Posted by jane on Saturday, June 13, 2009 at 7:28pm.
Do the second question first. (It is worded incorrectly I think because there is in fact no torque on the ball since the force goes right through its center. There is however a torque about that point 2 meters from the launch point) The torque is the same at the top and halfway down because the perpendicular distance from the force vector to the launch point is always 2 meters. The magnitude of the torque = m g (2)
Well, we better figure out how fast it is going when it is halfway down. You already answered for at the top, zero because velocity is zero.
a = - 9.8
v = 40 - 9.8 t
h = 0 + 40 t - 4.9 t^2
when h = 20
20 = 40 t - 4.9 t^2
4.9 t^2 - 40 t + 20 = 0
t = (1/9.8)[ 40 +/- sqrt(1600 -392)]
t = (1/9.8) [ 40 +/- 34.8 ]
t = 7.63 on the way down (the plus sign)
v = 40 - 9.8(7.63) = -34.8 m/s
angular momentum = m V x R
= m (-34.8)(horizontal distance which is 2)
= -69.6 * mass
THANK YOU SO MUCH!
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