T = mg T - mg = ma T = ma mg - T = ma A solid, frictionless cylindrical pulley h
ID: 1905283 • Letter: T
Question
T = mgT - mg = maT = mamg - T = ma A solid, frictionless cylindrical pulley has a radiusR= 0.400 m. A block is attached to a cord that is wrapped around the outside of the pulley. The block is falling with a downward acceleration of 5.40 m/s2, and the tension in the cord is 0.900 N. The free body diagrams for the pulley and the block are shown below. Note that the magnitudes of the two tensionsT1andT2are equal. (Just call the tensionT.) Also, "n" in the free body diagram of the pulley represents the normal force provided by the axle of the pulley, equal in magnitude to the weight of the pulley. Note that the normal force and the weight force do not produce torques on the pulley. (Why is this?)Explanation / Answer
a)
= a/R = 5.40/0.4 = 13.5 rad/s2
b)
= I
==> T R = I
==> I = TR/ = 0.900*0.4/13.5 = 0.0267 Kg.m2
c)
m g - T = ma
d)
m g - T = m a
m = T/(g - a) = 0.900/(9.8-5.4) = 0.205 Kg
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