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 m2, what is the angular acceleration of the cylinder.

Rotational dynamics – problems and solutions 1Known :

Torque (τ) = 2 N m

The moment of inertia (I) = 1 kg m2

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/s2

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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?

Rotational dynamics – problems and solutions 2Known :

Force (F) = 10 N

Radius of cylinder (R) = 0.2 m

The moment of inertia (I) = 1 kg m2

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/s2

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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 m2,. What is the angular acceleration of the cylinder.

Rotational dynamics – problems and solutions 3Known :

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 R2 = 1⁄2 (20)(0.2)2 = 1⁄2 (20)(0.04) = 0.4 kg m2

Angular acceleration of cylinder :

α = Στ / I = 2 / 0.4 = 5 rad/s2

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4. A 1-kg block hanging from a cord wrapped around a cylinder pulley. The moment of inertia of pulley is 1 kg m2 and the radius of pulley is 0.2 m. What is the angular acceleration of the pulley. Acceleration due to gravity is 10 m/s2.

Rotational dynamics – problems and solutions 4Known :

Moment of inertia of pulley (I) = 1 kg m2

Mass of block (m) = 1 kg

Acceleration due to gravity (g) = 10 m/s2

Weight (w) = m g = (1 kg)(10 m/s2) = 10 kg m/s2 = 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 m2

Angular acceleration :

α = Στ / I = 2 / 1 = 2 rad/s2

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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/s2.

Rotational dynamics – problems and solutions 5Known :

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/s2

Weight (w) = m g = (1 kg)(10 m/s2) = 10 kg m/s2 = 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 R2 = 1⁄2 (20)(0.2)2 = (10)(0.04) = 0.4 kg m2

The angular acceleration of the pulley :

α = Στ / I = 2 / 0.4 = 5 rad/s2

The free fall acceleration of the block :

a = R α = (0.2)(5) = 1 m/s2

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