A drum (see the figure) has a radius of 0.40 m and a moment of inertia of 2.2 kg

Question

A drum (see the figure) has a radius of 0.40 m and a moment of inertia of 2.2 kgm2. The frictional torque of the drum axle is 3.0 Nm. A 46 m length of rope is wound around the rim. The drum is initially at rest. A constant force is applied to the free end of the rope until the rope is completely unwound and slips off. At that instant, the angular velocity of the drum is 11 rad/s. The drum then decelerates and comes to a halt. In this situation, the constant force applied to the rope is closest to:

Explanation / Answer

frictional torque=frictional force*R=>frctional force=torque/R=3/0.4=7.5N

let the force applied =f

net force=f-7.5

net torque=I*angular accleration

(f-7.5)*0.4=2.2*angular accleration

angular accleration=0.182*(f-7.5)

let the total revolution be n

total length=2*pi*r*n=46

n=57.5/2*pi

total angle=115 radian

using newton equation

angular accleration=11^2/(2*115)=0.526 rad/s^2

0.182*(f-7.5)=0.526

we got f=10.4 N almost equal to 10 N

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