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 :Rotational motion – problems and solutions 1

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 53o = 0.8)

Known :

The axis of rotation at point DRotational motion – problems and solutions 2

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

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

F3 = 20 N and l3 = r1 sin θ = (10 cm)(sin 90o) = (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 53o = 0.8)

Known :

The axis of rotation at point D.Rotational motion – problems and solutions 3

Distance between F1 and the axis of rotation (rAD) = 40 cm = 0.4 m

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

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

F1 = 10 Newton

F2 = 10√2 Newton

F3 = 20 Newton

Sin 53o = 0.8

Wanted : The net torque

Solution :

The moment of the force 1

Στ1 = (F1)(rAD sin 53o) = (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)(rBD sin 45o) = (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)(rCD sin 90o) = (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 :Rotational motion – problems and solutions 4

Mass of A (mA) = 0.2 kg

Mass of B (mB) = 0.6 kg

Distance between A and the axis of rotation (rA) = 4 meters

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

Wanted : The moment of inertia of the system

Solution :

The moment of inertia of A

IA = (mA)(rA2) = (0.2)(4)2 = (0.2)(16) = 3.2 kg m2

The moment of inertia of B

IB = (mB)(rB2) = (0.6)(8)2 = (0.6)(64) = 38.4 kg m2

The moment of inertia of the system :

I = IA + IB = 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 gRotational motion – problems and solutions 6

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

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