# Rotational motion – problems and solutions

**Torque**

1. A beam 140 cm in length. There are three forces acts on the beam, F_{1} = 20 N, F_{2 }= 10 N, and F_{3} = 40 N with direction and position as shown in the figure below. What is the torque causes the beam rotates about the center of mass of the beam?

__Known :__

The center of mass located at the center of the beam.

Length of beam (l) = 140 cm = 1.4 meters

Force 1 (F_{1}) = 20 N, the lever arm 1 (l_{1}) = 70 cm = 0.7 meters

Force 2 (F_{2}) = 10 N, the lever arm 2 (l_{2}) = 100 cm – 70 cm = 30 cm = 0.3 meters

Force 3 (F_{3}) = 40 N, the lever arm 3 (l_{3}) = 70 cm = 0.7 meters

__Wanted :__ The magnitude of torque

__Solution :__

The torque 1 rotates beam clockwise, so assigned a negative sign to the torque 1.

τ_{1} = F_{1} l_{1} = (20 N)(0.7 m) = -14 N m

The torque 2 rotates beam counterclockwise, so assigned a positive sign to the torque 2.

τ_{2} = F_{2} l_{2} = (10 N)(0.3 m) = 3 N m

The torque 3 rotates beam clockwise, so assigned a positive sign to the torque 3.

τ_{3} = F_{3} l3 = (40 N)(0.7 m) = -28 N m

The net torque :

Στ = -14 Nm + 3 Nm – 28 Nm = – 42 Nm + 3 Nm = -39 Nm

The magnitude of the torque is 39 N m. The direction of rotation of the beam clockwise, so assigned a negative sign.

2. What is the net torque acts on the beam The axis of rotation at point D. (sin 53^{o} = 0.8)

__Known :__

The axis of rotation at point D

F_{1} = 10 N and l_{1 }= r_{1} sin θ = (40 cm)(sin 53^{o}) = (0.4 m)(0.8) = 0.32 meters

F_{2} = 10√2 N and l_{2 }= r_{2} sin θ = (20 cm)(sin 45^{o}) = (0.2 m)(0.5√2) = 0.1√2 meters

F_{3} = 20 N and l_{3} = r_{1} sin θ = (10 cm)(sin 90^{o}) = (0.1 m)(1) = 0.1 meters

__Wanted :__ The net torque

__Solution :__

τ_{1} = F_{1} l_{1} = (10 N)(0.32 m) = 3.2 Nm

(The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 1)

τ_{2} = F_{2} l_{2} = (10√2 N)( 0.1√2 m) = -2 Nm

(The torque 2 rotates beam clockwise so we assign negative sign to the torque 2)

τ_{3} = F_{2} l_{2} = (20 N)(0.1 m) = 2 Nm

(The torque 3 rotates beam counterclockwise so we assign positive sign to the torque 3)

__The net torque :__

Στ = τ_{1} – τ_{1} + τ_{3 }

Στ = 3.2 Nm – 2 Nm + 2 Nm

Στ = 3.2 Nm

3. What is the net torque if the axis of rotation at point D. (sin 53^{o }= 0.8)

__Known :__

The axis of rotation at point D.

Distance between F_{1} and the axis of rotation (r_{AD}) = 40 cm = 0.4 m

Distance between F_{2} and the axis of rotation (r_{BD}) = 20 cm = 0.2 m

Distance between F_{3} and the axis of rotation (r_{CD}) = 10 cm = 0.1 m

F_{1 }= 10 Newton

F_{2} = 10√2 Newton

F_{3} = 20 Newton

Sin 53^{o }= 0.8

__Wanted :__ The net torque

__Solution :__

__The moment of the force ____1__

Στ_{1} = (F_{1})(r_{AD} sin 53^{o}) = (10 N)(0.4 m)(0.8) = 3.2 N.m

(The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 1)

__The moment of the force 2__

Στ_{2 }= (F_{2})(r_{BD} sin 45^{o}) = (10√2 N)(0.2 m)(0.5√2) = -2 N.m

(The torque 2 rotates beam clockwise so we assign negative sign to the torque 2)

__The moment of the force 3__

Στ_{3} = (F_{3})(r_{CD} sin 90^{o}) = (20 N)(0.1 m)(1) = 2 N.m

(The torque 2 rotates beam counterclockwise so we assign positive sign to the torque 3)

__The net torque :__

Στ = Στ_{1} + Στ_{2 }+ Στ_{3 }

Στ = 3.2 – 2 + 2

Στ = 3.2 Newton meter

**The moment of inertia**

4. Length of wire = 12 m, l_{1} = 4 m. Ignore wire’s mass. What is the moment of inertia of the system.

__Known :__

Mass of A (m_{A}) = 0.2 kg

Mass of B (m_{B}) = 0.6 kg

Distance between A and the axis of rotation (r_{A}) = 4 meters

Distance between B and the axis of rotation (r_{B}) = 12 – 4 = 8 meters

__Wanted : __The moment of inertia of the system

__Solution :__

__The moment of inertia of A__

I_{A} = (m_{A})(r_{A}^{2}) = (0.2)(4)^{2} = (0.2)(16) = 3.2 kg m^{2}

__The moment of inertia of B__

I_{B }= (m_{B})(r_{B}^{2}) = (0.6)(8)^{2} = (0.6)(64) = 38.4 kg m^{2}

The moment of inertia of the system :

I = I_{A} + I_{B} = 3.2 + 38.4 = 41.6 kg m^{2}

5. A 6-N force is applied to a cord wrapped around a pulley of mass M = 5 kg and radius R = 20 cm. What is the angular acceleration of the pulley. The pulley is a uniform solid cylinder.

__Known :__

Force (F) = 6 Newton

Mass (M) = 5 kg

Radius (R) = 20 cm = 20/100 m = 0.2 m

__Wanted :__ Angular acceleration (α)

__Solution :__

__The moment of the force :__

τ = F R = (6 Newton)(0.2 meters) = 1.2 N m

__The moment of inertia for solid cylinder :__

I = 1/2 M R^{2}

I = 1/2 (5 kg)(0.2 m)^{2}

I = 1/2 (5 kg)(0.04 m^{2})

I = 1/2 (0.2)

I = 0.1 kg m^{2}.

The angular acceleration :

τ = I α

α = τ / I = 1.2 / 0.1 = 12 rad s^{-2 }

6. A block of mass = 4 kg hanging from a cord wrapped around a pulley of mass = 8 kg and radius R = 10 cm. Acceleration due to gravity is 10 ms^{-2 }. What is the linear acceleration of the block? The pulley is a uniform solid cylinder.

__Known :__

Mass of pulley (m) = 8 kg

Radius of pulley (r) = 10 cm = 0.1 m

Mass of block (m) = 4 kg

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

Weight (w) = m g = (4 kg)(10 m/s^{2}) = 40 kg m/s^{2} = 40 Newton

__Wanted :__ The free fall acceleration of the block

__Solution :__

The moment of inertia of the solid cylinder :

I = 1/2 M R^{2 }= 1/2 (8 kg)(0.1 m)^{2} = (4 kg)(0.01 m^{2}) = 0.04 kg m^{2 }

The moment of the force :

τ = F r = (40 N)(0.1 m) = 4 Nm

The angular acceleration :

Στ = I α

4 = 0.04 α

α = 4 / 0.04 = 100

The linear acceleration :

a = r α = (0.1)(100) = 10 m/s^{2}

7. A block with mass of m hanging from a cord wrapped around a pulley. If the free fall acceleration of the block is a m/s^{2}, what is the moment of inertia of the pulley..

__Known :__

weight = w = m g

Lever arm = R

The angular acceleration = α

The free fall acceleration of the block = a ms^{-2}

__Wanted:__ The moment of inertia of the pulley (I)

__Solution :__

The connection between the linear acceleration and the angular acceleration :

a = R α

α = a / R

The moment of inertia :

τ = I α

I = τ : α = τ : a / R = τ (R / a) = τ R a^{-1 }

**The angular momentum**

8. A 0.2-gram particle moves in a circle at a constant speed of 10 m/s. The radius of the circle is 3 cm. What is the angular momentum of the particle?

__Known :__

Mass of particle (m) = 0.2 gram = 2 x 10^{-4 }kg

Angular speed (ω) = 10 rad s^{-1}

Radius (r) = 3 cm = 3 x 10^{-2 }meters

__Wanted :__ The angular momentum of the particle

__Solution :__

The equation of the angular momentum :

L = I ω

*I = the angular momentum, I = the moment of inertia, ω = the angular speed *

The moment of inertia (for particle) :

I = m r^{2 }= (2 x 10^{-4 })(3 x 10^{-2})^{2 }= (2 x 10^{-4 })(9 x 10^{-4}) = 18 x 10^{-8}

The angular momentum :

L = I ω = (18 x 10^{-8})(10 rad s^{-1}) = 18 x 10^{-7} kg m^{2} s^{-1}