Question

A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleration α, the torque on the child is τ. The child moves to a position half way between the center and edge of the merry-go-round, and the angular acceleration increases to 2α. The torque on the child is now

Answer 1

Answer:

The torque on the child is now the same, τ.

Explanation:

• It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
• In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
• The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of  the mass of the child times the square of the distance to the center.
• When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

       $I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)$

• So, τ = 3/2*m*r²*α (2)

• When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

       $I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)$

• Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

       τ = 3/4*m*r² * (2α) = 3/2*m*r²

        same result than in (2), so the torque remains the same.