Q.A student sits on a rotating stool with arms outstretched, giving the student-plus-stool system a moment of inertia of $5\ \text{kg}\,\text{m}^2$ and an angular speed of $2\ \text{rad/s}$. The student then pulls the arms in, reducing the moment of inertia to $2\ \text{kg}\,\text{m}^2$. Find the new angular speed, assuming no external torque acts on the system.

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. …

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).

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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.

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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).

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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 st …

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 …

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.

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