Two masses m and 2m are connected by a light string going over a pulley (a disc) of mass 30m with radius r=0.1 m. The pulley is mounted in a vertical plane and is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is m/s. (…

Question

Two masses m and 2m are connected by a light string going over a pulley (a disc) of mass 30m with radius r=0.1 m. The pulley is mounted in a vertical plane and is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is m/s. (Assume the string does not slip and g=10 m/s².)

Accepted Answer

2. Use energy: the 2m falls through h while the m rises through the same h, and the pulley spins up. Net loss of potential energy: 2mgh-mgh=mgh. Kinetic energy gained by the two blocks: 1/2(2m)v²+1/2(m)v²=3/2mv². Pulley: I=1/2(30m)r²=15mr², and with no slipping ω=v/r, so 1/2Iω²=1/2(15mr²)(v²)/(r²)=(15)/2mv². Balance: mgh=3/2mv²+(15)/2mv²=9mv². v²=(gh)/9=((10)(3.6))/9=4. v=2 m/s. The radius cancels, as it must: only the pulley's mass and shape affect the answer.

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