Question

two children with same mass in a merry-go-round are at different distance from the center, child 'A' closer than 'b'. which one takes longer to complete one revolution?

Answer 1

Step-by-step explanation:

Let the distance of child A is r and the distance of child B is R.

Let the mass of each student is m.

According to the conservation of angular momentum

`I_(A)\omega _(A)=I_(B)\omega _(B)`

Let the time taken by A is TA and the time taken by B is TB.

`mr^(2)* (2\pi )/(T_(A))=mR^(2)* (2\pi )/(T_(B))`

`(T_(B))/(T_(A))}=\left ( (R^(2))/(r^(2)) \right )`

As R > r

So, TB > TA

Thus, the time taken by child B is more than the child A.

Answer 2

Both take the same time to complete one revolution

Step-by-step explanation:

Circular Motion

Imagine you and your friend are playing in the same merry-go-round. If he is sitting closer to the center that you, it won't make him move with respect to yourself. You are always in the same relative position from each other. If you complete a revolution, say, in 10 seconds, then your friend will also complete the revolution in 10 seconds.

Since both children are in different positions respect to the center and they complete the revolutions simultaneously, then they necessarily have different velocities. The closer to the center the child is, the less velocity he will have.

We are talking about the tangent velocity or vt. The other velocity is the angular velocity and that is the same for both children.