In the previous section, we obtained two results for rotation about a fixed axis:
• If the rotation is symmetric (symmetric objects rotating about an axis of symmetry), then:
♦ Eq,7.31: $\mathbf\small{\vec{L}=\vec{L}_z}$
• If the rotation is not symmetric, then
♦ Eq.7.30: $\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$
• In this section, we will derive the expression for torque acting on such bodies
1. We will consider the general case when the rotation is not symmetric
• That is., we will consider the Eq.7.30
2. $\mathbf\small{\vec{L}}$ is the angular momentum
• We know that, 'change in angular momentum' per unit time will give torque
• So $\mathbf\small{\frac{d \vec{L}}{dt} }$ will give the torque acting on the body
3. When we differentiate the left side, we must differentiate the terms on the right side also
• So we can write: $\mathbf\small{\frac{d \vec{L}}{dt}=\frac{d \vec{L}_z}{dt}+\frac{d \vec{L}_\bot}{dt}}$
4. The term on the left is torque
• So each of the two terms on the right side must be torques
5. The last term is related to torques which are not parallel to the axis of rotation
• We have seen that:
When the axis is fixed, those non-parallel torques have no effect
• So we need not consider the last term
6. We can write:
$\mathbf\small{\frac{d \vec{L}}{dt}=\frac{d \vec{L}_z}{dt}}$
• Let us differentiate the right side:
$\mathbf\small{\frac{d \vec{L}_z}{dt}=\frac{d (I|\vec{\omega}|\hat{k})}{dt}}$
7. In this differentiation, we are taking the ['change in $\mathbf\small{(I|\vec{\omega}|\hat{k})}$' per unit time] at an instant
• The body is rigid and the axis is fixed. So I does not change
• $\mathbf\small{\hat{k}}$ has a constant magnitude and direction. It also does not change
• So those two items can be taken outside. We get: $\mathbf\small{\frac{d \vec{L}_z}{dt}=I \hat{k}\frac{d (|\vec{\omega}|)}{dt}}$
8. But $\mathbf\small{\frac{d (|\vec{\omega}|)}{dt}}$ is the angular acceleration $\mathbf\small{\alpha}$
So we get: $\mathbf\small{\frac{d \vec{L}_z}{dt}=(I \alpha) \hat{k}}$
$\mathbf\small{\Rightarrow \vec{\tau}=(I \alpha) \hat{k}}$
$\mathbf\small{\Rightarrow |\vec{\tau}|=I \alpha}$
Thus we get the expression for the torque acting on a body rotating about a fixed axis
9. This expression is applicable to all rigid bodies rotating about a fixed axis. So we do not need to consider 'symmetric bodies in symmetric rotation' separately
• That means, we do not need to write the above steps for Eq.7.31
A 40 kg flywheel in the form of a uniform circular disc of 1 m radius is making 120 rpm. Determine the angular momentum
Solution:
1. A circular disc is a symmetric body. It is rotating about an axis of symmetry
• So we can use Eq.7.31: $\mathbf\small{\vec{L}=\vec{L}_z=I|\vec{\omega}|\,\hat{k}}$
2. The fly wheel rotates about a perpendicular axis passing through the center
• So $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{40\times 1^2}{2}=20\, \rm{kg\;m^2}}$
3. Given that $\mathbf\small{|\vec{\omega}|}$ = 120 rpm
• We have to convert it into rad s-1.
• 1 rotation = 2𝝅 rad
⇒ 120 rotations = 240𝝅 rad
⇒ 240𝝅 rad is covered in 60 s
• So angle covered in 1 s = $\mathbf\small{|\vec{\omega}|=\frac{240 \pi}{60}=4\pi\,\,\text{rad s}^{-1}}$
4. Thus we get:
Angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=20\times 4 \pi=251.2 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the fly wheel
5. Unit of angular momentum:
• Angular momentum is defined as the moment of linear momentum
• That is., we are multiplying the linear momentum by a distance
• The unit of linear momentum is kg ms-1
• So the unit of angular momentum is kg m2s-1
• Thus the required answer is: 251.2 kg m2s-1
Solved example 7.38
A wheel is rotating with an angular momentum of 3 kg m2s-1. A torque of 12 Nm is applied on the wheel for 4 s. What is the final angular momentum of the wheel?
Solution:
We have: Torque = Time rate of change of angular momentum
⇒ Torque = (Final angular momentum - Initial angular momentum)⁄time
⇒ 12 = (Final angular momentum - 3)⁄4
⇒ Final angular momentum = (48+3) = 51 kg m2s-1
Solved example 7.39
The diameter of a circular disc is 0.5 m and it's mass is 16 kg. What torque will increase it's angular velocity from zero to 120 rpm in 8 s?
Solution:
1. We have: $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{16\times 0.25^2}{2}=0.5\, \rm{kg\;m^2}}$
2. Initial angular velocity = 0
• So initial angular momentum = 0
3. Final angular velocity = 120 rpm = 4𝝅 rad s-1.
• So final angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=0.5\times 4 \pi=6.28 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the disc
4. We have: Torque = Time rate of change of angular momentum
⇒ Torque = (Final angular momentum - Initial angular momentum)⁄time
⇒ Torque = (6.28 - 0)⁄8 = 0.785 N m
• If the rotation is symmetric (symmetric objects rotating about an axis of symmetry), then:
♦ Eq,7.31: $\mathbf\small{\vec{L}=\vec{L}_z}$
• If the rotation is not symmetric, then
♦ Eq.7.30: $\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$
• In this section, we will derive the expression for torque acting on such bodies
1. We will consider the general case when the rotation is not symmetric
• That is., we will consider the Eq.7.30
2. $\mathbf\small{\vec{L}}$ is the angular momentum
• We know that, 'change in angular momentum' per unit time will give torque
• So $\mathbf\small{\frac{d \vec{L}}{dt} }$ will give the torque acting on the body
3. When we differentiate the left side, we must differentiate the terms on the right side also
• So we can write: $\mathbf\small{\frac{d \vec{L}}{dt}=\frac{d \vec{L}_z}{dt}+\frac{d \vec{L}_\bot}{dt}}$
4. The term on the left is torque
• So each of the two terms on the right side must be torques
5. The last term is related to torques which are not parallel to the axis of rotation
• We have seen that:
When the axis is fixed, those non-parallel torques have no effect
• So we need not consider the last term
6. We can write:
$\mathbf\small{\frac{d \vec{L}}{dt}=\frac{d \vec{L}_z}{dt}}$
• Let us differentiate the right side:
$\mathbf\small{\frac{d \vec{L}_z}{dt}=\frac{d (I|\vec{\omega}|\hat{k})}{dt}}$
7. In this differentiation, we are taking the ['change in $\mathbf\small{(I|\vec{\omega}|\hat{k})}$' per unit time] at an instant
• The body is rigid and the axis is fixed. So I does not change
• $\mathbf\small{\hat{k}}$ has a constant magnitude and direction. It also does not change
• So those two items can be taken outside. We get: $\mathbf\small{\frac{d \vec{L}_z}{dt}=I \hat{k}\frac{d (|\vec{\omega}|)}{dt}}$
8. But $\mathbf\small{\frac{d (|\vec{\omega}|)}{dt}}$ is the angular acceleration $\mathbf\small{\alpha}$
So we get: $\mathbf\small{\frac{d \vec{L}_z}{dt}=(I \alpha) \hat{k}}$
$\mathbf\small{\Rightarrow \vec{\tau}=(I \alpha) \hat{k}}$
$\mathbf\small{\Rightarrow |\vec{\tau}|=I \alpha}$
Thus we get the expression for the torque acting on a body rotating about a fixed axis
9. This expression is applicable to all rigid bodies rotating about a fixed axis. So we do not need to consider 'symmetric bodies in symmetric rotation' separately
• That means, we do not need to write the above steps for Eq.7.31
Now we will see some solved examples
Solved example 7.36
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Solution:
1. A solid cylinder is a symmetric body. It is rotating about an axis of symmetry
• So we can use Eq.7.31: $\mathbf\small{\vec{L}=\vec{L}_z=I|\vec{\omega}|\,\hat{k}}$
2. This solid cylinder rotates about it's axis
• So $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{20\times 0.25^2}{2}=0.625\, \rm{kg\;m^2}}$
3. Given that $\mathbf\small{|\vec{\omega}|}$ = 100 rad s-1
4. Thus we get:
Angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=0.625\times 100=62.5 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the cylinder
5. Unit of angular momentum:
• Angular momentum is defined as the moment of linear momentum
• That is., we are multiplying the linear momentum by a distance
• The unit of linear momentum is kg ms-1
• So the unit of angular momentum is kg m2s-1
• Thus the angular momentum is: 62.5 kg m2s-1
6. Next we have to find the kinetic energy
• We have: Eq.7.26: $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2}$
• Substituting the values, we get: $\mathbf\small{K=\frac{1}{2}\times 0.625 \times 100^2=3125\, \rm{J}}$
Solved example 7.37
Solved example 7.36
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Solution:
1. A solid cylinder is a symmetric body. It is rotating about an axis of symmetry
• So we can use Eq.7.31: $\mathbf\small{\vec{L}=\vec{L}_z=I|\vec{\omega}|\,\hat{k}}$
2. This solid cylinder rotates about it's axis
• So $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{20\times 0.25^2}{2}=0.625\, \rm{kg\;m^2}}$
3. Given that $\mathbf\small{|\vec{\omega}|}$ = 100 rad s-1
4. Thus we get:
Angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=0.625\times 100=62.5 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the cylinder
5. Unit of angular momentum:
• Angular momentum is defined as the moment of linear momentum
• That is., we are multiplying the linear momentum by a distance
• The unit of linear momentum is kg ms-1
• So the unit of angular momentum is kg m2s-1
• Thus the angular momentum is: 62.5 kg m2s-1
6. Next we have to find the kinetic energy
• We have: Eq.7.26: $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2}$
• Substituting the values, we get: $\mathbf\small{K=\frac{1}{2}\times 0.625 \times 100^2=3125\, \rm{J}}$
Solved example 7.37
Solution:
1. A circular disc is a symmetric body. It is rotating about an axis of symmetry
• So we can use Eq.7.31: $\mathbf\small{\vec{L}=\vec{L}_z=I|\vec{\omega}|\,\hat{k}}$
2. The fly wheel rotates about a perpendicular axis passing through the center
• So $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{40\times 1^2}{2}=20\, \rm{kg\;m^2}}$
3. Given that $\mathbf\small{|\vec{\omega}|}$ = 120 rpm
• We have to convert it into rad s-1.
• 1 rotation = 2𝝅 rad
⇒ 120 rotations = 240𝝅 rad
⇒ 240𝝅 rad is covered in 60 s
• So angle covered in 1 s = $\mathbf\small{|\vec{\omega}|=\frac{240 \pi}{60}=4\pi\,\,\text{rad s}^{-1}}$
4. Thus we get:
Angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=20\times 4 \pi=251.2 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the fly wheel
5. Unit of angular momentum:
• Angular momentum is defined as the moment of linear momentum
• That is., we are multiplying the linear momentum by a distance
• The unit of linear momentum is kg ms-1
• So the unit of angular momentum is kg m2s-1
• Thus the required answer is: 251.2 kg m2s-1
Solved example 7.38
A wheel is rotating with an angular momentum of 3 kg m2s-1. A torque of 12 Nm is applied on the wheel for 4 s. What is the final angular momentum of the wheel?
Solution:
We have: Torque = Time rate of change of angular momentum
⇒ Torque = (Final angular momentum - Initial angular momentum)⁄time
⇒ 12 = (Final angular momentum - 3)⁄4
⇒ Final angular momentum = (48+3) = 51 kg m2s-1
Solved example 7.39
The diameter of a circular disc is 0.5 m and it's mass is 16 kg. What torque will increase it's angular velocity from zero to 120 rpm in 8 s?
Solution:
1. We have: $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{16\times 0.25^2}{2}=0.5\, \rm{kg\;m^2}}$
2. Initial angular velocity = 0
• So initial angular momentum = 0
3. Final angular velocity = 120 rpm = 4𝝅 rad s-1.
• So final angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=0.5\times 4 \pi=6.28 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the disc
4. We have: Torque = Time rate of change of angular momentum
⇒ Torque = (Final angular momentum - Initial angular momentum)⁄time
⇒ Torque = (6.28 - 0)⁄8 = 0.785 N m
In the next section, we will see conservation of angular momentum
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