Question

A 1200 kg car is being raised over water by a winch. At the moment the car is 5.0 m above the water when the gearbox breaks. During the car's fall, there is no slipping between the massless rope, the pulley wheel, and the winch drum. The radius of the pulley is 30 cm and its mass is 15 kg. The radius of the drum is 80 cm and its mass is 500 kg. Approximating the pulley as a hoop and the drum as a uniform cylinder, what is the car's speed when it hits the water?

Answer #1

mass of car M_{1} = 1200 kg

height of the car at this instant above the water h = 5 m

radius of pulley r = 0.3 m

mass of pulley m = 15 kg

radius of drum R = 0.8 m

mass of drum M = 500 kg;

first calculate the moment of inertia for drum and pulley

I_{drum} = (1/2)MR^{2} (for cylinder)

I_{drum} = 0.5 x 500 x 0.8^{2}

**I _{drum} = 160 kg.m^{2}**

I_{pulley} = (2/3)mr^{2}

I_{pulley} = (2/3) x 15 x 0.3^{2}

**I _{pulley} = 0.9 kg.m^{2}**

Now use work energy theorem

loss in potential energy of car = gain in KE of the car + gain in rotational energy of the pulley and drum

M_{1}gh = (1/2)M_{1}V^{2} +
(1/2)I_{pulley} x w^{2}_{pulley} +
(1/2)I_{drum} x
w^{2}_{drum}

M_{1}gh = (1/2)M_{1}V^{2} +
(1/2)I_{pulley} x (V/r)^{2} + (1/2)I_{drum}
x (V/R)^{2} (since, w = V/r for pulley and V/R for
drum)

1200 x 9.8 x 5 = 0.5 x V^{2} [1200 +
0.9/0.3^{2} + 160/0.8^{2}]

**Speed of car just before striking the water V = 8.9749
m/s**

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