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.

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.

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.

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.

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

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.

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.

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

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.

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.

Using τ = Iα, we find α = τ/I. Assuming I = 1 kg·m², α = 5 rad/s². Time to stop = ω/α = 20 rad/s / 5 rad/s² = 4 s.

Using τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration. Assuming I = 25 kg·m², α = τ/I = 5/25 = 0.2 rad/s².

Angular deceleration = (final angular velocity - initial angular velocity) / time = (0 - 15) / 3 = -5 rad/s², so the magnitude is 5 rad/s².

Angular acceleration α = Torque/I. Assuming I = 6 kg·m², α = 3 N·m / 6 kg·m² = 0.5 rad/s².

Angular deceleration α = (ω_f - ω_i) / t = (0 - 20 rad/s) / 5 s = -4 rad/s², so the magnitude is 4 rad/s².

Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.

Torque = Force × Distance = 10 N × 2 m = 20 Nm.