# physics

posted by Anonymous

Can somebody please shows me how to get to the answer key,06H?
A thin, spherical shell of mass m and radius R rolls down a parabolic path PQR from height H without slipping (assume R ≪ H) as shown in the figure below. Path PQ is rough (and so the cylinder will roll on that path), whereas path QR is smooth, or frictionless (so the cylinder will only slide, not roll, in this region). Determine the height h above point Q reached by the cylinder on path QR. (Use the following as necessary: m, g, H, and R.)

The spherical shell starts on the left side of the parabola curve at point P then goes down to point Q then up to point R.H is from the floor to point P

1. bobpursley

PE going into the down side: mgH
KE is translational, and rotational or
mgH=1/2 m v^2 + 1/2 (2/3)mr^2*w^2
but when rolling, w=v/r
mgH=1/2 m v^2+1/3*mv^2
v^2=gh/(5/6) check me on that.
on the second side, rolling doesn't happen, the shell just spins, and how high it goes depends on the initial translational energy (1/2 m v^2=1/2 m gH*6/5)=.6mgH

so that inergy will equal how high it goes.

.6mgH=mgR
R=.6H I have no idea where the 06H came from. Now a question for you. I outlined all this on your post a couple of days ago.

2. Anonymous

PE going into the down side: mgH
KE is translational, and rotational or
mgH=1/2 m v^2 + 1/2 Imr^2*w^2
mgH=1/2 m v^2 + 1/2 (1/2)mr^2*w^2
but when rolling, w=v/r
mgH=1/2 m v^2+1/4*mv^2 =3/4mv^2
On the second side, rolling doesn t happen, the shell just spins, and how high it goes depends on the initial translational energy (1/2 m v^2= 1/2mgH*4/3)=2/3mgH. 2/3mgH=mgR . R=2/3H or .7H

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