In the system shown in the figure below, a m = 10.5 -kg mass is released from re

Question

In the system shown in the figure below, a m = 10.5-kg mass is released from rest and falls, causing the uniform M = 15.0-kg cylinder of diameter 29.0 cm to turn about a frictionless axle through its center. How far will the mass have to descend to give the cylinder245 J of kinetic energy?
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In the system shown in the figure below, a m = 10.5-kg mass is released from rest and falls, causing the uniform M = 15.0-kg cylinder of diameter 29.0 cm to turn about a frictionless axle through its center. How far will the mass have to descend to give the cylinder245 J of kinetic energy?

Explanation / Answer

Set potential energy equal to zero for the position in the figure. So (K+U)i = 0
(K+U)f = 1/2m*v^2-m*g*y +1/2*I*omega^2, but 1/2*i*omega^2 = 245J (given) From this we can determine the linear v of the edge of the cylinder. This is also the v of the mass.

I = 1/2*M*R^2 and omega = v/R

So 1/2*I*omega^2 = 1/2*1/2*15kg*v^2 = 245

This means that v^2 = 245/(1/4*15) = 65.34 m^2/s^2

Therefore the K of the mass = 1/2*m*v^2 = 1/2*15*65.34 = 490.05J

Now 490.05J + 245J -m*g*y = 0 This means that y = (735.05)/(9.8*15) = 5.0003m

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