Question

A girl stands on the rim of a merry-go-round of radius 2.0 m that is not moving. The rotational inertia of the merry-go-round (including the girl’s inertia) is 800 kg∙m2. She throws a rock of mass 0.5 kg horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 4.0 m/s. (a) What is the angular speed of the merry-go-round? (b) What is the speed of the girl?

Answer 1

As no external torques are acting on the system formed by the merry-go-round, the girl and the rock (all interaction are internal to the system), the angular momentum is conserved,

L0 = Lf

The task is to write the angular momentum before and after the stone is tossed.

Initial state:

The girl, the rock and the merry-go-around are at rest, i.e., the initial angular momentum is zero,

L0 = 0.

Final state:

The girl and the merry-go-around rotate with angular velocity ω. Meanwhile, the rock moves velocity v tangent to the rim of the circumference.

The total angular momentum in this configuration is,

The first two terms have the form where the moment of inertia for the girl is simply Ig=Mr2. The angular momentum of the rock is determined from the definition . In all cases the angular momentum is perpendicular to the plane of the diagram, consequently, we can drop the vector notation and use scalars.

Taking the positive direction exiting the page of the diagram the angular momentum of the rock is positive and those of the girl and the merry-go-around are negative. We substituted the vector product by its absolute value noticing that the angle between r(vector) and v(vector) is 900.

Equating the initial and final angular momentums,

Solving for the final angular velocity,

Put values to solve for ω = 0.5*2*4/800

The direction of the angular velocity is depicted in the diagram for a hypothetic direction of the rock. The rock and the girl move in opposite directions when the stone is launched.

b) The linear speed of the girl is computed from the relation between the angular and linear quantities in the circular motion,

Put values in the above equation:

vg=0.5*2*2*4/800