Rotational Kinetic Energy: Formula, Units & Examples

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Arpita Srivastava

Content Writer | Updated On - Apr 20, 2024

Rotational Kinetic Energy, also called Angular Kinetic Energy, is the energy caused by an object's rotation around an axis. A moving body's kinetic energy is determined by its mass and speed. It is part of the total kinetic energy.

  • When an object is worked on by applying a net force, it gains speed, which increases its kinetic energy.
  • A rigid body's energy is a type of energy that a moving body possesses as a result of its motion.
  • In the rotating kinetic energy, the moment of inertia acts as the mass, and the angular velocity acts as the linear velocity.
  • Rotating Fan Blades, Wheels of a Car and Wind Turbine Blades are some real-life examples of rotational kinetic energy.
  • Kinetic Energy is a scalar quantity that describes the energy of an object possessed by virtue of its motion. 
  • Mathematically, it can be represented as:

K= 1/2Iω2

Where,

  • KR : Rotational Kinetic energy,
  • I : moment of inertia,
  • ω: angular velocity.
 

Key Terms: Kinetic Energy, Rotation, Rotational Kinetic Energy, Velocity, Force, Rotational Power, Rotational Kinetic Energy Formula, Moment of Inertia


What is Rotational Kinetic Energy ?

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Rotational Kinetic Energy is the energy possessed by an object as a result of its rotation. The work-energy concept can express the equations for linear and rotational kinetic energy. 

  • It refers to the kinetic energy caused due to the rotation of an object.
  • The kinetic energy of a rotating object can be expressed using angular velocity, which is the same for the entire rigid body.
  • Rotational kinetic energy depends on how swiftly the object moves, the amount of mass it contains, and the location of the mass in relation to the spin. 
  • It is in direct proportion to the rotational inertia and the magnitude square of the angular velocity. 
  • The SI unit of rotational kinetic energy is the same as kinetic energy i.e. Joules (J). 

Note: Consider a constant torque given to a flywheel with a moment of inertia I and a constant force applied to a mass m, both of which begin at rest. Therefore:

  • The acceleration from Newton's second law is comparable to the final velocity divided by the time in the linear case.
  • The average velocity is half the final velocity, meaning that the block has the same kinetic energy as the work done on it.
  • The rotational work in the event of rotating motion, starting from rest, will be:

τ × θ

Newton's second law for rotation gives us the angular acceleration of the flywheel.

  • The final angular velocity divided by the time will equal the angular acceleration.
  • The average angular velocity will be half of the ultimate angular velocity. Furthermore, the rotational kinetic energy provided to the flywheel is comparable to the torque's work.

Rotational Kinetic Energy Formula

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Rotational kinetic energy formula can be used to determine the rotational kinetic energy of a rotating body. Inertia moments are denoted by the letter I” and are measured in kilograms per square meter.

  • Joules are the unit of kinetic energy (J).
  • One Joule is equal to one-kilogram meter squared per second squared (kgm2s-2) in other units.

Rotational Kinetic Energy = ½(moment of inertia) \(\times\) (angular velocity)² 

Here,

  • K = ½ Iω² 
  • K = kinetic energy (J = kg.m2/s2)
  • I = moment of inertia (kg.m2)
  • ω = angular velocity (radians/s)

Rotational Kinetic Energy Formula Example

Example. A flywheel is spinning at an angular velocity of 100 rad/s. If its moment of inertia is 0.2 kgm2 , what is its rotational kinetic energy?

Ans: Given, I = 0.2 kgm2 and ω = 200 rad/s

Use the formula for rotational kinetic energy: KE = 1/2Iω2

KE= 1/2 × 0.2 × (200)2 = 4000 Joules.


Rotational Kinetic Energy Derivation 

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Work is required to rotate things like grindstones or merry-go-rounds. The most basic rotating case is when the net force is applied perpendicular to the radius of a disc and stays perpendicular as the disc rotates.

  • The net work done is equal to the force multiplied by the arc length travelled since the force is perpendicular to the displacement:

⇒ net = (net Fs

  • We multiply and divide the right-hand side of the equation by r to bring torque and other rotating quantities into the equation, and we gather terms:

net = (r net F) Δs/r

We already know that r net F = net F τ and Δs/r = θ, thus

net = (net τ)θ

This equation is the expression for rotational work. It's extremely similar to the common definition of translational work, which is force times distance.

⇒ net W = Iαθ

We'll now solve one of the rotational kinematics equations. Let's begin with the equation:

ω2 = ωo2 + 2αθ

net W = ½ Iω2 - ½ Iωo

The work-energy theorem only applies to rotational motion in this equation. As you may recall, a network alters a system's kinetic energy. We define the term:

⇒ (½) Iω2  

As rotational kinetic energy KErot for an object with a moment of inertia I and an angular velocity ω

KErot = ½ Iω2 

The expression for rotational kinetic energy is identical to that for translational kinetic energy, with I representing m and ω representing v

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Rotational Power Formula 

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Torque and frequency of rotation determine the power of rotational systems. Power may be determined by knowing the torque and rotations per minute data. The formula for rotational power is:

P = [τ \(\times\) (2/Π) \(\times\) ​rpm]/60

Where,

  • P is the rotational power
  • τ is the torque
  • The rotations per minute (rpm) are a unit of measurement.

Rotational Kinetic Energy Dimensional Formula 

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Rotational kinetic energy is calculated as follows:

KR = ½ Iω2

Rotational kinetic energy, KR = ½ 

\(\therefore\) [Moment of inertia × (Angular velocity)2]

  • The dimensionless moment of inertia formula = M1L2T0
  • Angular velocity formula in dimensions = M0L0T-1

We may get, by substituting the dimensional formula for the moment of inertia and angular velocity, Rotational kinetic energy's dimensional formula = M1L2T-2


Units of Rotational Kinetic Energy 

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The table below shows the units of rotational kinetic energy in the SI & MKS systems:

Unit System Unit Value
SI Unit System Joules (J)
MKS Unit System kg.m2.s-2

Note: The connection between linear motion's energy and the result of rotational energy.

KTransitional = ½ mv2  

In a rotating system, the moment of inertia serves as the mass, whereas angular velocity serves as the linear velocity v.

Also Read:


Previous Year Questions

  1. A flywheel rotates with a uniform angular acceleration … 
  2. A drum of radius r and mass m … 
  3. A uniform thin bar of mass 6m and length 12L is bent … 
  4. From a circular disc of radius R and mass 9M … ​
  5. If a spherical ball rolls on a table without slipping the friction … 
  6. When a ceiling fan is switched on, it makes 10 revolutions … 
  7. A solid sphere and solid cylinder of identical radii approach … [JEE Mains 2019]
  8. A homogeneous solid cylindrical roller of radius … [JEE Mains 2019]
  9. A rod of length L has non-uniform linear mass density given … [JEE Mains 2020]
  10. A bead of mass mm stays at point P(a,b) on a wire bent … [JEE Mains 2020]

Things to Remember

  • Rotational kinetic energy is a type of kinetic energy that is calculated when an object rotates about its axis.
  • It is equivalent to linear kinetic energy and depends on the shape, mass and angular velocity.
  • The mechanical work done at the time of rotation is the torque time of the rotation angle, W = τθ
  • Rotational kinetic energy helps calculate the simple kinetic energy of an object while it is spinning.
  • If you see a skater skating on the ice, then it is caused due to this kinetic energy.

Sample Questions 

Ques. If the angular velocity is ω=7.29 × 10−5 rad per sec and the moment of inertia is 8.04 × 1037 kgm2, calculate the rotational kinetic energy? (2 marks)

Ans. The following are some of the parameters that are known:

Angular velocity, ω=7.29 × 10−5 rad per sec,

Moment of inertia, I=8.04 × 1037 kgm2

The rotational kinetic energy is Ek=1/2 Iω2

=1/2×8.04×1037 kgm2×(7.29×10−5rad/s)2

=2.13×1029J

As a result, the rotational kinetic energy is 2.13×1029J.

Ques. If the angular velocity is 100π rad per sec and the moment of inertia is 50 kgm2, calculate the rotational kinetic energy of an electric motor? (2 marks)

Ans. Angular velocity, omega = 100π rad per sec

Moment of inertia I=50kgm2

Ek=1/2 Iω2 is the rotational kinetic energy.

=1/2×100π rad/s×50kgm2

= 2500 J

As a result, the rotational kinetic energy is 2500 J.

Ques. Aspherical grindstone with a moment of inertia of 1600 kg.m2 rotates at 6 radians.s-1 angular velocity. What is a grindstone's rotational kinetic energy? (4 marks)

Ans. Moment of inertia,

I = 1600 kg.m2

Angular velocity,

w = 6 radians.s-1

The formula for rotational kinetic energy is as follows:

KR = ½ Iω2 

KR = ½ (1600kg.m2)(6 radians.s-¹)2

KR = ½ (1600kg.m2)(36 s-2

K = ½ (57600kg.m2.s-2)

KR = 28800 kg.m2.s-2

KR = 28800 J

As a result, the grindstone's rotational kinetic energy is 28800 J.

Ques. What are the rotational kinetic energy units? (1 mark)

Ans. The following are the SI and MKS units for rotational kinetic energy:

  • SI unit system: Joules (J)
  • MKS unit system: kg.m2.s-2

Ques. What effect does rotation have on a moving body's overall kinetic energy? (1 mark)

Ans. The kinetic energy of the body increases when it begins to rotate during linear motion. The sum of rotational and translational kinetic energy will equal the total kinetic energy.

Ques. What is the difference between rotational and linear kinetic energy? (2 marks)

Ans. Linear kinetic energy is similar to rotational kinetic energy. It is provided by,

KER = ½ Iω2 

The kinetic energy of a rigid body is given by, in the case of combined rotation and translation,

KE = ½ mv2com + ½ Icomω2 

Ques. Write the formula for rotational kinetic energy? (2 marks)

Ans. We now know that rotational kinetic energy is expressed as

KR = ½ Iω2 

The rotational mass or moment of inertia of a rotating object is denoted by I, while the angular speed is denoted by ω.

Ques. What is the relationship between rotational kinetic energy and angular momentum? (2 marks)

Ans. The rotational kinetic energy is given by,

KR = L2/2I

where I denote inertia and L denotes angular momentum.

Ques. What is the work energy principle? (3 marks)

Ans. According to the theorem of work-energy, the alteration in the kinetic energy of a body is equal to the net work performed by forces on it. It can be presented as: 

Wnet = K – Ko = ΔK

The net work done can be represented as Wnet

The final kinetic energy can be represented as K

The initial kinetic energy can be represented as K

When the initial kinetic energy is subtracted from the final kinetic energy we get the net work done.

The most widely used formula for work energy in order to solve mechanical problems is given below. It can be derived from the conservation of energy law.

Wnet =( ½ m*v2 final) - (½*m * v2 final)  

Ques. What is the law of conservation? (2 marks)

Ans. The law of conservation of energy states that energy can neither be created nor be destroyed. The total energy of an isolated system does not change. An isolated system can be explained as an assemblage of matter that has no interaction with the rest of the universe.

Ques. What is torque? (3 marks)

Ans. Torque is equivalent to force, and angle is equivalent to distance. Even though it was developed for a specific example, the equation net W = (net τ)θ is true in general. We must conduct some algebraic tricks once again to obtain an expression for rotational kinetic energy.

  • It's extremely similar to the common definition of translational work, which is force times distance.
  • Torque is equivalent to force, and angle is equivalent to distance. Even though it was developed for a specific example, the equation net W = (net τ)θ is true in general.

Ques. If the angular velocity is ω=7 × 10−5 rad per sec and the moment of inertia is 8 × 1037 kgm2, calculate the rotational kinetic energy? (2 marks)

Ans. The following are some of the parameters that are known:

Angular velocity, ω=7 × 10−5 rad per sec,

Moment of inertia, I=8 × 1037 kgm2

The rotational kinetic energy is Ek=1/2 Iω2

=1/2×8×1037 kgm2×(7×10−5rad/s)2

=3.92×1029J

As a result, the rotational kinetic energy is 3.92×1029J

Ques. If the angular velocity is 200π rad per sec and the moment of inertia is 50 kgm2, calculate the rotational kinetic energy of an electric motor? (2 marks)

Ans. Angular velocity, omega = 200π rad per sec

Moment of inertia I=50kgm2

Ek=1/2 Iω2 is the rotational kinetic energy.

=1/2×200π rad/s×50kgm2

= 25000 J

As a result, the rotational kinetic energy is 25000 J.


Also Check:

CBSE CLASS XII Related Questions

1.
A circular disc is rotating about its own axis. An external opposing torque 0.02 Nm is applied on the disc by which it comes rest in 5 seconds. The initial angular momentum of disc is

    • $0.1\,kgm^2s^{-1}$
    • $0.04\,kgm^2s^{-1}$
    • $0.025\,kgm^2s^{-1}$
    • $0.01\,kgm^2s^{-1}$

    2.
    A boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. He runs to the other, end. The distance through which the centre of mass of the boat boy system moves is

      • 0
      • 1 m

      • 2 m

      • 3 m

      3.
      A closely wound solenoid of \(2000 \) turns and area of cross-section \(1.6 × 10^{-4}\  m^2\), carrying a current of \(4.0 \ A\), is suspended through its centre allowing it to turn in a horizontal plane. 
      (a) What is the magnetic moment associated with the solenoid?
      (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of \(7.5 × 10^{-2}\  T\) is set up at an angle of \(30º\) with the axis of the solenoid?

          4.
          Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the to charges is the electric potential zero? Take the potential at infinity to be zero.

              5.
              A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

                  6.
                  (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. 
                  (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

                      CBSE CLASS XII Previous Year Papers

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