Posted by **dawn** on Monday, February 18, 2008 at 11:30am.

Figure 8-31 shows a ball with mass m = 1.0 kg attached to the end of a thin rod with length L = 0.56 m and negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle. (a) What initial speed must be given the ball so that it reaches the vertically upward position with zero speed? What then is its speed at (b) the lowest point and (c) the point on the right at which the ball is level with the initial point? (d) If the ball's mass were doubled, what would the answer to (a) be?

- physics -
**dawn**, Monday, February 18, 2008 at 11:31am
a thin rod, of length L = 1.6 m and negligible mass, that can pivot about one end to rotate in a vertical circle. A heavy ball of mass m = 9.2 kg is attached to the other end. The rod is pulled aside to angle è0 = 20° and released with initial velocity 0 = 0. What is the speed of the ball at the lowest point?

- physics -
**Granger**, Sunday, October 16, 2011 at 5:24pm
m=1kg

L=.56m

change in KE + change in PE = E

.5mv^2f + mghf = .5mv^2i + mghi

Even though this problem is a circle, the height variable in the potential energy equation shouldn't differ.

a)Use the bottom of the circle for h=0

mghi is canceled.

work = change in KE = .5mv^2f -.5mv^2i

You need the velocity

There is no KE initially so that equals 0. You're left with:

.5mv^2f=mghi solve for vf

(masses cancel out)

then plug v back into your intitial equation:

.5mv^2f-.5mv^2i= W

b) is just the negative of a

c) 0J because there is no KE and the change in PE = 0

d).5mv^2f=mghi solve for vf

because the masses aren't a factor in the final velocity, you can just go straight to the bottom equation

.5(2)mv^2f=mv^2f=W

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