A block (mass = 2.4 kg) is hanging from a massless cord that is wrapped around a

Question

A block (mass = 2.4 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 2.0 10-3 kg · m2), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.044 m during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.
? rad/s2
? N

Explanation / Answer

The mass of the block m = 2.4kg The moment of inertia of the pulley I = 2.0*10^-3 kg m^2 The radius of the pulley R = 0.044 m From Newton's laws        ma = mg - T then         T = mg -ma now from the rotation of the pulley        = I        TR = I    (mg - ma) R = I    mgR = I + maR     mgR = I + m(R) R     mgR = [ I + mR^2] therefore the angular acceleration         = mgR / I + mR^2             = (2.4kg)(9.8 m/s^2) (0.044m) / [ (2.0*10^-3 + (2.4kg)(0.044)^2]                        = 155.7 rad/s^2 then the acceleration            a = R = (0.044)(155.7) = 6.85 m/s^2 Now the tension in the string               T = (2.4kg)[9.8m/s^2 - 6.85m/s^2]                   = 7.08 N

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