In Fig. 11-42, one block has a mass M = 500 g, the other has mass m = 460 g, and
ID: 2062113 • Letter: I
Question
In Fig. 11-42, one block has a mass M = 500 g, the other has mass m = 460 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. When released from rest, the heavier block falls 85.5 cm in 2.80 s (without the cord slipping on the pulley).
(a) What is the magnitude of the block's acceleration?
___ m/s^2
(b) What is the tension in the part of the cord that supports the heavier block?
___ N
(c) What is the tension in the part of the cord that supports the lighter block?
___ N
(d) What is the magnitude of the pulley's angular acceleration?
___ rad/s^2
(e) What is its rotational inertia?
___ kg·m^2
Explanation / Answer
(a) What is the magnitude of the blocks’ acceleration? If you use the relation s=(1/2)at*t you will get the acc. as 0.06 m/s^2 (b) What is the tension in the part of the cord that supports the heavier block? 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 C) 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 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) = 0.3344 N r = 0.05m p = 6/5 substitute and get I which is 0.01393 kg-m^2