Question

In the figure shown a ring A is initially rolling without sliding with a velocity v on the horizontal surface of the body B(of same mass as A).All surfaces are smooth .B has no initial velocity.What will be the maximum height reached by A on B?

Answer 1

radius of ring is given???

Answer 2

I got it. See:
By conservation of energy: \[\frac{ 1 }{ 2 }mv ^{2} = mgh + \frac{ 1 }{ 2 }mv, ^{2}\]
And by conservation of momentum: \[mv = 2mv,\]
Solving these 2 equations we get: \[h = \frac{ v }{ 2 }\sqrt{\frac{ 3 }{ 2g }}\]

Answer 3

no the answer is v^2/4g

Answer 4

did u try it once again with my method???

Answer 5

yes :(

Answer 6

@DLS

Answer 7

I for ring about the surface can be found by parallel axis theorem.
\[\LARGE MR^2+\frac{MR^2}{2}=\frac{3MR^2}{2}\]
\[V=RW(Pure~Rolling)\]
Applying conservation of energy:
\[\frac{1}{2}mv^2+\frac{1}{2}IW^2=mgH(H~is~what~you~want)\]
\[\frac{1}{2}I \frac{v^2}{r^2}=mgH-\frac{1}{2}mv^2\]
\[\frac{1}{2} \times \frac{3}{2}M \cancel r^2 \frac{v^2}{\cancel r^2}=mgH-\frac{1}{2}mv^2\]
I can smell the answer :) let me know if you still don't get it.

Answer 8

don't confuse small m and capital M its the same things,sorry,M will cancel out..

Answer 9

Yash has given the answer to be v^2/4g.
Your method does not give this .

Answer 10

It is 3V^2/4G

Answer 11

He wrote it wrong,substitute 3v^2/4g in the 1st condition and solve for I

Answer 12

In this scenario there are 2 properties conserved.
1). linear momentum of the 2-body system in the horizontal direction.
2). energy is conserved

so from 1)
when the ring reaches maximum height, its relative velocity with respect to the bigger body would be 0 i.e. both would move with the same velocity at maximum height
so At maximum height,
let the combined velocity of the system by v
let the initial velocity of the ring be u
hence
m∗u=(m+M)∗v

v=m/(m+M)∗u


Thus we have the velocity of the system at ring's maximum height.

now lets apply energy conservation.
\[1/2∗m∗u^2=1/2∗(m+M)∗v^2+m∗g∗h\]



Now m = M (given)

hence,
\[u^2/2=u^2/4+gh\]

\[h=u^2/4g\]