Question

An object spins with an angular momentum of 8.9 kg*m^2/s. A torque on the object causes it to slow to a stop in 5.2 s. What is the angular impulse on the object?

i) 46.28 kgm^2/s
ii) 51.68 kgm^2/s
iii) 55.52 kgm^2/s
iv) 59.12 kgm^2/s

Answer 1

The angular impulse on an object that slowed to a stop with an initial angular momentum of 8.9 kg*m^2/s in 5.2 seconds is 46.28 kg*m^2/s, which corresponds to option (i).

Step-by-step explanation:

The question is asking for the angular impulse that caused an object with initial angular momentum of 8.9 kg*m^2/s to slow to a stop in 5.2 seconds without any other properties of the object being relevant. Angular impulse is the product of the torque applied to the object and the time over which it acts, and it is equal in magnitude to the change in angular momentum.

To find the angular impulse, we use the formula J = ΔL, where J is the angular impulse and ΔL is the change in angular momentum. Since the object stops, the final angular momentum is 0 kg*m^2/s, and hence the change in angular momentum is -8.9 kg*m^2/s (as it goes from 8.9 to 0). The negative sign indicates that the torque is acting opposite to the direction of the initial angular momentum to stop the object. Since the stopping time is 5.2 s, the magnitude of the angular impulse is:

J = | -8.9 kg*m^2/s | = 8.9 kg*m^2/s

To find the actual value, we multiply the change in angular momentum by the time:

J = 8.9 kg*m^2/s * 5.2 s = 46.28 kg*m^2/s

Therefore, the angular impulse on the object is 46.28 kg*m^2/s, which corresponds to option (i).