Another Solution check Please. (cylinder rolling down a slope)

[phalanx123]

Sorry here is another question that I am not sure of

1. A cylindrical hoop rests on a rough uniform incline. It is released and rolls
without slipping through a vertical distance h0. It then continues up a perfectly
smooth incline. What height does it reach?


Here is my solution

Because of the conservasion of energy

mgh0=1/2mv^2+1/2IOmega^2

since Omega=v/r and the moment of inertia of a cylindrical hoop is I=mr^2

therefore

mgh0=1/2mv^2+1/2mr^2*(v/r)^2
=mv^2

therefore final velocity is v=Square root(gh0)

when goin up the perfect smooth surface it looses the ability to roll, so the only energy is the translaional energy

so let the maximum height it reached be h

mgh=1/2mv^2=1/2mgh0

so h=1/20

Is this right? Thanks

 

[Doc Al]

Yes, looks good. (I presume you mean h = h0/2.)

when goin up the perfect smooth surface it looses the ability to roll

Just to be clear (I'm sure you know this), the hoop loses the ability to roll without slipping. Since there's no friction to exert a torque, the angular speed of the hoop remains constant as it slides up the incline.

 

[phalanx123]

Yes, looks good. (I presume you mean h = h0/2.)

Oops sorry about the typo, yes I mean1/2h0.

Since there's no friction to exert a torque, the angular speed of the hoop remains constant as it slides up the incline.

Is this constant angular speed as it slides up the incline equal to the final augular speed it obtained during the rolling down from the other slope?

I don't know if this is right, but is it because rolling and sliding happen at same time, so although it is rolling and we can see it (or can we?) but it doesn't contribute to the motion of the hoop up the incline? SO on a perfect smooth and level ground if we take away the translational motion the hoop will just roll at where it is without going anywhere? Also is that the other half of the initial total energy "lost" due to this sliding motion, the rotational energy is there but it is not contributing to the motion up the slope? Sorry about all those questions Thanks

 

[Doc Al]

Is this constant angular speed as it slides up the incline equal to the final augular speed it obtained during the rolling down from the other slope?

Absolutely.

I don't know if this is right, but is it because rolling and sliding happen at same time, so although it is rolling and we can see it (or can we?) but it doesn't contribute to the motion of the hoop up the incline? SO on a perfect smooth and level ground if we take away the translational motion the hoop will just roll at where it is without going anywhere?

Yes. With no friction, the rotational and translational motions are independent.

Also is that the other half of the initial total energy "lost" due to this sliding motion, the rotational energy is there but it is not contributing to the motion up the slope?

When friction is present, and something "rolls without slipping", the rotational and translational motions are coupled--they increase and decrease together. Without friction, there is no coupling. As the hoop rolls up the frictionless ramp, the only force acting on the hoop parallel to the ramp is gravity. Gravity reduces the translational speed, but not the rotational speed, since gravity exerts no torque on the hoop. So the rotational energy remains unchanged.