11.3. Pivoted Rotation + Linear Motion (11.3.20) In the diagram below, the coeff
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Question
11.3. Pivoted Rotation + Linear Motion (11.3.20) In the diagram below, the coefficient of kinetic friction between m1 and the incline is 0.3. m2 is moving down. The pulley has mass M= 4.0kg and radius R = 10cm. If m = 5.0kg and m2= 2.0kg, find the angular acceleration of the pulley. Assume the pulley to be a solid cylinder. (Hint 1: You will need to make a guess about whether the angular acceleration of the pulley is clockwise or counterclockwise. Set up acceleration vectors for m1 and m2 that are consistent with the assumption you have made about the motion of the pulley Hint 2: The tension in the piece of string connected to m1 is different than the tension in the piece of string connected to m2.)Explanation / Answer
assuming m1 moves upwards and m2 moves downwards, and acceleration of the system is a m/s^2
tension in string connecting m1 is T1
tension in string conencting m2 is T2.
then for m1:
T1-m1*g*sin(35)-0.3*m1*g*cos(35)=m1*a
T1-5*a=40.147....(1)
for m2:
m2*g-T2=m2*a
T2+2*a=19.6 ....(2)
for the pulley, as m2 is going downwards, the pulley will be rotating clockwise.
then (T2-T1)*R=moment of inertia*angular acceleration
assuming pulley to be solid cylinder, the moment of inertia=0.5*M*R^2= 0.02 kg.m^2
so (T2-T1)*0.1=0.02
T2-T1=0.2...(3)
putting equation 3 in equation 2,
T1+0.2+2*a=19.6 ...(4)
solving equation 1 and equation 4,
T1=25.328 N
a=-2.9639 m/s^2
angular acceleration=a/r=-29.639 rad/sec^2
hence the pulley is rotating in anticlockwise direction.
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