# Finding Angular Speed Using Energy Methods

## Homework Statement

You hang a thin hoop with radius R over a nail at the rim of the hoop. You displace it to the side (within the plane of the hoop) through an angle β from its equilibrium position and let it go. What is its angular position when it returns to its equilibrium position (use the gravitational potential energy equation).

## Homework Equations

K(1) + U(1) = K(2) + U(2)

U(1) = mgh

h = R - Rcosβ

K(2) = (1/2)mv^2 = (1/2)m(ω^2)R^2

## The Attempt at a Solution

0 + U(1) = K(2) + 0

mgh = (1/2)m(ω^2)R^2

mg(R - Rcosβ) = (1/2)m(ω^2)R^2

ω = √((2g(1 - cosβ)/R)

The answer in my textbook is exactly the same as my answer but without the 2. What could have happened to the 2?

## Homework Statement

You hang a thin hoop with radius R over a nail at the rim of the hoop. You displace it to the side (within the plane of the hoop) through an angle β from its equilibrium position and let it go. What is its angular position when it returns to its equilibrium position (use the gravitational potential energy equation).

## Homework Equations

K(1) + U(1) = K(2) + U(2)

U(1) = mgh

h = R - Rcosβ

K(2) = (1/2)mv^2 = (1/2)m(ω^2)R^2

## The Attempt at a Solution

0 + U(1) = K(2) + 0

mgh = (1/2)m(ω^2)R^2

mg(R - Rcosβ) = (1/2)m(ω^2)R^2

ω = √((2g(1 - cosβ)/R)

The answer in my textbook is exactly the same as my answer but without the 2. What could have happened to the 2?

The item in red is wrong.

You have used moment of inertia of hoop about the center.Whereas it should be about the rim attached to nail .

You need to use parallel axis theorem.The correct MI should be 2MR2 .

This would solve the mystery of the unwanted 2 appearing in your expression.

The item in red is wrong.

You have used moment of inertia of hoop about the center.Whereas it should be about the rim attached to nail .

You need to use parallel axis theorem.The correct MI should be 2MR2 .

This would solve the mystery of the unwanted 2 appearing in your expression.

Thank-you very much!

The item in red is wrong.

You have used moment of inertia of hoop about the center.Whereas it should be about the rim attached to nail .

You need to use parallel axis theorem.The correct MI should be 2MR2 .

This would solve the mystery of the unwanted 2 appearing in your expression.

Everything about your explanation makes perfect sense to me, but I just thought of a new reason why my first answer works too.

Imagine the hoop's center of mass in relation to the nail while the hoop swings through β. When I drew it out, the hoop's center of mass and the nail resembled a pendulum. The phantom string attaching the nail to the hoop's center of mass is R. So if I'm right, could the hoop's angular speed be calculated from

U(1) = K(2)

mgy = (1/2)mv^2 = (1/2)m(ω^2)(R^2)

ω = √((2g(1 - cosβ)/R),

which is still what I had?

The motion of hoop about the nail (point on the rim) can be thought in two ways.

1) Pure rotation about a fixed point on the rim .This is the original approach you have taken.Here kinetic energy is purely rotational given by (1/2)Iω2 , where I is the MI about the rim.

2)Translation motion of the CM + Rotation about the CM .

The KE is has two components (1/2)mvcm2 + (1/2)Iω2 , where I is the MI of the hoop about the center .

In post#3 you have only taken energy due to motion of CM (1/2mv2) into account .When you take into account the energy due to rotational motion , the factor of 2 disappears .

The motion of hoop about the nail (point on the rim) can be thought in two ways.

1) Pure rotation about a fixed point on the rim .This is the original approach you have taken.Here kinetic energy is purely rotational given by (1/2)Iω2 , where I is the MI about the rim.

2)Translation motion of the CM + Rotation about the CM .

The KE is has two components (1/2)mvcm2 + (1/2)Iω2 , where I is the MI of the hoop about the center .

In post#3 you have only taken energy due to motion of CM (1/2mv2) into account .When you take into account the energy due to rotational motion , the factor of 2 disappears .

But if this system really does act like a pendulum, why do we need the (1/2)Iω2 part?

The hoop is a rigid body not a point particle .If you insist on comparing the motion of hoop with a pendulum ,then you may consider it as a physical pendulum not a simple pendulum .

The hoop is a rigid body not a point particle .If you insist on comparing the motion of hoop with a pendulum ,then you may consider it as a physical pendulum not a simple pendulum .

Okay, I think it's sinking in. Does my answer work if the hoop's inside edge slips along the nail rather than rotating down the nail? I think that I was imagining a frictionless slide of the hoop that didn't rotate. In that case, I imagine the hoop sliding down a hill with a radius equal to the hoop's diameter. Does this make sense if the question would have specified that the hoop slips down to equilibrium with no rotation?

Sorry...I do not understand your setup .

Your answer would work only in case the body is not rotating around the CM .In other words if the translational speeds of every particle on the rigid body is same as that of CM ,then you may replace the body with a point mass located at the CM .

For example take the case of a ball on an incline .

Case 1) On a frictionless surface ,the ball just slides down .In this case it has only translational kinetic energy.

Case 2) On a surface with sufficient friction such that the ball rolls without slipping , the ball has translational kinetic energy as well as rotational kinetic energy .

In case 2) the ball has lower speed than case 1) on reaching the ground .

Sorry...I do not understand your setup .

Your answer would work only in case the body is not rotating around the CM .In other words if the translational speeds of every particle on the rigid body is same as that of CM ,then you may replace the body with a point mass located at the CM .

For example take the case of a ball on an incline .

Case 1) On a frictionless surface ,the ball just slides down .In this case it has only translational kinetic energy.

Case 2) On a surface with sufficient friction such that the ball rolls without slipping , the ball has translational kinetic energy as well as rotational kinetic energy .

In case 2) the ball has lower speed than case 1) on reaching the ground .

Wow, that is interesting, and thank-you so much for your help and patience!!!!