I don't know why my answer is wrong for this question. My answer is .8H. If you can help me,please do so because I got stuck at this question for a whole week. Please show me with steps. If you got ignore by the same question, I'm more ignore by it then you.

Question

A solid cylinder 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 friction-less (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 solid cylinder 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

bobpursley Monday, April 16, 2018 at

This is how I have tried to solve it base on what I had learned from a previous post.
mgh=1/2mv^2+1/2(1/4mr^2)*w^2
mgh==1/2mv^2+1/8mv^2
v^2=(gh)/(5/8)
(1/2 m v^2=1/2 m gH*8/5)=.8H

Answer 1

https://www.jiskha.com/display.cgi?id=1523916313

Answer 2

Is it possible you can put your name in the post, so we can track who asked what?

Answer 3

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
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=1/4 m gH*4/1)=2mgH. 2mgH=mgR . R=2H

Answer 4

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)=.6mgH. .6mgH=mgR . R=.6H

Answer 5

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

Answer 6

thank you