Rotational motion – problems and solutions

Torque

1. A beam 140 cm in length. There are three forces acts on the beam, F₁ = 20 N, F₂ = 10 N, and F₃ = 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 (F₁) = 20 N, the lever arm 1 (l₁) = 70 cm = 0.7 meters

Force 2 (F₂) = 10 N, the lever arm 2 (l₂) = 100 cm – 70 cm = 30 cm = 0.3 meters

Force 3 (F₃) = 40 N, the lever arm 3 (l₃) = 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.

τ₁ = F₁ l₁ = (20 N)(0.7 m) = -14 N m

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

τ₂ = F₂ l₂ = (10 N)(0.3 m) = 3 N m

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

τ₃ = F₃ 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.

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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

F₁ = 10 N and l₁ = r₁ sin θ = (40 cm)(sin 53^o) = (0.4 m)(0.8) = 0.32 meters

F₂ = 10√2 N and l₂ = r₂ sin θ = (20 cm)(sin 45^o) = (0.2 m)(0.5√2) = 0.1√2 meters

F₃ = 20 N and l₃ = r₁ sin θ = (10 cm)(sin 90^o) = (0.1 m)(1) = 0.1 meters

Wanted : The net torque

Solution :

τ₁ = F₁ l₁ = (10 N)(0.32 m) = 3.2 Nm

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

τ₂ = F₂ l₂ = (10√2 N)( 0.1√2 m) = -2 Nm

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

τ₃ = F₂ l₂ = (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 = τ₁ – τ₁ + τ₃

Στ = 3.2 Nm – 2 Nm + 2 Nm

Στ = 3.2 Nm

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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 F₁ and the axis of rotation (r_AD) = 40 cm = 0.4 m

Distance between F₂ and the axis of rotation (r_BD) = 20 cm = 0.2 m

Distance between F₃ and the axis of rotation (r_CD) = 10 cm = 0.1 m

F₁ = 10 Newton

F₂ = 10√2 Newton

F₃ = 20 Newton

Sin 53^o = 0.8

Wanted : The net torque

Solution :

The moment of the force 1

Στ₁ = (F₁)(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

Στ₂ = (F₂)(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

Στ₃ = (F₃)(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 = Στ₁ + Στ₂ + Στ₃

Στ = 3.2 – 2 + 2

Στ = 3.2 Newton meter

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The moment of inertia

4. Length of wire = 12 m, l₁ = 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_A²) = (0.2)(4)² = (0.2)(16) = 3.2 kg m²

The moment of inertia of B

I_B = (m_B)(r_B²) = (0.6)(8)² = (0.6)(64) = 38.4 kg m²

The moment of inertia of the system :

I = I_A + I_B = 3.2 + 38.4 = 41.6 kg m²

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Rotational dynamics

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 R²

I = 1/2 (5 kg)(0.2 m)²

I = 1/2 (5 kg)(0.04 m²)

I = 1/2 (0.2)

I = 0.1 kg m².

The angular acceleration :

τ = I α

α = τ / I = 1.2 / 0.1 = 12 rad s^-2

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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/s²

Weight (w) = m g = (4 kg)(10 m/s²) = 40 kg m/s² = 40 Newton

Wanted : The free fall acceleration of the block

Solution :

The moment of inertia of the solid cylinder :

I = 1/2 M R² = 1/2 (8 kg)(0.1 m)² = (4 kg)(0.01 m²) = 0.04 kg m²

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/s²

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/s², 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

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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 r² = (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 m² s^-1