An Atwood\'s machine consists of blocks of masses m 1 = 10.8 kg and m 2 = 22.0 k

Question

An Atwood's machine consists of blocks of masses m1 = 10.8 kg and m2 = 22.0 kg attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass M = 9.50 kg and radius r = 0.200 m. The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping.

(a) Why must the tension T2 be greater than the tension T1?

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(b) What is the acceleration of the system, assuming the pulley axis is frictionless?
m/s2

(c) Find the tensions T1 and T2.

Explanation / Answer

a)

The pulley (which has a non-zero moment of inertia) is accelerating. The only way for that to happen is for it to have a non-zero net torque. The only torques on the pulley are T1 and T2. Since it is accelerating clockwise as m2 drops, then the acceleration must be because T2 is greater than T1.

b)

m1 = 10.8kg, m2 = 22kg, M = 9.50kg, r = 0.200m

So, I = 1/2Mr^2 = ½*9.5*0.2^2 = 0.19 kg-m2

a = g(m2-m1)/(m1 + m2 + I/r^2)

= {9.81*(22-10.8)}/{10.8+22+(0.19/0.2^2)}

= 2.926 m/s2

c)

T1 = m1(g +a)

= 10.8(9.81 + 2.926) = 137.55N

T2 = m2(g -a)

= 22(9.81-2.926) = 151.45N

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