In the previous section, we saw conservation of angular momentum. In this section, we will see rolling motion
1. Consider a disc rolling on a horizontal surface (Fig.7.137 (a) below)
• Radius of the disc is R
■ Since it is rolling, it will have two types of motion:
(i) Translational motion
(ii) Rotational motion
2. Let us write all the available information about the rotational motion:
(i) The disc is a symmetric body
• So the rotation will be about an axis passing through the center C
• Also, this axis will be perpendicular to the plane of the disc (that is., plane of the computer screen)
(ii) The angular velocity of rotation is $\mathbf\small{\vec{\omega}}$
• All particles (except the particle at C) in the disc will be rotating with this same angular velocity
(iii) The particle at C will not have any rotational motion
3. Next we write all the available information about the translational motion:
(i) The particle at C does not have any rotational motion. But it does have translational motion
(ii) Since C is the center of mass (CM) of the disc, we will denote the translational velocity of C as $\mathbf\small{\vec{v}_{CM}}$
• This $\mathbf\small{\vec{v}_{CM}}$ is parallel to the surface on which the disc rolls
(iii) Since the disc is rigid, all particles will have the same translational velocity
4. Our next aim is to find this $\mathbf\small{\vec{v}_{CM}}$
• Consider fig.b, An arc AP is marked in magenta color on the periphery of the disc
• This arc subtends an angle of θ at the center C
5. The fig.b shows the instant at which A is in contact with the ground
6. The disc rolls towards the right
• Point A will lose contact with the surface
• As the rolling proceeds, new points on the arc will come into contact with the surface
7. Consider the instant at which P comes into contact with the surface
8. What is the linear distance covered by the disc between the following two instances:
♦ Instance mentioned in (5)
♦ Instance mentioned in (7)
Answer: Obviously, the linear distance will be equal to the length of the arc AP
(Note that, this will be true only if there is no slipping/skidding between the disc and the surface. For our present discussion, we assume that there is no such slipping/skidding)
9. We know how to find this length of arc
• We have: angle = Arc length⁄radius (Where angle is in radians)
• So Arc length AP = Linear distance traveled by the disc = Rθ.
10. Let 't' be the time duration between the two instances
• Then we can write two more information:
(i) The disc travels a linear distance of Rθ during a time 't' s
(ii) The disc turns through an angle θ during a time 't' s
11. Let us write the above equation again:
• Linear distance traveled by the disc = Rθ.
• Dividing both sides by 't', we get:
$\mathbf\small{\frac{\text{Linear distance traveled by the disc}}{t}=\frac{R\theta}{t}}$
• But $\mathbf\small{\frac{\text{Linear distance traveled by the disc}}{t}}$ is the linear velocity with which the disc travels during the time interval 't'
♦ 'Linear velocity of the disc' is the 'velocity of the center of mass'
♦ We denoted it as $\mathbf\small{\vec{v}_{CM}}$
• Also, $\mathbf\small{\frac{\theta}{t}}$ is the angular velocity with which the disc turns during the time interval 't'
♦ We denoted the angular velocity as $\mathbf\small{\vec{\omega}}$
■ Thus we get:
Eq.7.32: $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$
• Since it is a rigid body, every particle in the disc will have a translational velocity of $\mathbf\small{|\vec{v}_{CM}|}$ towards the right
12. This equation is similar to: $\mathbf\small{\vec{v}=R\,\vec{\omega}}$
• $\mathbf\small{\vec{v}}$ is the 'tangential velocity' of a particle at the periphery of a rotating body
• But $\mathbf\small{\vec{v}_{CM}}$ is the linear velocity with which the disc rolls
• We must clearly note the difference between the two
■ Eq.7.32 gives the condition for 'no slipping'
• Because if there is slipping, arc length will not be equal to the linear distance. The equation will not be valid any more
13. Let us see another interesting information:
• Fig.7.138(a) below shows 3 points in the disc:
(i) The bottom most point A
(ii) The center of mass C
(iii) The top most point B
• A and B pass through a vertical through C
14. First we will consider point B:
• The disc has an angular velocity of $\mathbf\small{|\vec{\omega}|}$
• Then the point B will be moving towards the right with a tangential velocity of $\mathbf\small{R\,|\vec{\omega}|}$
• Note that, this velocity will be parallel to the ground. That is., parallel to $\mathbf\small{|\vec{v}_{CM}|}$
• Also, this velocity is towards the right
15. We have seen that, due to translation, already every point in the disc has a velocity of $\mathbf\small{R|\vec{\omega}|}$ towards the right. For point B, this is shown in fig.b
• So for point B, the two velocities will add up
• Thus we can write:
Point B moves towards the right with a net velocity of $\mathbf\small{2R|\vec{\omega}|}$. This is shown in fig.c
16. But at the next instant, a succeeding particle takes over the position of B. This new particle will experience the same effect
■ So in general, we can write:
At any instant, the top most point of the disc will be travelling towards the right with a velocity of $\mathbf\small{2R|\vec{\omega}|}$
17. Next we will consider point C
• The point C is in the axis of rotation. So it will receive no contribution (towards velocity) from the rotation. Because, every point on the axis will be having zero rotation
• But C has translational velocity $\mathbf\small{|\vec{v}_{CM}|}$. It is the only velocity it has. This is shown in fig.b
■ Thus we can write:
Point C moves towards the right with a net velocity of $\mathbf\small{R|\vec{\omega}|}$. This is shown in fig.c
18. Finally we consider point A
• The disc has an angular velocity of $\mathbf\small{|\vec{\omega}|}$
• Then the point A will be moving towards the left with a tangential velocity of $\mathbf\small{R\,|\vec{\omega}|}$. This is shown in fig.a
• Note that, this velocity will be parallel to the ground. That is., parallel to $\mathbf\small{|\vec{v}_{CM}|}$
• Also, this velocity is towards the left
19. We have seen that, due to translation, already every point in the disc has a velocity of $\mathbf\small{R|\vec{\omega}|}$ towards the right. For point A, this is shown in fig.b
• So for point A, the two velocities will cancel each other
• Thus we can write:
Point A has zero net velocity. This is shown in fig.c
20. But at the next instant, the succeeding particle takes over the position of A, This new particle will experience the same effect
■ So in general, we can write:
At any instant, the bottom most point of the disc will be having zero net velocity
• That means, the point of contact (with the ground) of a rolling wheel/disc will be at rest
We will write this in steps:
1. A rolling body has both rotational and translational motion
2. We know that kinetic energy due to rotation is $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2}$
Where:
• I = Moment of inertia of the body about the axis of rotation
• $\mathbf\small{|\vec{\omega}|}$ = Magnitude of the angular velocity
3. Also we know that kinetic energy due to translation is $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
Where:
• m = mass of the body
• $\mathbf\small{|\vec{v}_{CM}|}$ = magnitude of the linear velocity
4. So the total kinetic energy possessed by a rolling body is given by:
Eq.7.33: $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2+\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
Solved example 7.44
A disc of mass 5 kg and radius 50 cm rolls on the ground with out slipping. It's linear velocity is 10 ms-1. What is the kinetic energy possessed by the disc?
Solution:
1. Kinetic energy due to rotation = $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2}$
(i) Moment of inertia (I) should be selected carefully
• A disc can have a moment of inertia about an axis which can be an extension of one of it's diameters
• It can also have a moment of inertia about an axis perpendicular to the disc. Obviously, a disc can roll only if the axis is perpendicular to the disc. So we must take the I corresponding to this axis
Thus we have: $\mathbf\small{I=\frac{1}{2}m\,R^2}$
(ii) Next we calculate $\mathbf\small{|\vec{\omega}|}$
• Given that the disc rolls without slipping. So the condition $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$ will be satisfied
• So we get: $\mathbf\small{|\vec{\omega}|=\frac{|\vec{v}_{CM}|}{R}}$
(iii) Substituting these expressions in (1), we get:
Kinetic energy due to rotation = $\mathbf\small{\frac{1}{2}\left(\frac{1}{2}m\,R^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{4}}$
2. Kinetic energy due to translation = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
3. So total kinetic energy = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{4}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{3m|\vec{v}_{CM}|^2}{4}}$
• Substituting the values, we get:
Total kinetic energy = $\mathbf\small{\frac{3(5\,\rm{kg})(10\,\rm{m\,s^{-1}})^2}{4}=375\;\rm{J}}$
• Recall that dimensions of energy is: ML2T-2.
Solved example 7.45
Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?
Solution:
• In this problem, we apply the law of conservation of energy
• The body is initially at rest at the top most point of the inclined plane. This is shown in fig.7.139 below:
• In that position, it has potential energy equal to mgh
• Since it is initially at rest, there is no initial kinetic energy
• When it reaches at the base, the total energy that it possesses must be equal to the total energy at top
• At the base, there is no potential energy. There is only kinetic energy
■ So we get:
Total kinetic energy of the body at the base = Total potential energy at the top
• Now we consider each body separately
1. Ring:
(i) Rotational kinetic energy
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(m\,R^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{2}}$
• Note that for a ring, I = mR2
• Also we have: $\mathbf\small{|\vec{\omega}|=\frac{|\vec{v}_{CM}|}{R}}$
(ii) Translational kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
(iii) So total kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2+\frac{m|\vec{v}_{CM}|^2}{2}=m|\vec{v}_{CM}|^2}$
(iv) potential energy = mgh
(iii) Equating the two energies, we get: $\mathbf\small{mgh=m|\vec{v}_{CM}|^2}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{gh}}$
2. Solid cylinder
(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 = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{4}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{3m|\vec{v}_{CM}|^2}{4}}$
(iv) potential energy = mgh
(v) Equating the two energies, we get: $\mathbf\small{mgh=\frac{3m|\vec{v}_{CM}|^2}{4}}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{\frac{4gh}{3}}}$
3. Solid sphere
(i) Rotational kinetic energy
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(\frac{2m\,R^2}{5}\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{5}}$
• Note that for a solid sphere, $\mathbf\small{I=\frac{2m\,R^2}{5}}$
(ii) Translational kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
(iii) So total kinetic energy = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{5}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{7m|\vec{v}_{CM}|^2}{10}}$
(iv) potential energy = mgh
(v) Equating the two energies, we get: $\mathbf\small{mgh=\frac{7m|\vec{v}_{CM}|^2}{10}}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{\frac{10gh}{7}}}$
4. We will now compare the results. We will use the subscripts r, c and s for the ring, solid cylinder and solid sphere respectively:
We have:
(i) $\mathbf\small{|\vec{v}_{CM(r)}|=\sqrt{gh}}$
(ii) $\mathbf\small{|\vec{v}_{CM(c)}|=\sqrt{\frac{4gh}{3}}}$
(iii) $\mathbf\small{|\vec{v}_{CM(s)}|=\sqrt{\frac{10gh}{7}}}$
• 4⁄3 = 1.333
• 10⁄7 = 1.429
• Thus we see that $\mathbf\small{\Rightarrow |\vec{v}_{CM(s)}|}$ has the greatest value
■ So we can write:
The sphere reaches the ground with the maximum velocity
Another method:
• We can derive a general formula (for the velocity at the base) for any body
1. For any body, we have: I = Mk2
• Where k is the radius of gyration of that body (Details here)
2. So for any body, the rotational kinetic energy
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(m\,k^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2}$
3. So total energy
= $\mathbf\small{\frac{1}{2}\left(m\,k^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2+\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
= $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2\left(1+\frac{k^2}{R^2} \right)}$
4. Equating this total energy to potential energy, we get:
$\mathbf\small{mgh=\frac{1}{2}m\,|\vec{v}_{CM}|^2\left(1+\frac{k^2}{R^2} \right)}$
5. From this, we get:
Eq.7.34: $\mathbf\small{|\vec{v}_{CM}|^2=\frac{2gh}{\left(1+\frac{k^2}{R^2} \right)}}$
• We see that, the velocity at the base is independent of the mass
• In an earlier chapter on pure translational motion also, we have proved that, the velocity of at the base of an inclined plane is independent of the mass
1. Consider a disc rolling on a horizontal surface (Fig.7.137 (a) below)
• Radius of the disc is R
 |
| Fig.7.137 |
(i) Translational motion
(ii) Rotational motion
2. Let us write all the available information about the rotational motion:
(i) The disc is a symmetric body
• So the rotation will be about an axis passing through the center C
• Also, this axis will be perpendicular to the plane of the disc (that is., plane of the computer screen)
(ii) The angular velocity of rotation is $\mathbf\small{\vec{\omega}}$
• All particles (except the particle at C) in the disc will be rotating with this same angular velocity
(iii) The particle at C will not have any rotational motion
3. Next we write all the available information about the translational motion:
(i) The particle at C does not have any rotational motion. But it does have translational motion
(ii) Since C is the center of mass (CM) of the disc, we will denote the translational velocity of C as $\mathbf\small{\vec{v}_{CM}}$
• This $\mathbf\small{\vec{v}_{CM}}$ is parallel to the surface on which the disc rolls
(iii) Since the disc is rigid, all particles will have the same translational velocity
4. Our next aim is to find this $\mathbf\small{\vec{v}_{CM}}$
• Consider fig.b, An arc AP is marked in magenta color on the periphery of the disc
• This arc subtends an angle of θ at the center C
5. The fig.b shows the instant at which A is in contact with the ground
6. The disc rolls towards the right
• Point A will lose contact with the surface
• As the rolling proceeds, new points on the arc will come into contact with the surface
7. Consider the instant at which P comes into contact with the surface
8. What is the linear distance covered by the disc between the following two instances:
♦ Instance mentioned in (5)
♦ Instance mentioned in (7)
Answer: Obviously, the linear distance will be equal to the length of the arc AP
(Note that, this will be true only if there is no slipping/skidding between the disc and the surface. For our present discussion, we assume that there is no such slipping/skidding)
9. We know how to find this length of arc
• We have: angle = Arc length⁄radius (Where angle is in radians)
• So Arc length AP = Linear distance traveled by the disc = Rθ.
10. Let 't' be the time duration between the two instances
• Then we can write two more information:
(i) The disc travels a linear distance of Rθ during a time 't' s
(ii) The disc turns through an angle θ during a time 't' s
11. Let us write the above equation again:
• Linear distance traveled by the disc = Rθ.
• Dividing both sides by 't', we get:
$\mathbf\small{\frac{\text{Linear distance traveled by the disc}}{t}=\frac{R\theta}{t}}$
• But $\mathbf\small{\frac{\text{Linear distance traveled by the disc}}{t}}$ is the linear velocity with which the disc travels during the time interval 't'
♦ 'Linear velocity of the disc' is the 'velocity of the center of mass'
♦ We denoted it as $\mathbf\small{\vec{v}_{CM}}$
• Also, $\mathbf\small{\frac{\theta}{t}}$ is the angular velocity with which the disc turns during the time interval 't'
♦ We denoted the angular velocity as $\mathbf\small{\vec{\omega}}$
■ Thus we get:
Eq.7.32: $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$
• Since it is a rigid body, every particle in the disc will have a translational velocity of $\mathbf\small{|\vec{v}_{CM}|}$ towards the right
12. This equation is similar to: $\mathbf\small{\vec{v}=R\,\vec{\omega}}$
• $\mathbf\small{\vec{v}}$ is the 'tangential velocity' of a particle at the periphery of a rotating body
• But $\mathbf\small{\vec{v}_{CM}}$ is the linear velocity with which the disc rolls
• We must clearly note the difference between the two
■ Eq.7.32 gives the condition for 'no slipping'
• Because if there is slipping, arc length will not be equal to the linear distance. The equation will not be valid any more
13. Let us see another interesting information:
• Fig.7.138(a) below shows 3 points in the disc:
(i) The bottom most point A
(ii) The center of mass C
(iii) The top most point B
 |
| Fig.7.138 |
14. First we will consider point B:
• The disc has an angular velocity of $\mathbf\small{|\vec{\omega}|}$
• Then the point B will be moving towards the right with a tangential velocity of $\mathbf\small{R\,|\vec{\omega}|}$
• Note that, this velocity will be parallel to the ground. That is., parallel to $\mathbf\small{|\vec{v}_{CM}|}$
• Also, this velocity is towards the right
15. We have seen that, due to translation, already every point in the disc has a velocity of $\mathbf\small{R|\vec{\omega}|}$ towards the right. For point B, this is shown in fig.b
• So for point B, the two velocities will add up
• Thus we can write:
Point B moves towards the right with a net velocity of $\mathbf\small{2R|\vec{\omega}|}$. This is shown in fig.c
16. But at the next instant, a succeeding particle takes over the position of B. This new particle will experience the same effect
■ So in general, we can write:
At any instant, the top most point of the disc will be travelling towards the right with a velocity of $\mathbf\small{2R|\vec{\omega}|}$
17. Next we will consider point C
• The point C is in the axis of rotation. So it will receive no contribution (towards velocity) from the rotation. Because, every point on the axis will be having zero rotation
• But C has translational velocity $\mathbf\small{|\vec{v}_{CM}|}$. It is the only velocity it has. This is shown in fig.b
■ Thus we can write:
Point C moves towards the right with a net velocity of $\mathbf\small{R|\vec{\omega}|}$. This is shown in fig.c
18. Finally we consider point A
• The disc has an angular velocity of $\mathbf\small{|\vec{\omega}|}$
• Then the point A will be moving towards the left with a tangential velocity of $\mathbf\small{R\,|\vec{\omega}|}$. This is shown in fig.a
• Note that, this velocity will be parallel to the ground. That is., parallel to $\mathbf\small{|\vec{v}_{CM}|}$
• Also, this velocity is towards the left
19. We have seen that, due to translation, already every point in the disc has a velocity of $\mathbf\small{R|\vec{\omega}|}$ towards the right. For point A, this is shown in fig.b
• So for point A, the two velocities will cancel each other
• Thus we can write:
Point A has zero net velocity. This is shown in fig.c
20. But at the next instant, the succeeding particle takes over the position of A, This new particle will experience the same effect
■ So in general, we can write:
At any instant, the bottom most point of the disc will be having zero net velocity
• That means, the point of contact (with the ground) of a rolling wheel/disc will be at rest
Kinetic energy of a rolling body
We will write this in steps:
1. A rolling body has both rotational and translational motion
2. We know that kinetic energy due to rotation is $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2}$
Where:
• I = Moment of inertia of the body about the axis of rotation
• $\mathbf\small{|\vec{\omega}|}$ = Magnitude of the angular velocity
3. Also we know that kinetic energy due to translation is $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
Where:
• m = mass of the body
• $\mathbf\small{|\vec{v}_{CM}|}$ = magnitude of the linear velocity
4. So the total kinetic energy possessed by a rolling body is given by:
Eq.7.33: $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2+\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
Solved example 7.44
A disc of mass 5 kg and radius 50 cm rolls on the ground with out slipping. It's linear velocity is 10 ms-1. What is the kinetic energy possessed by the disc?
Solution:
1. Kinetic energy due to rotation = $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2}$
(i) Moment of inertia (I) should be selected carefully
• A disc can have a moment of inertia about an axis which can be an extension of one of it's diameters
• It can also have a moment of inertia about an axis perpendicular to the disc. Obviously, a disc can roll only if the axis is perpendicular to the disc. So we must take the I corresponding to this axis
Thus we have: $\mathbf\small{I=\frac{1}{2}m\,R^2}$
(ii) Next we calculate $\mathbf\small{|\vec{\omega}|}$
• Given that the disc rolls without slipping. So the condition $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$ will be satisfied
• So we get: $\mathbf\small{|\vec{\omega}|=\frac{|\vec{v}_{CM}|}{R}}$
(iii) Substituting these expressions in (1), we get:
Kinetic energy due to rotation = $\mathbf\small{\frac{1}{2}\left(\frac{1}{2}m\,R^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{4}}$
2. Kinetic energy due to translation = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
3. So total kinetic energy = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{4}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{3m|\vec{v}_{CM}|^2}{4}}$
• Substituting the values, we get:
Total kinetic energy = $\mathbf\small{\frac{3(5\,\rm{kg})(10\,\rm{m\,s^{-1}})^2}{4}=375\;\rm{J}}$
• Recall that dimensions of energy is: ML2T-2.
Solved example 7.45
Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?
Solution:
• In this problem, we apply the law of conservation of energy
• The body is initially at rest at the top most point of the inclined plane. This is shown in fig.7.139 below:
 |
| Fig.7.139 |
• Since it is initially at rest, there is no initial kinetic energy
• When it reaches at the base, the total energy that it possesses must be equal to the total energy at top
• At the base, there is no potential energy. There is only kinetic energy
■ So we get:
Total kinetic energy of the body at the base = Total potential energy at the top
• Now we consider each body separately
1. Ring:
(i) Rotational kinetic energy
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(m\,R^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{2}}$
• Note that for a ring, I = mR2
• Also we have: $\mathbf\small{|\vec{\omega}|=\frac{|\vec{v}_{CM}|}{R}}$
(ii) Translational kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
(iii) So total kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2+\frac{m|\vec{v}_{CM}|^2}{2}=m|\vec{v}_{CM}|^2}$
(iv) potential energy = mgh
(iii) Equating the two energies, we get: $\mathbf\small{mgh=m|\vec{v}_{CM}|^2}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{gh}}$
2. Solid cylinder
(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 = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{4}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{3m|\vec{v}_{CM}|^2}{4}}$
(iv) potential energy = mgh
(v) Equating the two energies, we get: $\mathbf\small{mgh=\frac{3m|\vec{v}_{CM}|^2}{4}}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{\frac{4gh}{3}}}$
3. Solid sphere
(i) Rotational kinetic energy
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(\frac{2m\,R^2}{5}\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{5}}$
• Note that for a solid sphere, $\mathbf\small{I=\frac{2m\,R^2}{5}}$
(ii) Translational kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
(iii) So total kinetic energy = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{5}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{7m|\vec{v}_{CM}|^2}{10}}$
(iv) potential energy = mgh
(v) Equating the two energies, we get: $\mathbf\small{mgh=\frac{7m|\vec{v}_{CM}|^2}{10}}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{\frac{10gh}{7}}}$
4. We will now compare the results. We will use the subscripts r, c and s for the ring, solid cylinder and solid sphere respectively:
We have:
(i) $\mathbf\small{|\vec{v}_{CM(r)}|=\sqrt{gh}}$
(ii) $\mathbf\small{|\vec{v}_{CM(c)}|=\sqrt{\frac{4gh}{3}}}$
(iii) $\mathbf\small{|\vec{v}_{CM(s)}|=\sqrt{\frac{10gh}{7}}}$
• 4⁄3 = 1.333
• 10⁄7 = 1.429
• Thus we see that $\mathbf\small{\Rightarrow |\vec{v}_{CM(s)}|}$ has the greatest value
■ So we can write:
The sphere reaches the ground with the maximum velocity
Another method:
• We can derive a general formula (for the velocity at the base) for any body
1. For any body, we have: I = Mk2
• Where k is the radius of gyration of that body (Details here)
2. So for any body, the rotational kinetic energy
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(m\,k^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2}$
3. So total energy
= $\mathbf\small{\frac{1}{2}\left(m\,k^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2+\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
= $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2\left(1+\frac{k^2}{R^2} \right)}$
4. Equating this total energy to potential energy, we get:
$\mathbf\small{mgh=\frac{1}{2}m\,|\vec{v}_{CM}|^2\left(1+\frac{k^2}{R^2} \right)}$
5. From this, we get:
Eq.7.34: $\mathbf\small{|\vec{v}_{CM}|^2=\frac{2gh}{\left(1+\frac{k^2}{R^2} \right)}}$
• We see that, the velocity at the base is independent of the mass
• In an earlier chapter on pure translational motion also, we have proved that, the velocity of at the base of an inclined plane is independent of the mass
In the next section, we will see relation between rolling motion and friction
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