Question

Suppose you are standing on the edge of a spinning platform and step off at right angles to the edge (radially outward). Now consider it the other way. You are standing on the ground next to a spinning carousel and you step onto the platform at right angle to the edge (radially inward). Then,
A. There is no change in rotational speed of the carousel in either situation.
B. There is a change in rotational speed in the first situation but not the second.
C. There is a change in rotational speed in the second situation but not the first.
D. There is a change in rotational speed in both instances

Answer 1

Hint: This question is an easy application of the Law of conservation of angular momentum. Law of conservation of angular momentum states that as long as their sum stays continuous, any of the individual angular moments will shift. This law is similar to retaining linear momentum when the external force on a device is zero.

Complete step by step answer:
Angular momentum is the counterpart of momentum in linear motion for a moving solid object. The momentum is the mass of the object time velocity for linear motion. As the latter is a vector, so is linear momentum. We would use the same physical quantities of moment of inertia for rotational motion. I, angular velocity $\omega $, and angular momentum L
So,
\[\vec{L}=I\vec{\omega }~\]
Now,
In all cases, the moment of inertia is altered. The moment of inertia is diminished in the first case as the spinning mass is diminished, thus increasing the angular velocity, while in the second case it is the other way round. Thus, in both cases, there is a shift in rotational speed. We use the concept of angular momentum conservation.
\[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\]

So, the correct answer is “Option C”.

Note: Vector quantities are angular velocity and angular momentum and have both magnitude and direction. The angular velocity direction and the angular momentum direction are perpendicular to the rotating axis.