In the figure, block 1 has mass m 1 = 470 g, block 2 has mass m 2 = 560 g, and t

Question

In the figure, block 1 has mass m1 = 470 g, block 2 has mass m2 = 560 g, and the pulley is on a frictionless horizontal axle and has radius R = 5.4 cm. When released from rest, block 2 falls 80 cm in 5.0 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T2 (the tension force on the block 2) and (c) tension T1 (the tension force on the block 1)? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s2.

The figure is just a pulley with block 1 on side with a rope going around the pulley to block 2 on the other side. It has an arrow pointing in the upwards direction on block 1 and an arrow pointing down by block number 2.

Explanation / Answer

Sum forces on m1

T1 - m1*g = m1*a

or T1 = m1*g + m1*a

sum forces on m2

m2*g - T2 = m2*a

or T2 = m2*g - m2*a

sum torques about the pulley

T2*r - T1*r = I* = I*a/r

or (m2*g - m2*a)*r - (m1*g + m1*a)*r = I*a/r

but a is determined from the kinematic data

y = 1/2*a*t^2 so a = 2*y/t^2 = 2*0.80m/5^2 = 6.4 x 10^-2 m/s^2

so I = r/a*((m2*g - m2*a)*r - (m1*g + m1*a)*r)

I = 0.054/6.4x10^-2*(0.560*9.8 - 0.560*6.4x10^-2)*0.054 -(0.470*9.8 + 0.470*6.4x10^-2)*0.054)

I = 0.04000 kg-m^2

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