The system shown below is released from rest. The m b = 29 kg block is 2 m above

Question

The system shown below is released from rest. The mb = 29 kg block is 2 m above the ledge. The pulley is a uniform disk with a radius of 10 cm and mass m = 4 kg. Assume that the string does not slip on the pulley.

1)

(a) Find the speed of the 29 kg block just before it hits the ledge.
m/s

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2)

(b) Find the angular speed of the pulley at that time.
rad/s

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3)

(c) Find the tensions in the strings.
N (left)

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4)

N (right)

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5)

(d) Find the time it takes for the 29 kg block to reach the ledge.

Explanation / Answer

1] Writing force equation on blocks , m1 is mb, m2 is 20 kg, m is mo

m1g - T1 = m1a

T2 - m2g = m2a ,

now writing torque equation on pully,

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

T1-T2 = 0.5 ma

adding this to first two equation,

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

a = (m1-m2)*g/ (m1+m2+0.5m) = [29-20]*9.8/[29+20+0.5*4] = 1.7294 m/s^2

now using third equation of motion, v = sqrt(2as) = sqrt(2*1.7294*2) = 2.63 m/s answer

2] angular speed w = v/r = 2.63/0.10 = 26.3 rad/s answer

3] Tleft = T2 = m2g + m2a = 20*(9.8+1.7294) = 230.6 N answer

4] Tright = T1 = m2g -m2a = 29* (9.8 - 1.7294) = 234.05 N answer

5] t = sqrt(2h/a) = sqrt(2*2/1.7294) = 1.5208 s answer

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