Q.A spherical water balloon is rotating at 60 rpm. In the course of time, 48.8 % of its water leaks out. With what frequency will the remaining balloon rotate now? Neglect all non-conservative forces.

Angular momentum and mechanical energy are conserved; linear momentum is not. Answer (A).

When a body moves under the gravitational attraction of another (central) body:

• The force is central (directed along the line joining them), so torque about the centre is zero -> angular momentum is conserved.

• Gravity is conservative, so total mechanical energy is conserved.

• But gravity is a net external force on the moving body, continuously changing its velocity direction (as in an orbit); hence its linear momentum keeps changing and is NOT conserved.

Motion under a central force stays in a plane because angular momentum is conserved. Answer (B).

A central force always points toward (or away from) a fixed centre, so its torque about that centre is zero (r and F are parallel/anti-parallel, r x F = 0).

Zero torque means the angular momentum L = r x p is constant in both magnitude and direction. Since L is perpendicular to the plane containing r and v, a fixed L direction forces r and v to remain in one fixed plane throughout the motion.

Kepler's second law (the radius vector sweeps equal areas in equal times) is a direct geometric consequence of angular momentum being conserved in a planet's orbit around the Sun.

Gravitational force on a planet from the Sun always acts along the line joining them (a central force), so it produces zero torque about the Sun. With zero net torque, the planet's angular momentum L = m v r sin(theta) about the Sun stays constant throughout the orbit.

The rate at which the radius vector sweeps out area, dA/dt, can be shown to equal L/(2m), which is constant since L is constant. This constancy of dA/dt is exactly Kepler's second law (equal areas in equal times).

Angular momentum in uniform circular motion is conserved only about the centre of the circle, because the centripetal force passes through that point and produces zero torque there.

For a particle moving in a circle of radius r with constant speed, the only force acting is the centripetal force F, which is always directed radially inward, straight at the centre of the circle.

Torque about any point O is tau = r' x F, where r' is the position vector from O to the particle. If O is the centre of the circle, r' is exactly along the same (or opposite) line as F, so r' x F = 0 at every instant -- zero torque, hence angular momentum about the centre stays constant (both in magnitude, m v r, and direction, perpendicular to the plane of motion).

About any OTHER point (inside, outside, or on the circumference but not the centre), the position vector from that point to the particle is generally NOT parallel to F, so the torque is nonzero and the angular momentum about that point changes as the particle moves -- it is not conserved.

The gravitational force of the Sun on a planet always acts along the line joining the Sun and the planet (a central force), so it produces zero torque about the Sun.

Since $\vec\tau = \dfrac{d\vec L}{dt}=\vec r\times \vec F$, and $\vec F$ is always anti-parallel to $\vec r$ (radial), $\vec\tau=0$, so the angular momentum $\vec L$ of the planet about the Sun stays constant. This conservation of angular momentum is exactly what produces Kepler's second law (equal areas in equal times).

Planetary motion is a direct illustration of conservation of angular momentum, since gravity is a central force.

The Sun's gravitational pull on a planet always points from the planet towards the Sun, i.e. along the position vector $\vec r$ of the planet (measured from the Sun). Torque is $\vec\tau = \vec r \times \vec F$; since $\vec F$ is anti-parallel to $\vec r$, this cross product is zero, so the torque on the planet about the Sun is always zero. Because $\vec\tau = d\vec L/dt = 0$, the angular momentum $\vec L$ of the planet about the Sun remains constant throughout its orbit. This is exactly Kepler's second law (equal areas in equal times), which is a geometric statement of angular-momentum conservation.

Halving Earth's radius shortens the day to 6 hours.

Earth spins with angular momentum L = Iω. With no external torque and unchanged mass, L is conserved.

Moment of inertia of a solid sphere I = (2/5)MR^2 ∝ R^2. If R → R/2, then I → I/4.

Conserving L: I·ω = (I/4)·ω' ⇒ ω' = 4ω. Since the period T = 2π/ω, T' = T/4 = 24/4 = 6 hours.