# KE of a hoop rolling within a tube

• davon806
In summary, the professor attempted to solve a problem where the hoop had to rotate about its center at ##-R\dot\theta/a## in order to restore it to rolling contact. However, if you make the hoop rotate at rate ##\omega## about its center then the difference from the rigid arm model is a rotation rate of ##\omega-\dot\theta##. The KE calculation in the attachment seems a crazy approach to me, but I agree with haruspex that if you allow the center of mass coordinates to rotate with the arm then the second form is correct.
davon806

## Homework Statement

Please see the attached
In this example my professor is trying to calculate the total KE of the hoop,and let since the velocity at the contact point of the hoop and tube is zero we have v = vt + aω.This is equivalent to
(R-a)Θ. = - aω.

However,he later said that it was wrong since the no-slip condition should be RΘ. = aω but he did not give any explanation.Though it makes sense if you think about it geometrically,I think both of them are correct but I am not sure which one is correct?

## The Attempt at a Solution

I am not sure,the equation T = T' + 1/2MV^2 where S and S' are an inertial frame and the centre of momentum frame,V is the speed of S' wrt S,holds under Galilean transformation.I think the set of Cartesian coordinates at CM of hoop is somewhat rotating wrt to the observer/lab frame so that it is not a Galilean transformation and we couldn't calculate the KE in the way as shown in the attached?

#### Attachments

• COM.png
94.9 KB · Views: 396
davon806 said:

## Homework Statement

Please see the attached
In this example my professor is trying to calculate the total KE of the hoop,and let since the velocity at the contact point of the hoop and tube is zero we have v = vt + aω.This is equivalent to
(R-a)Θ. = - aω.

However,he later said that it was wrong since the no-slip condition should be RΘ. = aω but he did not give any explanation.Though it makes sense if you think about it geometrically,I think both of them are correct but I am not sure which one is correct?

## The Attempt at a Solution

I am not sure,the equation T = T' + 1/2MV^2 where S and S' are an inertial frame and the centre of momentum frame,V is the speed of S' wrt S,holds under Galilean transformation.I think the set of Cartesian coordinates at CM of hoop is somewhat rotating wrt to the observer/lab frame so that it is not a Galilean transformation and we couldn't calculate the KE in the way as shown in the attached?
Your professor was right the first time. I think I can see how he got the second expression. If you treat the hoop as attached rigidly to an arm rotating about the centre of the cylinder, the outermost part of it slides along the cylinder wall at ##R\dot\theta##. So he thought the hoop would have to rotate about its centre at ##-R\dot\theta/a## to restore it to rolling contact. But attaching it to the rigid arm made the hoop itself rotate at rate ##\dot\theta##. If, instead, we make the hoop rotate at rate ##\omega## about its centre then the difference from the rigid arm model is a rotation rate of ##\omega-\dot\theta##. Thus the no slip condition is ##R\dot\theta+a(\omega-\dot\theta)=0##.

The KE calculation in the attachment seems a crazy approach to me. What is the hoop's instantaneous centre of rotation? What is its moment of inertia about that point?

davon806
I agree with haruspex. If you allow the center of mass coordinates to rotate with the arm you get the second form. If you don't allow the COM coords to rotate then the hoop rotates one fewer times and the first form is correct. The problem expressly says the COM coords do not rotate, so the first version is correct.

davon806

## 1. What is the formula for calculating the KE of a hoop rolling within a tube?

The formula for calculating the KE of a hoop rolling within a tube is KE = ½ * m * v², where m is the mass of the hoop and v is the velocity of the hoop.

## 2. How does the radius of the hoop affect its KE while rolling within a tube?

The radius of the hoop has a direct effect on its KE while rolling within a tube. A larger radius will result in a larger KE, as it covers a greater distance with each rotation, while a smaller radius will result in a smaller KE.

## 3. Does the size of the tube have an impact on the KE of a rolling hoop?

Yes, the size of the tube has an impact on the KE of a rolling hoop. A larger tube will result in a larger KE, as it allows for a greater distance for the hoop to travel, while a smaller tube will result in a smaller KE.

## 4. What is the relationship between the speed of the hoop and its KE while rolling within a tube?

The speed of the hoop and its KE while rolling within a tube are directly proportional. This means that as the speed of the hoop increases, so does its KE, and vice versa.

## 5. Can the KE of a hoop rolling within a tube be negative?

No, the KE of a hoop rolling within a tube cannot be negative. According to the formula, the mass and velocity are squared, resulting in a positive value. Even if the velocity is negative, it will be squared, making it positive.

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