# Question about a rigid body and energy

• Clara Chung
In summary, the small sphere does not rotate the same number of times as it rolls around inside the larger sphere. The angle of rotation for the small sphere is represented by ##\theta## and is equal to ##\frac{b-a}{a}\phi##, where ##\phi## is the angle representing how far round the large sphere the small sphere has rolled.
Clara Chung

## The Attempt at a Solution

I have posted the solution, but I don't really understand about the second term in the equation.
Shouldn't the angular velocity of the ball be (b dθ/dt) /a instead of ( (b-a) dθ/dt )/a in the solution?

#### Attachments

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if i am not wrong the nonslip condition implies that
##
v_{cm} = \omega a\\
v_{cm} = (b-a)\dot \theta
##
hence the equation

Clara Chung
vishnu 73 said:
if i am not wrong the nonslip condition implies that
##
v_{cm} = \omega a\\
v_{cm} = (b-a)\dot \theta
##
hence the equation
Thank you, but I still don't understand. The ball touches the sphere. If the angle displace is \theta, the ball's circumference undergone a distance of b \theta.

i think it is any rigid body motion can be described as translation and rotation about a point
i think in principle any other can be used but getting its translation velocity will be hard(i think not sure) but for cm it is quite easy it is always b-a away from centre hence the bottom most point is translating at the velcoity
vcm forward but rotating backward at omega r you set them equal to each other for non slip

vishnu 73 said:
i think it is any rigid body motion can be described as translation and rotation about a point
i think in principle any other can be used but getting its translation velocity will be hard(i think not sure) but for cm it is quite easy it is always b-a away from centre hence the bottom most point is translating at the velcoity
vcm forward but rotating backward at omega r you set them equal to each other for non slip
I am not sure if vcm = wa

when the ball is rolling , the velocity of point in contact with the sphere can be described as translation of the center of mass and the rotation about cm the sum of the two must be zero .
the translation is vcm and the rotation velocity about cm is w times distance which is w a
i could not find a link for ball rolling within ball but i found one for ball rolling on top of r
here vcm is r+R away and at 3:20 he write w = v/r

Clara Chung said:
Thank you, but I still don't understand. The ball touches the sphere. If the angle displace is \theta, the ball's circumference undergone a distance of b \theta.

You need to draw a good diagram of the situation and look very closely at how many times the small sphere rotates as it moves all the way round the inside of the larger sphere. For example, you could take the small sphere to have 1/4 of the radius of the large sphere.

PeroK said:
You need to draw a good diagram of the situation and look very closely at how many times the small sphere rotates as it moves all the way round the inside of the larger sphere. For example, you could take the small sphere to have 1/4 of the radius of the large sphere.
Sorry I just can't get the idea, maybe I am too stupid.
If the centre of the small ball travel by theta. The circumference of the small ball has traveled a distance of bθ.
Therefore the angle traveled by the small ball is Φ=bθ/a.
By differentiating, I can get
w=dΦ/dt = b/a dθ/dt.

Clara Chung said:
Sorry I just can't get the idea, maybe I am too stupid.
If the centre of the small ball travel by theta. The circumference of the small ball has traveled a distance of bθ.
Therefore the angle traveled by the small ball is Φ=bθ/a.
By differentiating, I can get
w=dΦ/dt = b/a dθ/dt.

It's not a question of being stupid. But, that is nothing to do with what I posted. It's not a question of differentiation. It's a question of figuring out a tricky physical scenario.

I'll give you some help:

PeroK said:
You need to draw a good diagram of the situation and look very closely at how many times the small sphere rotates as it moves all the way round the inside of the larger sphere. For example, you could take the small sphere to have 1/4 of the radius of the large sphere.

In this scenario, the small sphere does not rotate 4 times as it rolls around inside the larger sphere. You need to figure out how many times it does rotate and why.

PS in the above diagram, mark the point on the small sphere which starts at the bottom, touching the large sphere. Imagine the small sphere rolling thru one quarter of the large sphere. Where is the mark on the small sphere? How much has the small sphere rotated in rolling 1/4 of the way round the large sphere?

PeroK said:
PS in the above diagram, mark the point on the small sphere which starts at the bottom, touching the large sphere. Imagine the small sphere rolling thru one quarter of the large sphere. Where is the mark on the small sphere? How much has the small sphere rotated in rolling 1/4 of the way round the large sphere?
Does the point land on the rightmost of the sphere?

Clara Chung said:
Does the point land on the rightmost of the sphere?

Yes. And that's less than one rotation of the small sphere.

Last edited:
Clara Chung
PeroK said:
Yes. And that's more than one rotation of the small sphere.
Does it rotate 3/4 cycle (less than one rotation)? What does it mean by a rotation, is it when the point is again in the buttom position?

Clara Chung said:
Does it rotate 3/4 cycle (less than one rotation)? What does it mean by a rotation, is it when the point is again in the buttom position?

Yes, less than one rotation. What that means is that you have one angle representing how far round the large sphere the small sphere has rolled. Let's call this ##\phi##. And another angle representing the angular rotation of the small sphere. Let's call this ##\theta##.

In this case, we have ##\theta = 3 \phi##. And, in general, we have ##\theta = \frac{b-a}{a}\phi##.

You also know how far the centre of mass of the small sphere has moved. It is moving in a circle of radius ##4-1## units. And, in general, of radius ##b-a##.

These two things give you the KE of the rolling sphere.

PS I tried to get ##\phi## and ##\theta## the right way round, but of course the book has them the other way round. Apologies.

Last edited:
timetraveller123 and Clara Chung
PeroK said:
Yes, less than one rotation. What that means is that you have one angle representing how far round the large sphere the small sphere has rolled. Let's call this ##\phi##. And another angle representing the angular rotation of the small sphere. Let's call this ##\theta##.

In this case, we have ##\theta = \frac34 \phi##. And, in general, we have ##\theta = \frac{b-a}{a}\phi##.

You also know how far the centre of mass of the small sphere has moved. It is moving in a circle of radius ##4-1## units. And, in general, of radius ##b-a##.

These two things give you the KE of the rolling sphere.

@PeroK
i was thinking about the problem as a sum of translation and rotation

so first i imagined it being frictionless in this case when when the centre is displaced theta the initial point of contact also is rotated by theta from the final point of contact

hence the point which is always is contact with the sphere has moved by b theta but the smaller circle will have appeared to have rotated by a theta
when friction is turned on friction will make the smaller circle turn an additional angle phi to make the translational and rotational distance equal
so b theta = a theta + a phi differentiating yields the necessary expression
is this valid to do it?

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vishnu 73 said:
@PeroK
i was thinking about the problem as a sum of translation and rotation

so first i imagined it being frictionless in this case when when the centre is displaced theta the initial point of contact also is rotated by theta from the final point of contact
View attachment 231468

hence the point which is always is contact with the sphere has moved by b theta but the smaller circle will have appeared to have rotated by a theta
when friction is turned on friction will make the smaller circle turn an additional angle phi to make the translational and rotational distance equal
so b theta = a theta + a phi differentiating yields the necessary expression
is this valid to do it?

Yes, I think that's another way to look at it.

timetraveller123
ok thanks

## 1. What is a rigid body?

A rigid body is a physical object that maintains its shape and size regardless of external forces acting on it. This means that the object does not deform or change its shape when subjected to forces, such as twisting, bending, or stretching.

## 2. How is energy related to a rigid body?

Energy is a fundamental concept in physics that is closely related to a rigid body. A rigid body possesses both potential energy and kinetic energy. Potential energy is the energy an object has due to its position or configuration, while kinetic energy is the energy an object has due to its motion.

## 3. Can a rigid body have zero energy?

No, a rigid body cannot have zero energy. This is because even if the body is stationary, it still possesses potential energy due to its position in a gravitational field. Additionally, the molecules and atoms that make up the rigid body are always in motion, meaning the body also has kinetic energy.

## 4. How does the conservation of energy apply to a rigid body?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. This means that in a closed system, the total amount of energy remains constant. In the case of a rigid body, the total energy (potential + kinetic) remains constant as long as there are no external forces acting on the body.

## 5. How does the concept of work apply to a rigid body?

Work is defined as the product of force and displacement. In the case of a rigid body, work is done when a force acts on the body and causes a displacement. The amount of work done is equal to the force multiplied by the distance over which it acts. This work can either increase or decrease the energy of the rigid body, depending on the direction of the force and displacement.

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