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# a disk of mass m and radius r, a hoop of mass 2m and radius r, and a ball of mass m and radius 2r are rolling without slipping. the hoop can be treated as a thin ring and the ball should be modeled as a hollow sphere.

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## A disk of mass M and radius R, a hoop of mass 2M and radius R, and a ball of mass M and radius 2R are rolling without slipping. The hoop can be treated as a thin ring and the ball should be modeled as a hollow sphere. Part (a) The objects are rolling on a

Answer to: A disk of mass M and radius R, a hoop of mass 2M and radius R, and a ball of mass M and radius 2R are rolling without slipping. The hoop...

Moment of inertia

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## Question:

A disk of mass M and radius R, a hoop of mass 2M and radius R, and a ball of mass M and radius 2R are rolling without slipping. The hoop can be treated as a thin ring and the ball should be modeled as a hollow sphere.

Part (a) The objects are rolling on a flat surface with the same linear speed. Which have the same angular speed?

Part (b) The objects are rolling on a flat surface with the same angular speed. Which have the same linear speed?

Part (c) Which of the objects has the smallest moment of inertia?

Part (d) The objects are placed at the top of an incline and released from rest. Which one is first to reach the bottom of the incline?

## Rolling Motion:

There is a particular case of combined translational and rotational motion such that there is no slippage at the bearing point on the surface. This is what is known as pure rolling. This condition is met if there is a link between linear and angular velocity:

v = ω R v=ωR

a)

Since there is no slippage, linear speed and angular velocity are made through the following expression:

v = ω R ω = v R v=ωRω=vR

Those with the same radius (disk, hoop) have the same angular speed.

b) The objects are rolling on a flat surface with the same angular speed. Those with the same radius (disk, hoop) have the same linear speed.

c)

The moment of inertia of a disk is:

I d = M R 2 2 Id=MR22

The moment of inertia of a hoop is:

I h = 2 M R 2 Ih=2MR2

The moment of inertia of a ball is:

I b = 8 3 M R 2 Ib=83MR2

The object with the smallest moment of inertia is the disk.

d)

Bodies move in the plane a distance equal to:

Δ x = v 0 t + a t 2 2 Δ x = a t 2 2 Δx=v0t+at22Δx=at22

The time it takes to reach the base depends on the distance and the acceleration. But the distance that the three travel is the same. Let's analyze the acceleration of the center of mass.

Newton's second law is:

F g x = m a g sin θ = a Fgx=magsin⁡θ=a

The acceleration of the center of mass is the same for the three bodies. They will arrive at the same time.

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The moment of inertia is a representation of the distribution of a rotating object and the amount of mass it contains. Learn about rolling motion and the moment of inertia, measuring the moment of inertia, and the theoretical value.

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## Rolling without slipping problems (video)

David explains how to solve problems where an object rolls without slipping. Current time:0:00Total duration:15:00

Torque, moments, and angular momentum

## Rolling without slipping problems

David explains how to solve problems where an object rolls without slipping. Created by David SantoPietro.

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## Torque, moments, and angular momentum

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## Want to join the conversation?

Posted 6 years ago. Direct link to Ninad Tengse's post “At 13:10 isn't the height...”

At 13:10 isn't the height 6m?

• CLayneFarr 6 years ago

Posted 6 years ago. Direct link to CLayneFarr's post “No, if you think about it...”

No, if you think about it, if that ball has a radius of 2m. So when the ball is touching the ground, it's center of mass will actually still be 2m from the ground. It's true that the center of mass is initially 6m from the ground, but when the ball falls and touches the ground the center of mass is again still 2m from the ground. That means the height will be 4m. If you work the problem where the height is 6m, the ball would have to fall halfway through the floor for the center of mass to be at 0 height.

Tuan Anh Dang 4 years ago

Posted 4 years ago. Direct link to Tuan Anh Dang's post “I could have sworn that j...”

I could have sworn that just a couple of videos ago, the moment of inertia equation was I=mr^2, but now in this video it is I=1/2mr^2. Please help, I do not get it.

• Andrew M 4 years ago

Posted 4 years ago. Direct link to Andrew M's post “depends on the shape of t...”

depends on the shape of the object, and the axis around which it is spinning.

ananyapassi123 5 years ago

Posted 5 years ago. Direct link to ananyapassi123's post “At 14:17 energy conservat...”

At 14:17 energy conservation is used which is only applicable in the absence of non conservative forces. However, isn't static friction required for rolling without slipping?

• Johnny Dollard 2 years ago

Posted 2 years ago. Direct link to Johnny Dollard's post “In this case, conservatio...”

In this case, conservation of energy can be used because no energy is lost to friction. The static friction force is actually what causes the object to rotate, so the energy is conserved as rotational kinetic energy.

James 5 years ago

Posted 5 years ago. Direct link to James's post “02:56; At the split secon...”

02:56; At the split second in time v=0 for the tire in contact with the ground. How is it, reference the road surface, the exact opposite point on the tire (180deg from base) is exhibiting a v>0? Surely the finite time snap would make the two points on tire equal in v? Unless the tire is flexible but this seems outside the scope of this problem...

• shreyas kudari 5 years ago

Posted 5 years ago. Direct link to shreyas kudari's post “I have a question regardi...”

I have a question regarding this topic but it may not be in the video. Suppose a ball is rolling without slipping on a surface( with friction) at a constant linear velocity.

We know that there is friction which prevents the ball from slipping. Why doesn't this frictional force act as a torque and speed up the ball as well?The force is present. It can act as a torque.

Also consider the case where an external force is tugging the ball along. In this case, my book (Barron's) says that friction provides torque in order to keep up with the linear acceleration.

In the first case, where there's a constant velocity and 0 acceleration, why doesn't friction provide

torque?

Can someone please clarify this to me as soon as possible?

Thanks a lot!!

#APphysicsCMechanics

• V_Keyd 5 years ago

Posted 5 years ago. Direct link to V_Keyd's post “If the ball is rolling wi...”

If the ball is rolling without slipping at a constant velocity, the point of contact has no tendency to slip against the surface and therefore, there is no friction. If the ball were skidding and rolling, there would have been a friction force acting at the point of contact and providing a torque in a direction for increasing the rotational velocity of the ball. This increase in rotational velocity happens only up till the condition V_cm = R.ω is achieved. So friction force will act and will provide a torque only when the ball is slipping against the surface and when there is no external force tugging on the ball like in the second case you mention.

In the second case, as long as there is an external force tugging on the ball, accelerating it, friction force will continue to act so that the ball tries to achieve the condition of rolling without slipping.

Posted 5 years ago. Direct link to Anjali Adap's post “I really don't understand...”

## A ring (R), a disc (D), a solid sphere (S) and a hollow sphere (H) with thin walls , all having the same mass and radii, start together from rest at the top of an inclined plane and roll down without slipping.

Click here👆to get an answer to your question ✍️ A ring (R), a disc (D), a solid sphere (S) and a hollow sphere (H) with thin walls , all having the same mass and radii, start together from rest at the top of an inclined plane and roll down without slipping. Question

## A ring (R), a disc (D), a solid sphere (S) and a hollow sphere with thin walls (H), all having the same mass and radii, start together from rest at the top of an inclined plane and roll down without slipping.

This question has multiple correct options

A

B

C

D

## All of them will have the same kinetic energy at the bottom of the incline.

Hard Open in App Solution Verified by Toppr

Correct options are C) and D)

The moment of inertia of ring, disc, solid sphere and hollow sphere are mr

2 , 2 1 ​ mr 2 , 5 2 ​ mr 2 and 3 2 ​ mr 2

respectively. As the moment of inertia is the resistance to rotational motion we see that the object with least moment of inertia will reach the bottom first. Thus the order of reaching the bottom is S,D,HandR.

The gain in total kinetic energy is equal to loss in gravitational potential energy which in turn in same for all 4 objects and equal to mgh, where h is height of inclined plane.

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