Speed of block as cylinder reaches bottom of circular track.

[takando12]

Homework Statement

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.

Homework Equations

conservation of momentum.
v_{centre of mass}= v₁m₁ + v₂m₂/m₁ +m₂.

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?

 

[gneill]

mg(h+1.1) -mgh= 1/2m(v₂²-v₁²)

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

 

[takando12]

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

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.

 

[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.

 

[gneill]

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.

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...

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.

I've altered the title accordingly.

 

[takando12]

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.

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.

 

[haruspex]

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.

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?

 

[COM]

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

 

[COM]

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

can relative velocity come into picture?

 

[COM]

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

 

[TSny]

Welcome to PF!

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

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?

 

[jbriggs444]

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

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?