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

- physics -
**dawn**, Monday, February 18, 2008 at 11:31ama 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 5:24pmm=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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