In the previous section, we saw the theorem of parallel axes. In this section we will see kinematics of rotational motion about a fixed axis
• In the fig.7.126 below, a rigid body rotates about a fixed axis
• This axis is taken as the z-axis of the reference frame
• The body is given some transparency so that, particles inside it can be seen
• Three particles in the body are highlighted. They are shown as red, yellow and green spheres
♦ The yellow sphere is closer to the axis than the red sphere
♦ The green sphere is situated exactly on the axis
• We have seen this situation in an earlier section in fig.7.10
• There we noted the characteristics of rotational motion:
■ If a rigid body is in rotation about an axis, then:
• Every particle of the body, which lies on the axis will be stationary
• Each of the other particles will be rotating in it's own circular path
♦ The center of that circular path lies on the axis
♦ Radius of that circular path = Distance of the particle from the axis
• That circular path lies on a plane
♦ This plane is perpendicular to the axis
■ The following two points are also important for our present discussion:
1. At any instant:
• Linear velocity of red sphere > Linear velocity of yellow sphere
• That is., $\mathbf\small{\vec{v}_{Red}>\vec{v}_{Yellow}}$
2. At any instant:
• Angular velocity of red sphere
= Angular velocity of yellow sphere
= Angular velocity of the whole body
• That is., $\mathbf\small{\vec{\omega}_{Red}=\vec{\omega}_{Yellow}=\vec{\omega}_{(Whole\;Body)}}$
• The red plane is the plane
2. When the initial reading (t1) in the stopwatch is zero, the particle is at A
• We need to specify the location A in terms of angles
• For that, we must have a reference direction
• The reference direction is denoted as x'
♦ x' is parallel to the x-axis
♦ x' lies in the red plane
3. Now we can specify the location A:
• It is at an angular distance of θ0 from x'
■ So we can write:
• When the initial reading (t1) in the stopwatch is zero, the angular displacement of the particle is θ0
• In other words:
The initial angular displacement of the particle is θ0
4. When the final reading (t2) in the stopwatch is t, the particle is at B
• B is at an angular distance of θ from x’
■ So we can write:
• When the final reading (t2) in the stopwatch is t, the angular displacement of the Particle is θ
• In other words:
The final angular displacement of the Particle is θ
5. Now the change in angular displacement = Δθ = (θ - θ0)
• This change occurred in a time interval of Δt = (t2-t1) = (t – 0) = t
• So average angular velocity = $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$
■ We call it ‘average angular velocity’ because, the angular velocity with which the particle travels from A to B may not be uniform
• This situation occurs especially when there is angular acceleration
6. If we want instantaneous angular velocity, the time interval (t2-t1) must be very small
• In such a situation, we use calculus
• The instantaneous angular velocity is given by $\mathbf\small{\frac{d \theta}{d t}}$
7. Note that, the angular velocity (instantaneous or average) possessed by the particle is the 'same angular velocity possessed by the whole body'
8. We have seen that, the angular velocity (denoted as $\mathbf\small{\vec{\omega}}$ ) is a vector (details here)
• In our present case, we deal with rotation about a fixed axis
• That means the direction of $\mathbf\small{\vec{\omega}}$ is either upwards or downwards along that fixed axis
• 'upwards or downwards' can be distinguished by using either positive or negative sign
• So for rotation about a fixed axis, we do not need to treat $\mathbf\small{\omega}$ as a vector
• The role played by linear velocity $\mathbf\small{v}$ is played by angular velocity $\mathbf\small{\omega}$
• The role played by linear acceleration $\mathbf\small{a}$ is played by angular acceleration $\mathbf\small{\alpha}$
• The role played by linear displacement $\mathbf\small{x}$ is played by angular displacement $\mathbf\small{\theta}$
■ We have 3 kinematic equations in linear motion:
$\mathbf\small{v=v_0+at}$
$\mathbf\small{x=x_0+v_0 t+\frac{1}{2}at^2}$
$\mathbf\small{v^2=v_0^2+2ax}$
■ Correspondingly, we have 3 kinematic equations in rotational motion with uniform acceleration:
$\mathbf\small{\omega=\omega_0+\alpha t}$
$\mathbf\small{\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2}$
$\mathbf\small{\omega^2=\omega_0^2+2\alpha (\theta - \theta_0)}$
Solved example 7.32
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the acceleration to be uniform? (ii) How many revolutions does the engine make during this time?
Solution:
1. Initial angular velocity = 1200 rpm = 1200 revolutions in one minute
(This is the angular velocity at the instant when the stop watch is started)
• 1 revolution is 2π radians
• So 1200 revolutions = (1200 × 2π) radians = 2400π radians
• So the motor wheel travels an angular distance of 2400π radians in one minute
• That means, the motor wheel travels an angular distance of $\mathbf\small{\frac{2400 \pi}{60}=40 \pi}$ radians in one second
• So initial angular velocity ω0= 40π radians/sec
2.In the same way, final angular velocity ω = 3120 rpm = 104π radians/sec
3. Given that, the time during which this change in angular velocity occurs = 16 s
4. We will use the equation: $\mathbf\small{\omega=\omega_0+\alpha t}$
• Substituting the values, we get: $\mathbf\small{104 \pi=40 \pi+\alpha \times 16}$
⇒ α = 4π radians/sec2.
• This is the answer for part (i)
5. We want the angular distance traveled by the wheel during those 16 seconds
• We will use the equation: $\mathbf\small{\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2}$
• We do not want the 'angular distance traveled' before the stopwatch is started. So θ0 = 0
• Substituting the values, we get: $\mathbf\small{\theta=0+40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2=1152 \pi}$
6. So we get:
• The angular distance traveled during those 16 seconds = 1152π radians
• 1 revolution is 2π radians
• So number of revolutions = $\mathbf\small{\frac{1152 \pi}{2\pi}}$ = 576 revolutions
• In the fig.7.126 below, a rigid body rotates about a fixed axis
• This axis is taken as the z-axis of the reference frame
 |
| Fig.7.126 |
• Three particles in the body are highlighted. They are shown as red, yellow and green spheres
♦ The yellow sphere is closer to the axis than the red sphere
♦ The green sphere is situated exactly on the axis
• We have seen this situation in an earlier section in fig.7.10
• There we noted the characteristics of rotational motion:
■ If a rigid body is in rotation about an axis, then:
• Every particle of the body, which lies on the axis will be stationary
• Each of the other particles will be rotating in it's own circular path
♦ The center of that circular path lies on the axis
♦ Radius of that circular path = Distance of the particle from the axis
• That circular path lies on a plane
♦ This plane is perpendicular to the axis
■ The following two points are also important for our present discussion:
1. At any instant:
• Linear velocity of red sphere > Linear velocity of yellow sphere
• That is., $\mathbf\small{\vec{v}_{Red}>\vec{v}_{Yellow}}$
2. At any instant:
• Angular velocity of red sphere
= Angular velocity of yellow sphere
= Angular velocity of the whole body
• That is., $\mathbf\small{\vec{\omega}_{Red}=\vec{\omega}_{Yellow}=\vec{\omega}_{(Whole\;Body)}}$
Based on the above discussion, we can write the steps to find angular velocity:
1. The 'path of the red sphere', as well as the 'plane of the red sphere' are shown in fig.7.127 below:
• The red circle is the path
1. The 'path of the red sphere', as well as the 'plane of the red sphere' are shown in fig.7.127 below:
 |
| Fig.7.127 |
2. When the initial reading (t1) in the stopwatch is zero, the particle is at A
• We need to specify the location A in terms of angles
• For that, we must have a reference direction
• The reference direction is denoted as x'
♦ x' is parallel to the x-axis
♦ x' lies in the red plane
3. Now we can specify the location A:
• It is at an angular distance of θ0 from x'
■ So we can write:
• When the initial reading (t1) in the stopwatch is zero, the angular displacement of the particle is θ0
• In other words:
The initial angular displacement of the particle is θ0
4. When the final reading (t2) in the stopwatch is t, the particle is at B
• B is at an angular distance of θ from x’
■ So we can write:
• When the final reading (t2) in the stopwatch is t, the angular displacement of the Particle is θ
• In other words:
The final angular displacement of the Particle is θ
5. Now the change in angular displacement = Δθ = (θ - θ0)
• This change occurred in a time interval of Δt = (t2-t1) = (t – 0) = t
• So average angular velocity = $\mathbf\small{\frac{\Delta \theta}{\Delta t}}$
■ We call it ‘average angular velocity’ because, the angular velocity with which the particle travels from A to B may not be uniform
• This situation occurs especially when there is angular acceleration
6. If we want instantaneous angular velocity, the time interval (t2-t1) must be very small
• In such a situation, we use calculus
• The instantaneous angular velocity is given by $\mathbf\small{\frac{d \theta}{d t}}$
7. Note that, the angular velocity (instantaneous or average) possessed by the particle is the 'same angular velocity possessed by the whole body'
8. We have seen that, the angular velocity (denoted as $\mathbf\small{\vec{\omega}}$ ) is a vector (details here)
• In our present case, we deal with rotation about a fixed axis
• That means the direction of $\mathbf\small{\vec{\omega}}$ is either upwards or downwards along that fixed axis
• 'upwards or downwards' can be distinguished by using either positive or negative sign
• So for rotation about a fixed axis, we do not need to treat $\mathbf\small{\omega}$ as a vector
■ The rotational motion about a fixed axis is analogous to linear motion
• The role played by linear acceleration $\mathbf\small{a}$ is played by angular acceleration $\mathbf\small{\alpha}$
• The role played by linear displacement $\mathbf\small{x}$ is played by angular displacement $\mathbf\small{\theta}$
■ We have 3 kinematic equations in linear motion:
$\mathbf\small{v=v_0+at}$
$\mathbf\small{x=x_0+v_0 t+\frac{1}{2}at^2}$
$\mathbf\small{v^2=v_0^2+2ax}$
■ Correspondingly, we have 3 kinematic equations in rotational motion with uniform acceleration:
$\mathbf\small{\omega=\omega_0+\alpha t}$
$\mathbf\small{\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2}$
$\mathbf\small{\omega^2=\omega_0^2+2\alpha (\theta - \theta_0)}$
Now we will see a solved example:
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the acceleration to be uniform? (ii) How many revolutions does the engine make during this time?
Solution:
1. Initial angular velocity = 1200 rpm = 1200 revolutions in one minute
(This is the angular velocity at the instant when the stop watch is started)
• 1 revolution is 2π radians
• So 1200 revolutions = (1200 × 2π) radians = 2400π radians
• So the motor wheel travels an angular distance of 2400π radians in one minute
• That means, the motor wheel travels an angular distance of $\mathbf\small{\frac{2400 \pi}{60}=40 \pi}$ radians in one second
• So initial angular velocity ω0= 40π radians/sec
2.In the same way, final angular velocity ω = 3120 rpm = 104π radians/sec
3. Given that, the time during which this change in angular velocity occurs = 16 s
4. We will use the equation: $\mathbf\small{\omega=\omega_0+\alpha t}$
• Substituting the values, we get: $\mathbf\small{104 \pi=40 \pi+\alpha \times 16}$
⇒ α = 4π radians/sec2.
• This is the answer for part (i)
5. We want the angular distance traveled by the wheel during those 16 seconds
• We will use the equation: $\mathbf\small{\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2}$
• We do not want the 'angular distance traveled' before the stopwatch is started. So θ0 = 0
• Substituting the values, we get: $\mathbf\small{\theta=0+40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2=1152 \pi}$
6. So we get:
• The angular distance traveled during those 16 seconds = 1152π radians
• 1 revolution is 2π radians
• So number of revolutions = $\mathbf\small{\frac{1152 \pi}{2\pi}}$ = 576 revolutions
In the next section, we will see dynamics of rotational motion about a fixed axis
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