The diagram shows a block on a horizontal frictionless surface. Attached to the

Question

The diagram shows a block on a horizontal frictionless surface. Attached to the block is a string that hangs over the outside of a pulley (that has mass). On the other end of the string is a hanging mass. The hanging mass is 0.780 kg and the block is 2.48 kg. The pulley has a radius of 0.0710 m. The block is released from rest. Once the block has moved 1.36 m toward the pulley it is traveling at a speed of 1.68 m/s. Assuming that there is no work done by non-conservative forces, what is the pulley

Explanation / Answer

For the free body diagram from the hanging mass, we have a Tension acting up and a weight acting down. The sum of the forces = ma, so

m1g - T1 = m1a

Solve for T1 and get T1 = m1g - m1a

The free body diagram for the mass on the table has only the tension pulling in the direction of motion, so

T2 = m2a

The difference in the tensions causes a torque on the pully

The two formulas for torque are = FL and = I, so FL = I

(T1 - T2)(r) = I

= a/r, so substitute

(m1g - m1a - m2a)r = Ia/r

We can find a using vf2 = vo2 + 2ad

(1.68)2 = (0) + (2)(a)(1.36)

a = 1.04 m/s2

Finally, [(.780)(9.8) - (.780)(1.04) - (2.48)(1.04)](.071) = (I)(1.04)/(.071)

I = .0206 kg m2

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