Roational Motion Physics

1. Physics Rotational Motion Formulas
2. Rotational Motion Physics Notes
3. Rotational Motion Physics Definition
4. Rotational Motion Physics Definition

Physics » Circular and Rotational Motion; Complete JEE Main/Advanced Course and Test Series OFFERED PRICE: Rs. 15,900 View Details; Revision Notes on Circular& Rotational Motion Circular Motion:- Uniform Circular Motion:-Circular motion is said to the uniform if the speed. ¹Physics types usually describe rotational speed, ω, in terms of the number of 'radians' turned in a unit of time. There are a little more than 6 radians in a full rotation (2π radians, to be exact). When a direction is assigned to rotational speed, we call it rotational velocityty.). Rotational velocity is a vector whose magnitude is the. Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Making Connections: Rotational Motion Dynamics Dynamics for rotational motion is completely analogous to linear or translational dynamics. Dynamics is concerned with force and mass and their effects on motion. For rotational motion, we will find direct analogs to force and mass that behave just as we would expect from our earlier experiences. Sep 10, 2020 Making Connections: Rotational Motion Dynamics Dynamics for rotational motion is completely analogous to linear or translational dynamics. Dynamics is concerned with force and mass and their effects on motion. For rotational motion, we will find direct analogs to force and mass that behave just as we would expect from our earlier experiences.

Physics Rotational Motion Formulas

The Dynamics of Rotational Motion Calculator will calculate:

1. Torque in terms of moment of inertia and angular acceleration
2. Angular momentum in terms of moment of inertia and angular velocity
3. Rotational work in terms of torque and angular displacement
4. Rotational kinetic energy in terms of moment of inertia and angular velocity
5. Rotational power in terms of torque and angular velocity

Please note that the formula for each calculation along with detailed calculations are available below. As you enter the specific factors of each dynamics of rotational motion calculation, the Dynamics Of Rotational Motion Calculator will automatically calculate the results and update the Physics formula elements with each element of the dynamics of rotational motion calculation. You can then email or print this dynamics of rotational motion calculation as required for later use.

We hope you found the Dynamics Of Rotational Motion Calculator useful with your Physics revision, if you did, we kindly request that you rate this Physics calculator and, if you have time, share to your favourite social network. This allows us to allocate future resource and keep these Physics calculators and educational material free for all to use across the globe. We believe everyone should have free access to Physics educational material, by sharing you help us reach all Physics students and those interested in Physics across the globe.

Related Physics Sections with Tutorials

Dynamics of Rotational Motion Calculator Limitations

If any of the quantities required within the Dynamics of Rotational Motion Calculator are not explicitly given, please use the an alternative calculator to calculatoe them. For example, you can use the Moment of Inertia Calculator to find the value of Moment of Inertia in different situations if it is not given. Likewise, you can use the Uniformly Accelerated (Decelerated) Circular Motion Calculator to find angular acceleration, angular velocity and angular displacement if the values are missing from your problem or equation.

Calculating torque by applying Newton's Second Law in Rotational Motion - Formula and Calculation

Angular momentum in rotational motion Formula and Calculation

Work in rotational motion Formula and Calculation

Rotational kinetic energy Formula and Calculation

Rotational power Formula and Calculation

Rotation Physics Tutorials associated with the Uniform Motion Calculator

The following Physics tutorials are provided within the Rotation section of our Free Physics Tutorials. Each Rotation tutorial includes detailed Rotation formula and example of how to calculate and resolve specific Rotation questions and problems. At the end of each Rotation tutorial you will find Rotation revision questions with a hidden answer that reveals when clicked. This allows you to learn about Rotation and test your knowledge of Physics by answering the test questions on Rotation.

Physics Calculators

You may also find the following Physics calculators useful.

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.

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

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

The moment of inertia

Rotational Motion Physics Notes

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

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

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

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

Rotational Motion Physics Definition

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 :

Rotational Motion Physics Definition

L = I ω = (18 x 10^-8)(10 rad s^-1) = 18 x 10^-7 kg m² s^-1