is because of the action of the force F
♦ Because of the F, the body will be moving with an acceleration
• So $\mathbf\small{\frac{\Delta s}{\Delta t}}$ will give us the average velocity
• That means: $\mathbf\small{\frac{\Delta s}{\Delta t}=\frac{v_1+v_2}{2}}$
7. So the result in (5) becomes: $\mathbf\small{\frac{0.5m(v_2^2-v_1^2)}{\Delta t}=F\frac{(v_1+v_2)}{2}}$
$\mathbf\small{\Rightarrow \frac{0.5m(v_2+v_1)(v_2-v_1)}{\Delta t}=F\frac{(v_1+v_2)}{2}=0.5F(v_1+v_2)}$
$\mathbf\small{\Rightarrow \frac{0.5m(v_2-v_1)}{\Delta t}=0.5F}$
$\mathbf\small{\Rightarrow \frac{m(v_2-v_1)}{\Delta t}=F}$
8. But $\mathbf\small{\frac{(v_2-v_1)}{\Delta t}}$ is the acceleration a
• So the result in (7) becomes: $\mathbf\small{m\,a=F}$
• This is Newton's second law of motion
Thus, starting with the 'work done', we reached Newton's second law. We did it in the case of linear motion. Let us see if it is possible for rotational motion also:
1. A rigid body in rotational motion, has an initial angular velocity of ω1 rad s-1
• The moment of inertia of the body about the axis of rotation is I
• A torque 𝝉 acts on it for a time duration of Δt seconds
♦ As a result, the angular velocity of the body increases to ω2
• During the Δt seconds, the body undergoes an angular displacement of Δθ
2. Work done by the torque
= Torque × angular displacement
= $\mathbf\small{\tau \times \Delta \theta}$
3. The body is rigid. So there is no internal motion of particles. All the external work will be utilized for increasing the kinetic energy of the body
• We have:
Kinetic energy of a rotating body = $\mathbf\small{\frac{1}{2}I \omega^2=0.5I\,\omega^2}$ (see Eq.7.26)
4. Now, increase in kinetic energy = $\mathbf\small{0.5I\,\omega_2^2-0.5I\,\omega_1^2=0.5I(\omega_2^2-\omega_1^2)}$
• Equating this to the result in (2), we get: $\mathbf\small{0.5I(\omega_2^2-\omega_1^2)=\tau \times \Delta \theta}$
5. Dividing both sides by Δt, we get: $\mathbf\small{\frac{0.5I(\omega_2^2-\omega_1^2)}{\Delta t}=\frac{\tau \times \Delta \theta}{\Delta t}=\tau \times\frac{\Delta \theta}{\Delta t}}$
6. $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$ normally gives angular velocity
• But in our present case, the angular velocity is not uniform
♦ This is because of the action of the torque 𝝉
♦ Because of the 𝝉, the body will be rotating with an acceleration
• So $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$ will give us the average angular velocity
• That means: $\mathbf\small{\frac{\Delta \theta}{\Delta t}=\frac{\omega_1+\omega_2}{2}}$
7. So the result in (5) becomes: $\mathbf\small{\frac{0.5I(\omega_2^2-\omega_1^2)}{\Delta t}=\tau\frac{(\omega_1+\omega_2)}{2}}$
$\mathbf\small{\Rightarrow \frac{0.5I(\omega_2+\omega_1)(\omega_2-\omega_1)}{\Delta t}=\tau\frac{(\omega_1+\omega_2)}{2}=0.5\tau(\omega_1+\omega_2)}$
$\mathbf\small{\Rightarrow \frac{0.5I(\omega_2-\omega_1)}{\Delta t}=0.5\tau}$
$\mathbf\small{\Rightarrow \frac{I(\omega_2-\omega_1)}{\Delta t}=\tau}$
8. But $\mathbf\small{\frac{(\omega_2-\omega_1)}{\Delta t}}$ is the angular acceleration 𝜶
• So the result in (7) becomes:
Eq.7.29: $\mathbf\small{I\,\alpha=\tau}$
9. '$\mathbf\small{I\,\alpha=\tau}$' is analogous to '$\mathbf\small{m\,a=F}$' of linear motion
• '$\mathbf\small{I\,\alpha=\tau}$' is called the Newton's second law for rotation about a fixed axis
• Thus, in the case of rotation (about a fixed axis) also, starting with the 'work done', we reached Newton's second law
1. A rigid body in rotational motion, has an initial angular velocity of ω1 rad s-1
• The moment of inertia of the body about the axis of rotation is I
• A torque 𝝉 acts on it for a time duration of Δt seconds
♦ As a result, the angular velocity of the body increases to ω2
• During the Δt seconds, the body undergoes an angular displacement of Δθ
2. Work done by the torque
= Torque × angular displacement
= $\mathbf\small{\tau \times \Delta \theta}$
3. The body is rigid. So there is no internal motion of particles. All the external work will be utilized for increasing the kinetic energy of the body
• We have:
Kinetic energy of a rotating body = $\mathbf\small{\frac{1}{2}I \omega^2=0.5I\,\omega^2}$ (see Eq.7.26)
4. Now, increase in kinetic energy = $\mathbf\small{0.5I\,\omega_2^2-0.5I\,\omega_1^2=0.5I(\omega_2^2-\omega_1^2)}$
• Equating this to the result in (2), we get: $\mathbf\small{0.5I(\omega_2^2-\omega_1^2)=\tau \times \Delta \theta}$
5. Dividing both sides by Δt, we get: $\mathbf\small{\frac{0.5I(\omega_2^2-\omega_1^2)}{\Delta t}=\frac{\tau \times \Delta \theta}{\Delta t}=\tau \times\frac{\Delta \theta}{\Delta t}}$
6. $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$ normally gives angular velocity
• But in our present case, the angular velocity is not uniform
♦ This is because of the action of the torque 𝝉
♦ Because of the 𝝉, the body will be rotating with an acceleration
• So $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$ will give us the average angular velocity
• That means: $\mathbf\small{\frac{\Delta \theta}{\Delta t}=\frac{\omega_1+\omega_2}{2}}$
7. So the result in (5) becomes: $\mathbf\small{\frac{0.5I(\omega_2^2-\omega_1^2)}{\Delta t}=\tau\frac{(\omega_1+\omega_2)}{2}}$
$\mathbf\small{\Rightarrow \frac{0.5I(\omega_2+\omega_1)(\omega_2-\omega_1)}{\Delta t}=\tau\frac{(\omega_1+\omega_2)}{2}=0.5\tau(\omega_1+\omega_2)}$
$\mathbf\small{\Rightarrow \frac{0.5I(\omega_2-\omega_1)}{\Delta t}=0.5\tau}$
$\mathbf\small{\Rightarrow \frac{I(\omega_2-\omega_1)}{\Delta t}=\tau}$
8. But $\mathbf\small{\frac{(\omega_2-\omega_1)}{\Delta t}}$ is the angular acceleration 𝜶
• So the result in (7) becomes:
Eq.7.29: $\mathbf\small{I\,\alpha=\tau}$
9. '$\mathbf\small{I\,\alpha=\tau}$' is analogous to '$\mathbf\small{m\,a=F}$' of linear motion
• '$\mathbf\small{I\,\alpha=\tau}$' is called the Newton's second law for rotation about a fixed axis
• Thus, in the case of rotation (about a fixed axis) also, starting with the 'work done', we reached Newton's second law
Now we will see some solved examples
Solved example 7.33
A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in the fig.7.135 below. The fly wheel is mounted on a horizontal axle with frictionless bearings
(a) Compute the angular acceleration of the wheel
(b) Find the work done by the pull when 2 m of the chord is unwound
(c) Find also the kinetic energy of the wheel at this point
(d) compare answers of parts (b) and (c)
Solution:
1. Moment of inertia (I) of the fly wheel = $\mathbf\small{\frac{MR^2}{2}=\frac{(20)(0.2)^2}{2}=0.4\,\text{kg m}^2}$
(I of a circular disc about a perpendicular axis at center)
2. Torque $\mathbf\small{\tau}$ acting on the wheel:
• The tension in the cord will pull a particle at the periphery of the wheel. So force on that particle will be 25 N
• This force will be tangential to the fly wheel
• We know this:
Perpendicular distance between a tangent from the center of circle (axis of rotation) = radius of the circle
• So we get:
$\mathbf\small{\tau}$ = Force × perpendicular distance from center
= Force × radius = 25 × 0.2 = 5 Nm
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting known values, we get: $\mathbf\small{5=0.4\,\alpha}$
$\mathbf\small{\Rightarrow \alpha=\frac{5}{0.4}=12.5\,\text{rad s}^{-2}}$
• This is the answer for part (a)
4. Perimeter of the fly wheel = 2𝞹R = 2𝞹(0.2) = 0.4𝞹 m
• So 0.4𝞹 m of the circumference will cover 2𝞹 radians
• Then 1 m of the circumference will cover: $\mathbf\small{\frac{2 \pi}{0.4 \pi}=5}$ radians
• So 2 m of the cord will cover: 10 radians
■ Thus, when 2 m of the cord is unwound, the fly wheel will turn through 10 rad
5. We have: Work done = 𝞽 × dθ = 5 × 10 = 50 joules
• This is the answer for part (b)
6. Change in kinetic energy = $\mathbf\small{0.5I(\omega_2^2-\omega_1^2)}$
• Given that, the fly wheel starts from rest. So $\mathbf\small{\omega_1}$ = 0
• Thus, change in kinetic energy = $\mathbf\small{0.5I \omega_2^2}$
7. We have: $\mathbf\small{\omega_2^2=\omega_1^2+2 \alpha \theta}$
• Substituting the values, we get: $\mathbf\small{\omega_2^2=0^2+2(12.5)(10)=250}$
8. Substituting the values in (6), we get:
• Change in kinetic energy = $\mathbf\small{0.5(0.4)(250)}$ = 50 joules
• This is the answer for part (c)
9. Comparing (b) and (c), we find that, the answers are the same
• That means:
Work done by the external torque = Increase in kinetic energy of the body
• We obtained this equality because, no work is lost against friction
• This is the answer for part (d)
Solved example 7.34
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
Solution:
1. Standard axis of symmetry of a cylinder is it's axis
• I of a hollow cylinder about it's axis = $\mathbf\small{MR^2}$
• I of a solid sphere about an axis passing through it's center = $\mathbf\small{\frac{2MR^2}{5}}$
(Significance of 'passing through center' can be seen here )
2. Given that, the torques acting are equal. Let us denote it as $\mathbf\small{\tau}$
• Given that the masses are same. Let us denote it as M
• Given that the radii are same. Let us denote it as R
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting the values for the hollow cylinder, we get: $\mathbf\small{\tau=MR^2\,\alpha_C}$
♦ Where $\mathbf\small{\alpha_C}$ is the angular acceleration of the cylinder
• Substituting the values for the solid sphere, we get: $\mathbf\small{\tau=\frac{2MR^2}{5}\,\alpha_S}$
♦ Where $\mathbf\small{\alpha_S}$ is the angular acceleration of the sphere
4. Since the torques are equal, we can equate them:
• $\mathbf\small{MR^2\,\alpha_C=\frac{2MR^2}{5}\,\alpha_S}$
$\mathbf\small{\Rightarrow \alpha_C=\frac{2}{5}\,\alpha_S}$
5. We see that, $\mathbf\small{\alpha_C}$ is only a fraction of $\mathbf\small{\alpha_S}$.
• That means $\mathbf\small{\alpha_C}$ is less than $\mathbf\small{\alpha_S}$
• So after a given time, the sphere will acquire a greater angular speed
Solved example 7.35
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
1. I of a hollow cylinder about it's axis = $\mathbf\small{MR^2}$
2. Torque $\mathbf\small{\tau}$ acting on the wheel:
• The tension in the rope will pull a particle at the periphery of the cylinder
• So force on that particle will be 30 N
• This force will be tangential to the cylinder
• Perpendicular distance between a tangent from the center of circle (axis of rotation) = radius of the circle
• So we get: $\mathbf\small{\tau}$ = Force × perpendicular distance from center
= Force × radius = 30 × 0.4 = 12 Nm
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting the values, we get: $\mathbf\small{12=(3)(0.4)^2\,\alpha}$
$\mathbf\small{\Rightarrow \alpha=}$ 25 rad s-2
• This is the answer for part (i)
4. The cylinder starts from rest
• So initial angular velocity $\mathbf\small{\omega_0=0}$
5. Let us find the angular velocity after any convenient interval of time, say 2 s
• We have: $\mathbf\small{\omega=\omega_0+\alpha \, t}$
• Substituting the values, we get: $\mathbf\small{\omega=0+(25) \, (2)}$ = 50 rad s-1
6. So at the instant when the stop watch shows 2 seconds, the cylinder will be rotating with an angular speed of 50 rad s-1
• Every particle in the cylinder will be rotating with the angular speed of 50 rad s-1 at that instant
• Any particle at the periphery will also be rotating with the angular speed of 50 rad s-1 at that instant
7. Now, we use the relation between linear velocity and angular velocity
• We have: $\mathbf\small{v=r\;\omega}$
• We apply it to a particle at the periphery: v = 0.4 × 50 = 20 ms-1.
• So, when the stop watch shows 2 seconds, any particle at the periphery will be moving with a linear speed of 20 ms-1.
8. Given that, there is no slip between the cylinder and the rope
• So the rope will also be moving with a linear speed of 20 ms-1
• We will use the relation $\mathbf\small{v=v_0+at}$
• The rope also started from rest. So $\mathbf\small{v_0=0}$
• Substituting the values, we get: 20= 0 + a × 2
⇒ a = 10 ms-2.
Solved example 7.33
A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in the fig.7.135 below. The fly wheel is mounted on a horizontal axle with frictionless bearings
(a) Compute the angular acceleration of the wheel
(b) Find the work done by the pull when 2 m of the chord is unwound
(c) Find also the kinetic energy of the wheel at this point
(d) compare answers of parts (b) and (c)
 |
| Fig.7.135 |
Solution:
1. Moment of inertia (I) of the fly wheel = $\mathbf\small{\frac{MR^2}{2}=\frac{(20)(0.2)^2}{2}=0.4\,\text{kg m}^2}$
(I of a circular disc about a perpendicular axis at center)
2. Torque $\mathbf\small{\tau}$ acting on the wheel:
• The tension in the cord will pull a particle at the periphery of the wheel. So force on that particle will be 25 N
• This force will be tangential to the fly wheel
• We know this:
Perpendicular distance between a tangent from the center of circle (axis of rotation) = radius of the circle
• So we get:
$\mathbf\small{\tau}$ = Force × perpendicular distance from center
= Force × radius = 25 × 0.2 = 5 Nm
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting known values, we get: $\mathbf\small{5=0.4\,\alpha}$
$\mathbf\small{\Rightarrow \alpha=\frac{5}{0.4}=12.5\,\text{rad s}^{-2}}$
• This is the answer for part (a)
4. Perimeter of the fly wheel = 2𝞹R = 2𝞹(0.2) = 0.4𝞹 m
• So 0.4𝞹 m of the circumference will cover 2𝞹 radians
• Then 1 m of the circumference will cover: $\mathbf\small{\frac{2 \pi}{0.4 \pi}=5}$ radians
• So 2 m of the cord will cover: 10 radians
■ Thus, when 2 m of the cord is unwound, the fly wheel will turn through 10 rad
5. We have: Work done = 𝞽 × dθ = 5 × 10 = 50 joules
• This is the answer for part (b)
6. Change in kinetic energy = $\mathbf\small{0.5I(\omega_2^2-\omega_1^2)}$
• Given that, the fly wheel starts from rest. So $\mathbf\small{\omega_1}$ = 0
• Thus, change in kinetic energy = $\mathbf\small{0.5I \omega_2^2}$
7. We have: $\mathbf\small{\omega_2^2=\omega_1^2+2 \alpha \theta}$
• Substituting the values, we get: $\mathbf\small{\omega_2^2=0^2+2(12.5)(10)=250}$
8. Substituting the values in (6), we get:
• Change in kinetic energy = $\mathbf\small{0.5(0.4)(250)}$ = 50 joules
• This is the answer for part (c)
9. Comparing (b) and (c), we find that, the answers are the same
• That means:
Work done by the external torque = Increase in kinetic energy of the body
• We obtained this equality because, no work is lost against friction
• This is the answer for part (d)
Solved example 7.34
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.
Solution:
1. Standard axis of symmetry of a cylinder is it's axis
• I of a hollow cylinder about it's axis = $\mathbf\small{MR^2}$
• I of a solid sphere about an axis passing through it's center = $\mathbf\small{\frac{2MR^2}{5}}$
(Significance of 'passing through center' can be seen here )
2. Given that, the torques acting are equal. Let us denote it as $\mathbf\small{\tau}$
• Given that the masses are same. Let us denote it as M
• Given that the radii are same. Let us denote it as R
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting the values for the hollow cylinder, we get: $\mathbf\small{\tau=MR^2\,\alpha_C}$
♦ Where $\mathbf\small{\alpha_C}$ is the angular acceleration of the cylinder
• Substituting the values for the solid sphere, we get: $\mathbf\small{\tau=\frac{2MR^2}{5}\,\alpha_S}$
♦ Where $\mathbf\small{\alpha_S}$ is the angular acceleration of the sphere
4. Since the torques are equal, we can equate them:
• $\mathbf\small{MR^2\,\alpha_C=\frac{2MR^2}{5}\,\alpha_S}$
$\mathbf\small{\Rightarrow \alpha_C=\frac{2}{5}\,\alpha_S}$
5. We see that, $\mathbf\small{\alpha_C}$ is only a fraction of $\mathbf\small{\alpha_S}$.
• That means $\mathbf\small{\alpha_C}$ is less than $\mathbf\small{\alpha_S}$
• So after a given time, the sphere will acquire a greater angular speed
Solved example 7.35
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping.
Solution:
1. I of a hollow cylinder about it's axis = $\mathbf\small{MR^2}$
2. Torque $\mathbf\small{\tau}$ acting on the wheel:
• The tension in the rope will pull a particle at the periphery of the cylinder
• So force on that particle will be 30 N
• This force will be tangential to the cylinder
• Perpendicular distance between a tangent from the center of circle (axis of rotation) = radius of the circle
• So we get: $\mathbf\small{\tau}$ = Force × perpendicular distance from center
= Force × radius = 30 × 0.4 = 12 Nm
3. We have: $\mathbf\small{\tau=I\,\alpha}$
• Substituting the values, we get: $\mathbf\small{12=(3)(0.4)^2\,\alpha}$
$\mathbf\small{\Rightarrow \alpha=}$ 25 rad s-2
• This is the answer for part (i)
4. The cylinder starts from rest
• So initial angular velocity $\mathbf\small{\omega_0=0}$
5. Let us find the angular velocity after any convenient interval of time, say 2 s
• We have: $\mathbf\small{\omega=\omega_0+\alpha \, t}$
• Substituting the values, we get: $\mathbf\small{\omega=0+(25) \, (2)}$ = 50 rad s-1
6. So at the instant when the stop watch shows 2 seconds, the cylinder will be rotating with an angular speed of 50 rad s-1
• Every particle in the cylinder will be rotating with the angular speed of 50 rad s-1 at that instant
• Any particle at the periphery will also be rotating with the angular speed of 50 rad s-1 at that instant
7. Now, we use the relation between linear velocity and angular velocity
• We have: $\mathbf\small{v=r\;\omega}$
• We apply it to a particle at the periphery: v = 0.4 × 50 = 20 ms-1.
• So, when the stop watch shows 2 seconds, any particle at the periphery will be moving with a linear speed of 20 ms-1.
8. Given that, there is no slip between the cylinder and the rope
• So the rope will also be moving with a linear speed of 20 ms-1
• We will use the relation $\mathbf\small{v=v_0+at}$
• The rope also started from rest. So $\mathbf\small{v_0=0}$
• Substituting the values, we get: 20= 0 + a × 2
⇒ a = 10 ms-2.
In the next section, we will see angular momentum
