one block has mass M = 500.00 g, the other has mass m = 460.00 g, and the pulley

Question

one block has mass M = 500.00 g, the other has mass m = 460.00 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.0000 cm. When released from rest, the more massive block falls 81.200 cm in 2.6000 s (without the cord slipping on the pulley, do not treat the pulley as a uniform disk). (Give your answers to five significant figures. Use these rounded values in subsequent calculations.) (a) What is the magnitude of the block's acceleration? (b) What is the tension in the part of the cord that supports the heavier block? (c) What is the tension in the part of the cord that supports the lighter block? (d) What is the magnitude of the pulley's angular acceleration? (e) What is its rotational inertia?

Explanation / Answer

As the pulley has mass and radius is given, it will have moment of inertia. So, the tension on both the strings will be different, even though the blocks will have same acceleration. If you use the relation s=(1/2)at*t you will get the acc. as 0.06 m/s^2 The equation for 0.5kg block is 0.5 * a = 0.5 g - T1 ( where T1 is the tension in that string and a is the acceleration i.e, 0.06 T1 = 4.87 N The equation for 0.46kg block is 0.46 * 0.06 = T2 - 0.46 g ( T2 is the tension in the other part of the string) T2 = 4.5356 N also, a= r *(p) where p is the angular acceleration of the pulley. therefore, p = a/r ( take all the units in S.I. system) These tensions provide a torque to the pulley. As it rotates in anticlockwise direction ( if the 0.5 kg block is on your left ), there is a net anticlockwise torque acting. torque = force * distance of the force from the axis = ( T1-T2 ) * r but torque is also equal to ( I * p ) where I is the moment of inertia and p is the angular acceleration. you now know ( T1-T2) = r = 0.05m p = 6/5 substitute and get the value

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