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 m-3.6 kg and the right box has a mass m2 1.6 kg. The pulley has mass m - 3.2 kg and radius R 0.15 m. 1) What is the linear acceleration of the left box? (up is the positive direction) m's Submit You currently have O submissions for this question. Only 2 submission are allowed. You can make 2 more submissions for this question. 2) What is the angular acceleration of the pulley? (let counter-clockwise be positive) rad/s Submit You currently have 0 submissions for this question, Only 2 subnission are allowed. Ybu can make 2 more submissions for this question. 3) what is the tension in the string between the left mass and the pulley? N Submit You currently have 0 submissions for this question. Only 2 submission are allowed. You can make 2 more submisslons for this question. 4) What is the tension in the string between the right mass and the pulley? N Submi You currently have O submissions for this question. Only 2 submission are allowed. You can make 2 more submissions for this question. 5) The boxes accelerate for a time t 1.65 s what distance does each box move in the time 1.65 s? m Submit Ybu currently have 0 submissions for this question. Only 2 submission are allowed. You can make 2 more submissions for this question. 6) What is the magnitude of the velocity of the boxes after the time 1.65 s? mrs Submit You currently have O submissions for this question. Only 2 submission are allowed. You can make 2 more submissions for this question. 7) What is the magnitude of the final angular speed of the pulley? rad/s Submit You currently have 0 submissions for this question. Only 2 subnission are ollowed. Ybu can make 2 more submissions for this question.

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)

= (3.6-1.6)*9.8/(3.6+1.6+0.5*3.2) = 2.882 m/s^2

downward acceleration is negative , so a = - 2.882 m/s^2 answer

2] angular acceleration alpha = a/r = 2.882/0.15 = 19.213 rad/s^2

3] T1 = m1g-m1a = 3.6*(9.8-2.882) = 24.9 N

4] T2 = m2g+m2a =  1.6*(9.8+2.882) = 20.29 N

5] distance s = 0.5 at^2 = 0.5*2.882*1.65^2 = 3.923 m

6] v = at = 2.882*1.65 = 4.755 m/s

7] w = v/r = 4.755/0.15 = 31.7 rad/s

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