The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.

The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.

The disk will reach the bottom first because it has a smaller moment of inertia compared to the ring.

Using the formula ω = ω₀ + αt, we have ω = 10 + 2*5 = 20 rad/s.

Angular momentum L = Iω, where I = (1/2)MR^2 for a disk.

Angular momentum L = Iω = (1/2)MR^2ω, where I = (1/2)MR^2 for a disk.

The moment of inertia I of a disk about its axis is (1/2)MR^2. Therefore, the kinetic energy K.E. = (1/2)Iω^2 = (1/2)(1/2)MR^2ω^2 = (1/4)MR^2ω^2.

At the bottom, the total energy is converted into translational and rotational energy. The translational energy is 2/3 of the total energy.

At the bottom, the total kinetic energy is split equally between translational and rotational for a disk, hence the ratio is 1:1.

Using conservation of energy, potential energy at height h converts to kinetic energy at the bottom. The speed is √(2gh).

The relationship between linear speed and angular speed for rolling without slipping is given by ω = v/R.

The moment of inertia I of a disk is proportional to r^2, so if the radius is doubled, I becomes 4I. Thus, angular momentum L = Iω becomes 4Iω.

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

Angular momentum L = Iω. If the radius is doubled, the moment of inertia increases by a factor of 4, thus L = 4Iω.

Angular momentum L = Iω, and if the radius is doubled, the moment of inertia I becomes 4I, thus L = 4Iω.

To conserve angular momentum, if the radius is doubled, the angular velocity must be halved.

The linear velocity v = rω. If the radius is doubled, to maintain the same v, the angular velocity must remain ω.

Torque (τ) = Force (F) × Distance (r) = 20 N × 0.8 m = 16 Nm.

Torque (τ) = F × d = 20 N × 1 m = 20 Nm.

Torque (τ) = Force (F) × Distance (r) = 50 N × 1 m = 50 Nm.

Torque = Force × Distance = 50 N × 1.2 m = 60 Nm.

Angular momentum is conserved; it remains constant, but her angular velocity increases.

By conservation of angular momentum, if the moment of inertia decreases, the angular velocity must increase.

Angular momentum remains the same; however, her angular velocity increases due to a decrease in moment of inertia.

Pulling her arms in decreases her moment of inertia, causing her rotational speed to increase to conserve angular momentum.

By conservation of angular momentum, pulling arms in decreases moment of inertia, thus increasing angular velocity.

From Newton's second law for rotation, τ = Iα, thus α = τ/I.

Using the equation ω_f = ω + αt, where α = τ/I, we get ω_f = ω + (τ/I)t.

First convert rpm to rad/s: 1000 rpm = (1000 * 2π)/60 rad/s. Then use α = (ωf - ωi)/t = (0 - (1000 * 2π)/60) / 10.

Angular deceleration α = (final angular speed - initial angular speed) / time = (0 - 20 rad/s) / 5 s = -4 rad/s².