Need help with a qn on rotational motion.

[Wen]

A hoop pf radius r and mass m is rolling without slipping with velocity v towards a step of height h on a horizontal surface. Assume that it does not rebound and no slipping occur at the point of contact when the hoop roll up, what is the minimum velocity needed for the hoop to roll up?



I tried to use conservation of energy to solve:
Ek(transl.)i +EK(rotat.)i=mgh+(Torque.change in angle of rotation)
1/2 mv^2 + 1/2 I w^2 = mgh + T.ditre
...
1/2 mv^2 + 1/2 mr^2(v^2/r^2)= mgh + m(r^2).(angu. acele)(ditre)
v^2=3/2 gh

this is far from the correct answ of 2r(gh)^0.5/(2r-h)

Please help me

 

[Tide]

The correct answer doesn't look quite right either. If h = r then the hoop won't make it up the step at any speed.

In any case, only the component of velocity perpendicular to a line joining the corner of the step and the center of the hoop can contribute to lifting the hoop.

 

[quicknote]

Hello,

Thank you for reaching out for help with your question on rotational motion. It looks like you have made some good progress in your solution and have correctly identified the use of conservation of energy to solve the problem.

One thing to keep in mind when solving rotational motion problems is that the kinetic energy of rotation is not simply 1/2 I w^2, but rather 1/2 I w^2 + 1/2 mv^2, where I is the moment of inertia and w is the angular velocity. This is because the object has both translational and rotational kinetic energy.

In this case, the moment of inertia for a hoop is I=mr^2, so the correct expression for the kinetic energy of rotation is 1/2 mr^2w^2. Substituting this into your conservation of energy equation, we get:

1/2 mv^2 + 1/2 mr^2w^2 = mgh + m(r^2)(angular acceleration)(distance)

Next, we need to find the relationship between the angular velocity and the linear velocity of the hoop. Since the hoop is rolling without slipping, we know that the linear velocity is equal to the angular velocity multiplied by the radius, or v=w*r.

Substituting this into our equation, we get:

1/2 mv^2 + 1/2 mr^2(v^2/r^2) = mgh + m(r^2)(angular acceleration)(distance)

1/2 mv^2 + 1/2 mv^2 = mgh + m(r^2)(angular acceleration)(distance)

mv^2 = 2mgh + m(r^2)(angular acceleration)(distance)

Now, we need to find the relationship between the angular acceleration and the linear acceleration. The linear acceleration at the point of contact is equal to the angular acceleration multiplied by the radius, or a=angular acceleration*r.

Substituting this into our equation, we get:

mv^2 = 2mgh + m(r^2)(angular acceleration)(distance)

mv^2 = 2mgh + m(r^2)(a)(distance)

Next, we need to find the relationship between the linear acceleration and the distance traveled. Since the hoop is rolling without slipping, we know that the distance traveled is equal to the circumference of the hoop, or 2πr.

Substituting this into our equation, we get:

mv^2