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# Mechanics Problem watch

1. http://img81.imageshack.us/img81/2522/untitled21ap.jpg

What I've done so far:

Using conservation of momentum:
mu = (m+M)V
V = mu/(m+M)

Change in kinetic energy:
1/2mu² - 1/2(m+M)(mu/(m+M)) = 1/2mu² - 1/2mu = 1/2mu(u-1)

For the next part: I assumed the equation of motion would be:
T + (m+M)g = [(m+M)V²]/R

But does the velocity remain constant?
2. (Original post by SunGod87)
Change in kinetic energy:
1/2mu² - 1/2(m+M)(mu/(m+M)) = 1/2mu² - 1/2mu = 1/2mu(u-1)
You forgot to square that.

Now, the velocity certainly doesn't remain constant as the particle is changing direction. So let's find the velocity v when the particle is at the top using conservation of energy:
0.5(m+M)V^2 = (m+M)g(2R) + 0.5(m+M)v^2
=> v^2 = V^2 - 4gR

So we have:
T = (m+M)v^2/R - (m+M)g
= (m+M)(v^2 - gR)/R
= (m+M)(V^2 - 5gR)/R, as required.

The string remains taught as long as T>=0, that is, as long as:
V^2 >= 5gR
(mu/(m+M))^2 >= 5gR

So the minimum value of u is:
(m+M)/m * sqrt(5gR)

And finally:
T_u - (m+M)g = (m+M)V^2/R
= (m+M)(5gR)/R
= 5(m+M)g
=> T_u = 6(m+M)g
3. Thanks very much

For some reason I was assuming at the top all of the energy was potential and none kinetic, silly me.

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Updated: April 11, 2006
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