Two boxes hang from a solid disk pulley that is free to rotate as the blocks ris

Question

Two boxes hang from a solid disk pulley that is free to rotate as the blocks rise/fall. The left box has a mass m1 = 4.1 kg and the right box has a mass m2 = 1.6 kg. The pulley has mass m3 = 2.9 kg and radius R = 0.13 m.

1)

What is the linear acceleration of the left box? (up is the positive direction)
m/s2

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2)

What is the angular acceleration of the pulley? (let counter-clockwise be positive)
rad/s2

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3)

What is the tension in the string between the left mass and the pulley?
N

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4)

What is the tension in the string between the right mass and the pulley?
N

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5)

The boxes accelerate for a time t = 1.51 s.

What distance does each box move in the time 1.51 s?
m

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6)

What is the magnitude of the velocity of the boxes after the time 1.51 s?
m/s

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7)

What is the magnitude of the final angular speed of the pulley?

Explanation / Answer

1] Writing force equation on blocks

m1g - T1 = m1a

T2 - m2g = m2a , now writing torque equation on pully,

(T1-T2)r = i alpha where i = 0.5m3r^2 and alpha = a/r

T1-T2 = 0.5 m3a

adding this to first two equation,

m1g - m2g = (m1+m2+0.5m3)a

a = (m1-m2)*g/ (m1+m2+0.5m3)

= (4.1-1.6)*9.8/(4.1+1.6+0.5*2.9) = 3.4266 m/s^2

downward acceleration is negative, so a = -3.4266 m/s^2 answer

2] angular acceleration alpha = a/r = 3.4266/0.13 = 26.36 rad/s^2

3] T1 = m1g-m1a = 4.1*(9.8-3.4266) = 26.13 N

4] T2 = m2g+m2a =  1.6*(9.8+3.4266) = 21.16 N

5] distance s = 0.5 at^2 = 0.5*3.4266*1.51^2 = 3.906 m

6] v = at = 3.906*1.51 = 5.898 m/s

7] w = v/r = 5.898/0.13 = 45.369 rad/s

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