Asked by Anonymous
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.
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
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
Answers
Answered by
bobpursley
https://www.jiskha.com/display.cgi?id=1523916313
Answered by
bobpursley
Is it possible you can put your name in the post, so we can track who asked what?
Answered by
Einstein
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
Answered by
Einstein
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
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
Answered by
Einstein
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
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
Answered by
Cinnamon Roll (Izuku)
thank you
There are no AI answers yet. The ability to request AI answers is coming soon!