A pencil of mass m and length l lies at rest on a frictionless table. You push o
ID: 1998491 • Letter: A
Question
A pencil of mass m and length l lies at rest on a frictionless table. You push on it at its midpoint (perpendicular to it), with a constant force F for a time t. Find the final speed and the distance traveled. Verify that the work you do equals the final kinetic energy. Assume that you apply the same force F for the same time t as above, but that you now apply it at one of the pencil's ends (perpendicular to the pencil). Assume that t is small, so that the pencil doesn't have much time to rotate (this means that you can assume that your force is always essentially perpendicular to the pencil, as far as the torque is concerned). Find the final CM speed, the final angular speed, and the distance your hand moves. Verify that the work you do equals the final kinetic energy.Explanation / Answer
a) acceleration a = F/m
v = at = Ft/m
s = 0.5*at^2
= 0.5F/m*t^2
Work done = Fs = F*0.5F/m*t^2
= 0.5 F^2t^2/m
= 0.5 m (Ft/m)^2......substituting value from first equation
= 0.5 m v^2
= kinetic energy change
Hence proved
b) Final CM speed = Ft/m
Torque about CM = Fl/2
angular acceleration alpha = [Fl/2]/i
final angular speed = [(Fl/2)/i]t
distance hand moved = 0.5at^2 = 0.5 Ft^2/m
work done = F.s + Torque.theta
= 0.5 F^2t^2/m + Fl/2*0.5* ([Fl/2]/i)*t^2
= 0.5 m v^2 + 0.5*i*([Fl/2]/i)^2*t^2
= 0.5 mv^2 + 0.5 iw^2
= translational KE + rotational KE
= total KE
Hence proved
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