A ball is tied to one end of a string. The other end of the string is fixed. The

Question

A ball is tied to one end of a string. The other end of the string is fixed. The ball isset in motion around a vertical circle without friction. At the top of the circle, theball has a speed of gR, as shown in the figure below. At what angle should the string be cut so that the ball will travel through the center of the circle?

PS. The answer is 100.6 I need to show work.

Energy = 1/2 mgR + mg(2R)

B/C Energy is conserved

Energy = mgR(5/2) J

Please Help ME!!!

A ball is tied to one end of a string. The other end of the string is fixed. The ball isset in motion around a vertical circle without friction. At the top of the circle, theball has a speed of root gR, as shown in the figure below. At what angle ? should the string be cut so that the ball will travel through the center of the circle? PS. The answer is 100.6 I need to show work. Energy = 1/2 mgR + mg(2R) B/C Energy is conserved Energy = mgR(5/2) J Please Help ME!!!

Explanation / Answer

at given angle suppose it has tangential velocity v

then horizontal component = v*cos(180-) = -vcos()

vertical component = vsin(180-) = vsin()

Rcos(-90) = -vcos()*T --> Rsin()=-vcos()*T

Rsin(-90) = vsin()T+.5gT^2 ---> -Rcos()=vsin()T+.5gT^2

put T from the first equation into second

top it has KE = .5mRg
if it is at angle as shown then it is [R-Rsin(-90)] below the top point = R(1+sin())

loss in PE =mgR(1+sin())

KE at this point = .5mRg + mgR(1+sin())

we will get v from this equation

equating v from both and solving will give the answer  

I hope this may help you some what

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