As shown in the sketch, a 16 kg, 3.95 m pole is balanced vertically on its tip.

Question

As shown in the sketch, a 16 kg, 3.95 m pole is balanced vertically on its tip. Then it starts to fall with no slipping at its lower end. What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]

V= 10.77 m/s

If on the upper end of the pole there is a ball which has a mass of Mpole/6, what is the speed of the ball (or the upper end of the pole) just before it hits the ground? (Assume the radius of the ball can be ignore compared with the pole)

V= ?

I was able to find the initial velocity of the stick without the ball, but am unable to find the velocity with the ball.

Explanation / Answer

solved a similar question already with different numbers.. i put the question and the answer.. hope it helps!! :) kindly rate a lifesaver :) As shown in the sketch, a 17 kg, 3.25 m pole is balanced vertically on its tip. Then it starts to fall with no slipping at its lower end. If on the upper end of the pole there is a ball which has a mass of Mpole/4, what is the speed of the ball (or the upper end of the pole) just before it hits the ground? (Assume the radius of the ball can be ignore compared with the pole) ??? m/s Picture http://www.webassign.net/userimages/2414%20a9p8.jpg?db=v4net&id=172251 ANSWER: Wok done by gravity = Change in KE of the system Change in KE of the system = (1/2) I ?2 I = Mpole * L2/3 + (Mpole/4) * L2 = 0.5833 Mpole * L2 Wok done by gravity = (M pole + Mball)gh h = change on COM height New COM position(X): M pole x1 + Mball * x2 = (M pole + Mball)* X 17*3.25/2 + (17/4)*3.25 = (17 + (17/4)) X X =1.95 m h = change on COM height = 1.95 m (1/2) I ?2 = (M pole + Mball)gh Solving we have ? = 2.78 rad/s Velocity of the ball = L? = 9.05 m/s

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