A 980 kg car is held in place by an unrealiable winch. The gearbox of the winch

Question

A 980 kg car is held in place by an unrealiable winch. The gearbox of the winch breaks and at that moment the car-drum system free-falls from rest, shown in the figure. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. Find the speed of the car as it hits the ground. The moment of inertia of the winch drum is 258 kg m2 and that of the pulley is 50 kg m2. The radius of the drum is 90 cm and the radius of the pulley is 3 cm and the acceleration due to gravity is 9.81 m/s2. Answer in units of m/s.

Explanation / Answer

Given that the mass of car is m = 980 kg Moment of inertia of drum is I1 = 258 kg.m2 Moment of inertia of pully is I2 = 50 kg.m2 Radius of drum is R = 99 cm = 0.99 m Radius of puley is r = 3.0 cm = 0.03 m ------------------------------------------------------------- Apply conservation of energy to the system,           mgh = (1/2)I112+ (1/2)I222 + (1/2)mv2           mgh = (1/2)I1(v/R)2+ (1/2)I2(v/r)2+ (1/2)mv2           2mgh = I1(v/R)2+ I2(v/r)2+ mv2 v2 =  2mgh / [  I1/R2+ I2/r2+ m ]     = 2*980kg*9.8m/s2*8m / [258 kg.m2/ (0.99m)2 +50 kg.m2/ (0.03m)2 + 980kg]     = 2.705 m/s Then we get       v = 2.705 m/s          = 1.644 m/s           Moment of inertia of pully is I2 = 50 kg.m2 Radius of drum is R = 99 cm = 0.99 m Radius of puley is r = 3.0 cm = 0.03 m ------------------------------------------------------------- Apply conservation of energy to the system,           mgh = (1/2)I112+ (1/2)I222 + (1/2)mv2           mgh = (1/2)I1(v/R)2+ (1/2)I2(v/r)2+ (1/2)mv2           2mgh = I1(v/R)2+ I2(v/r)2+ mv2 v2 =  2mgh / [  I1/R2+ I2/r2+ m ]     = 2*980kg*9.8m/s2*8m / [258 kg.m2/ (0.99m)2 +50 kg.m2/ (0.03m)2 + 980kg]     = 2.705 m/s Then we get       v = 2.705 m/s          = 1.644 m/s                     mgh = (1/2)I1(v/R)2+ (1/2)I2(v/r)2+ (1/2)mv2           2mgh = I1(v/R)2+ I2(v/r)2+ mv2 v2 =  2mgh / [  I1/R2+ I2/r2+ m ]     = 2*980kg*9.8m/s2*8m / [258 kg.m2/ (0.99m)2 +50 kg.m2/ (0.03m)2 + 980kg]     = 2.705 m/s Then we get       v = 2.705 m/s          = 1.644 m/s                     2mgh = I1(v/R)2+ I2(v/r)2+ mv2 v2 =  2mgh / [  I1/R2+ I2/r2+ m ]     = 2*980kg*9.8m/s2*8m / [258 kg.m2/ (0.99m)2 +50 kg.m2/ (0.03m)2 + 980kg]     = 2.705 m/s Then we get       v = 2.705 m/s          = 1.644 m/s          

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