Using τ = Iα, we have α = τ/I = 20 N·m / 10 kg·m² = 2 rad/s².

Rotational kinetic energy is given by KE = 1/2 I ω^2. If I is doubled and ω remains constant, KE also doubles.

Angular acceleration α = τ/I. If I is doubled and τ remains constant, α is halved.

According to Newton's second law for rotation, τ = Iα, if I increases and τ is constant, α must decrease.

The moment of inertia of a disc is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is quadrupled.

The moment of inertia of a disk is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is 4 times the original.

Angular momentum L = Iω; if radius is doubled, moment of inertia I increases by a factor of 4, hence L quadruples.

Linear velocity v = rω. If r is halved and ω remains constant, v also halves.

Moment of inertia I is proportional to the square of the radius, so halving the radius reduces I to one-fourth.

Linear velocity v = rω; if r is halved and ω remains constant, v is halved.

The moment of inertia of a solid disk is I = 1/2 MR^2. If R is doubled, I becomes 1/2 M(2R)^2 = 2MR^2, which is 4 times the original.

Torque = Force × Distance; if Torque is doubled and Distance is constant, Force must also double.

If the torque is zero, it means that the forces are acting along the same line, resulting in no rotational effect.

If the torque is zero, it means that the line of action of the forces passes through the pivot point.

The net torque is zero because the forces are equal and opposite, resulting in no rotational effect.

The net torque is zero because the forces are equal and opposite, producing no rotational effect.

The net torque is zero because the forces are equal and opposite, producing no rotational effect.

Torque is directly proportional to the force applied; hence, the 10 N force produces more torque.

Torque is greater when the force is applied further away from the pivot, given equal magnitudes.

The net torque is zero because the torques produced by the two forces cancel each other out.

The moment of inertia of the bottom disk is MR^2, and the top disk contributes an additional MR^2 due to the parallel axis theorem, giving a total of 2MR^2.

The moment of inertia depends only on the mass distribution and shape, not on the angular velocity.

Since L = Iω, if L is doubled and I remains constant, then ω must also double.

Torque (τ) = F × r = 30 N × 2 m = 60 Nm.

To balance the seesaw, the other child must exert an equal torque of 30 N·m.

Total angular momentum of the system is the vector sum of individual angular momenta: L_total = L1 + L2.

Angular momentum L = Iω; if one has twice the moment of inertia, it will have twice the angular momentum at the same angular velocity.

Angular momentum L = Iω; since IA = 2IB and ωA = ωB, LA = 2LB.

Rotational kinetic energy is associated with the rotation of the object in rolling motion.

The moment of inertia of a composite body can be calculated using the parallel axis theorem.