A playground merry-go-round has a radius R and a rotational inertia I. When the merry-go-round is at rest, a child with mass m runs with speed v along a line tangent to the rim and jumps on. The angular velocity of the merry-go-round is then:

The angular velocity of the merry-go-round is .

Explanation:

After a careful reading of the statement, we notice that merry-go-round-child system is a system with no external force exerting on it, such that the Principle of Angular Momentum Conservation can applied to analyze the system:

(Eq. 1)

Where:

- Mass of the child, measured in kilograms.

- Initial speed of the child, measured in meters per second.

- Radius of the playground merry-go-round, measured in meters.

- Moment of inertia, measured in kilograms per square meter.

- Angular velocity of the merry-go-round-child system, measured in radians per second.

Then, we clear the final angular speed:

The angular velocity of the merry-go-round is .

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.