The figure below shows a ball with mass m = 0.311 kg attached to the end of a th

Question

The figure below shows a ball with mass m = 0.311 kg attached to the end of a thin rod with length L = 0.507 m and negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle. The rod is held horizontally as shown and then given enough of a downward push to cause the ball to swing down and around and just reach the vertically up position, with zero speed there.

(a) What initial speed must be given the ball so that it reaches the vertically upward position with zero speed?
1      m/s

(b) What then is its speed at the lowest point
2      m/s

(c) What then is its speed at the point on the right level with the initial point
3      m/s If the ball's mass were doubled, would the answers to (a) through (c) increase, decrease, or remain the same?  

Explanation / Answer

L = 0.507m m = 0.311kg

hf = 0.507 m

vi = ? , vf = 0

Initial height of the ball (at the bottom of the vertical h = 0): h = L - L*cos(theta)

= L(1 - cos(theta))

Now, for the ball to reach the high point, the Kinetic Energy must be transfered into Potential Energy.

PE at the Top: PE = m*g*hf

So the KE needed at the low point = PE KE = PE 1/2*m*v^2 = m*g*hf

v = sqrt(2*g*hf)= V = sqrt ( 2* 9.8 * 0.507) =3.15 m/s<==== answer

b Now, the intial Energy must equal that Energy,

so at the point of release you have some PE and some KE.

Energy = m*g*h + 1/2*m*vi^2 = m*g*hf

solve for vi

vi = sqrt(v0^2 + 2*g*h)) = 4.44 m/s <==== answer

for a and c The reason the velocity is the same at both points is because of the height being the same and energy is conservered.

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