Chapter 10, Problem 051 In the figure, block 1 has mass m1- 467 g, block 2 has m

Question

Chapter 10, Problem 051 In the figure, block 1 has mass m1- 467 g, block 2 has mass m2 -534 g, and the pulley is on a frictionless horizontal axle and has radius R-4.92 cm. When released from rest, block 2 falls 74.8 cm in 4.67 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. m1 (a) Number (b) Number (c) Number (d) Number Units Units Units Units

Explanation / Answer

a)

Applying one of the kinematic equations to block 2, we get the acceleration of the block.

d = (1/2)at2 a = 2d/t2 = (2*0.748)/4.67^2 = 0.0686 m/s2

b) Newton’s 2nd law applied to the blocks yield tension T2

m2*a = m2*g – T2 =====> T2 = m2*(g - a) = 0.534*(9.81 - 0.0686) = 5.2 N

c) Newton’s 2nd law applied to the blocks yield tension T1

m1*a = T1 – m1*g =====> T1 = m1*(a + g) = 0.467*(0.0686 + 9.81) = 4.61 N

d) Since the cord does not slip on the pulley, the tangential acceleration of the pulley is equal to the acceleration of the blocks. The angular acceleration of the pulley is

R = at = a ======> = a/R = 0.0686/0.0492 = 1.394 rad/s2

e) Applying Newton’s 2nd law to the pulley, the rotational inertia of the pulley is

I = T2*R – T1*R =======> I = (T2 - T1)R/ = 0.0208 kg·m2

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