A light piece of rope is wrapped around a frictionless heavy pulley (disk, M = 1

Question

A light piece of rope is wrapped around a frictionless heavy pulley (disk, M = 1.8 kg, R = 30 cm, I = ½ MR2) that can rotate freely about a horizontal axis going through its center. A small block with mass m = 325 g is attached to the open end of the rope. The block is released from rest to fall. (a) Apply Newton’s 2nd law to the pulley (rotation) and to the block (translation) and combine the resulting equations to calculate the acceleration of the falling block. (Hint: Do not forget to use a = R?. The positive sense of the x-axis is shown in the figure.) (b) Calculate the tension in the string

Explanation / Answer

net torque acting on the pulley is Tnet = I*alpha

T*R = I*alpha

alpha = a/R

T*R = (1/2)*M*R^2*(a/R) = (1/2)*M*R*a

Tension T = (1/2)*M*a.................(1)

net force on the hanging mass is Fnet = mg-T = ma.........(2)

substituting (1) in (2) we get

mg-(1/2)*M*a = ma

mg = ma + (1/2)*M*a

mg = [m+(M/2)]*a

accelaration a = mg/[m+(M/2)]

B) from (1) we get

Tension T = (1/2)*M*m*g/[m+(M/2)]

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