# Rotational dynamics – problems and solutions

1. A force F applied to a cord wrapped around a cylinder pulley. The torque is 2 N m and the moment of inertia is 1 kg m^{2}, what is the angular acceleration of the cylinder.

__Known :__

Torque (τ) = 2 N m

The moment of inertia (I) = 1 kg m^{2}

__Wanted:__ The angular acceleration of the cylinder

__Solution :__

Στ = I α

*Στ **= net torque, I = moment of inertia, **α = angular acceleration*

Angular acceleration of cylinder :

α = Στ / I = 2 / 1 = 2 rad/s^{2}

2. A force F applied to a cord wrapped around a cylinder pulley. The magnitude of the force is 10 N, the radius of the cylinder is 0.2 m and the moment of inertia is 1 kg m2, What is the angular acceleration of the cylinder?

__Known :__

Force (F) = 10 N

Radius of cylinder (R) = 0.2 m

The moment of inertia (I) = 1 kg m^{2}

__Wanted:__ The angular acceleration of the cylinder.

__Solution :__

τ = F R

*τ** = torque, F = force, R = radius of cylinder*

Torque :

τ = F R = (10 N)(0.2 m) = 2 N m

Στ = I α

*Στ **= net torque, I = moment of inertia, **α = angular acceleration*

Angular acceleration of cylinder :

α = Στ / I = 2 / 1 = 2 rad/s^{2}

3. A force F applied to a cord wrapped around a cylinder pulley. The magnitude of force is 10 N, the radius of cylinder is 0.2 m and the mass of cylinder is 20 kg m^{2,}. What is the angular acceleration of the cylinder.

__Known :__

Force (F) = 10 N

Radius of cylinder (R) = 0.2 m

Mass of cylinder (M) = 20 kg

__Wanted :__ Angular acceleration of cylinder

__Solution :__

τ = F R = (10 N)(0.2 m) = 2 N m

Moment of inertia :

I = 1⁄2 M R^{2 }= 1⁄2 (20)(0.2)^{2 }= 1⁄2 (20)(0.04) = 0.4 kg m^{2}

Angular acceleration of cylinder :

α = Στ / I = 2 / 0.4 = 5 rad/s^{2}

4. A 1-kg block hanging from a cord wrapped around a cylinder pulley. The moment of inertia of pulley is 1 kg m^{2} and the radius of pulley is 0.2 m. What is the angular acceleration of the pulley. Acceleration due to gravity is 10 m/s^{2}.

__Known :__

Moment of inertia of pulley (I) = 1 kg m^{2 }

Mass of block (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Weight (w) = m g = (1 kg)(10 m/s^{2}) = 10 kg m/s^{2} = 10 N

Radius of pulley (R) = 0.2 m

__Wanted :__ Angular acceleration

__Solution :__

Torque :

τ = F R = w R = (10 N)(0.2 m) = 2 N m

Moment of inertia :

I = 1 kg m^{2}

Angular acceleration :

α = Στ / I = 2 / 1 = 2 rad/s^{2}

5. A 1-kg block hanging from a cord wrapped around a cylinder pulley. The mass of pulley is 20 kg and the radius of pulley is 0,2 m. What is the angular acceleration of the pulley and the free fall acceleration of the block. Acceleration due to gravity is 10 m/s^{2}.

__Known :__

Mass of pulley (M) = 20 kg

Radius of pulley (R) = 0,2 m

Mass of block (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Weight (w) = m g = (1 kg)(10 m/s^{2}) = 10 kg m/s^{2} = 10 N

__Wanted :__ the angular acceleration of the pulley and the free fall acceleration of the block.

__Solution :__

The torque :

τ = F R = w R = (10 N)(0.2 m) = 2 N m

The moment of inertia of cylinder pulley :

I = 1⁄2 M R^{2 }= 1⁄2 (20)(0.2)^{2 }= (10)(0.04) = 0.4 kg m^{2}

The angular acceleration of the pulley :

α = Στ / I = 2 / 0.4 = 5 rad/s^{2}

The free fall acceleration of the block :

a = R α = (0.2)(5) = 1 m/s^{2}