# Speed of block as cylinder reaches bottom of circular track.

1. Oct 3, 2015

### takando12

1. The problem statement, all variables and given/known data
A block of mass M=2kg with a semicircular track of radius R=1.1m rests on a horizontal frictionless surface.A uniform cylinder of radius r=10cm and mass m=1kg is released from rest from the top most point .The cylinder slips on the semicircular frictionless track.The speed of the block when the cylinder reaches the bottom of the track is:(g=10m/s).
A)√10/3 B)√4/3 C)√5/2 D)√10.

2. Relevant equations
conservation of momentum.
vcentre of mass= v1m1 + v2m2/m1 +m2.

3. The attempt at a solution
I am not very sure how to solve it. I tried to calculate the speed of the cylinder at the bottom.
Consider the the bottom of the semicircular track to be h m above the surface. v1=0m/s
mg(h+1.1) -mgh= 1/2m(v22-v12)
2*10*1.1=v22.
v2=√22 m/s.
then using conservation of momentum, I tried to find the velocity of the block but the answer doesn't match.
How do I solve this?

2. Oct 3, 2015

### Staff: Mentor

Can you explain your KE subexpression: 1/2m(v22-v12) ?

3. Oct 3, 2015

### takando12

Sir,
I thought the change in the potential energy as it goes from the top to the bottom of the track will become the kinetic energy. Here v1=0m/s as it started from rest. Please do correct me if I am wrong.

4. Oct 3, 2015

### takando12

And please do forgive me but the heading of the question is wrong. It is not the speed of the cylinder when it reaches the bottom , but the speed of the track when the cylinder reaches the bottom.

5. Oct 3, 2015

### Staff: Mentor

Yes, energy conservation is applicable. What I'm wondering about is how you account for the fact that the KE will be split between the cylinder and block... there won't be just one velocity involved, and the initial velocity won't be one of them...
I've altered the title accordingly.

6. Oct 3, 2015

### takando12

Oh so the kinetic energy splits between the block and the cylinder? I have no idea how to proceed.
The only other formula we were taught is the shift in the centre of mass position. Here the shift will be zero as no external force is applied. But even if we do find the displacement of the centre of mass of the block , we don't have time to solve for speed.

7. Oct 3, 2015

### haruspex

The constancy of the X coordinate of the mass centre is directly related to another conservation law. What other conservation laws do you know that might be relevant?

8. Aug 20, 2016

### COM

even after applying conservation of momentum and energy, answer is not matching with any of the options......what to do??

9. Aug 20, 2016

### COM

can relative velocity come into picture????

10. Aug 20, 2016

### COM

i wonder what can be the use of the cylinder's radius??????????????

11. Aug 20, 2016

### TSny

Welcome to PF!
What is the change in height of the cylinder from the point of release to the when the cylinder is at the bottom of the track?

Last edited: Aug 21, 2016
12. Aug 20, 2016

### jbriggs444

Is there a drawing? The picture I have in mind is that the release point for the cylinder is with its center level with the top surface of the block. How much below that position can the cylinder reach?

Now, you say that you have applied conservation of momentum. How have you applied it? What equation can you write to express it?