Rotational Kinetic Energy of a Point Mass: Detailed Explanations


The rotational kinetic energy of the object is equal to the energy released to do the work to apply the torque.

The rotational kinetic energy depends upon the bulk of mass and density, configuration, and the axis of rotation of a point mass. The rotational kinetic energy of a point mass is the product of mass and the square of the radius of the mass.

Does a Point Mass have Rotational Kinetic Energy?

The point mass spin about its axis of rotation and has some mass and thus the velocity.

The point mass rotating at its axis will be spinning continuously with some energy called rotational kinetic energy.

The rotational kinetic energy of the point mass depends upon the work required to give the torque on the point mass and the angular velocity acquired by the point mass.

Relationship between Mass and Rotational Kinetic Energy

The rotational kinetic energy of the point mass is the kinetic energy of the point mass about the center of mass and is given as,

K.E = 1/2 ω2

Where I is a moment of inertia of a point mass

ω is an angular velocity of the point mass

Linear and Rotational motion of bicycle wheel; Image Credit: Pixabay

The moment of inertia is the total mass of all the particles constituting the object and the tendency of the mass to resist the angular velocity of the body. Here, we are concerned about the point mass, and the moment of inertia of a point mass is

I = mr2

Where m is a mass

And r is the radius of a point mass

Hence, the relation between the mass and the rotational kinetic energy is

K.E = 1/2 mr2 ω2

What is the Ratio of Rotational and Center of Mass Kinetic Energies?

The velocity of the matter depends upon the kinetic energy of the matter in the motion.

The ratio of rotational and center of mass kinetic energies is directly related to the ratio of the entire mass around the center in the matter to the center of mass.

Rotational motion of ferries wheel having fixed center of mass; Image Credit: Pixabay

The rotational kinetic energy is the total kinetic energy of all the masses constituting the matter around the center of mass and is given as K.E = 1/2 ω2 and the center of mass kinetic energy is the kinetic energy of the mass at the center and is given as K.E = 1/2 mv2 .

Hence the ratio of the rotational and center of mass kinetic energies is

KErot/KEcm = 1/2 ω2/ 1/2 mv2

KErot/KEcm = mr2/mv2

The angular velocity is the ratio of the velocity of a particle with respect to the distance from the center of the mass, and therefore the velocity is related to the angular velocity of the particle by the relation v= ωr .

Substituting this in the above relation we have,

KErot/KEcm = mr2/mv2 = M/m

The total kinetic energy is the sum of the kinetic energy of the center of mass and the rotational kinetic energy. The ratio of rotational and center of mass kinetic energies is directly proportional to the respective mass associated with the particular motion.

How to Calculate Rotational Kinetic Energy of Earth?

We can calculate the rotational kinetic energy of Earth by using the formula K.E = 1/2 ω2 .

By calculating the moment of inertia of the Earth and the angular acceleration of the Earth, we can find the value of the rotational kinetic energy of the Earth.

What is the Rotational Kinetic Energy of Earth on its Axis?

The rotational kinetic energy of the Earth depends upon its axis of rotation and its mass of the Earth.

The rotational kinetic energy of the Earth is equal to one-half times the square of the angular velocity of the Earth and its moment of inertia.

Earth axis of rotation; Image Credit: Pixabay

Since Earth is nearly a compact spherical in shape the moment of inertia of the Earth on its axis is

I = 2/5 MR2

I = 2/5 * 6* 1024 * 6.4 * (106)2

I = 98.3 * 1036 kgm2

We can calculate the angular velocity of the Earth which is the angle traced by the Earth in completing one rotation in a particular time interval. Hence, the angular velocity of the Earth is,

ω = 2πrad/24 hr

ω = 2πrad/24 * 60 * 60 = 7.3 * 10-5 rad/s

Now, the rotational kinetic energy of the Earth is

KErot = 1/2 ω2

Substituting all the above values that we have calculated in this equation, we have

KErot = 1/2 * 98.3 * 1036 kgm2 * (7.3 * 10-5 rad/s)2

KErot = 2.6 * 1029 Joules

The rotational kinetic energy of the Earth on its axis is

2.6 * 1029 Joules .

How to Find Rotational Kinetic Energy from Torque?

The object rotates when the equivalent amount of torque is applied to the edges of the object.

Hence, the rotational kinetic energy of the object is equal to the work done on applying the torque.

The energy required to do the work on giving the torque will be converted into the rotational kinetic energy of the body, that is

W𝜏 = ∫W𝜏 = KErot .

The work done is equal to the torque incident on the body,

W𝜏 = ∫W𝜏

rotational kinetic energy of a point mass
Rotation of spinning top due to constant angular momentum; Image Credit: Pixabay

The torque applied to the body is equal to the product of the force and the displacement of the object due to this external force. Suppose ‘r’ is the displacement of the object then

𝜏 = F * r

The tangential acceleration of the object due to torque is defined as the change in the velocity in time.

at = dv/dt

We know that the angular velocity is related to the velocity as

v = ωr

Hence,

at = dω/dt

at = rα

Where α is the angular acceleration.

We also know that F= mat

at = f/m

rα = F/m

therefore F = αmr

Now we can write the equation of torque as 𝜏 =α mr2

The moment of inertia is the product of the total mass and the square of its distance from the center of mass. That is,

l=mr2

Hence,

𝜏 = Iα

How to Calculate the Kinetic Energy of a Rotating Body in Terms of Angular Momentum?

The angular momentum of an object is the angular velocity of the object which is spinning around on its axis of rotation.

By definition, the angular momentum is a product of the moment of inertia and the angular velocity of the object given by the relation L=Iω .

We know that the rotational kinetic energy E is

E = 1/2 Iω2

The moment of inertia in terms of the angular momentum becomes

I = L/ω

Using this in the above equation,

E = 1/2 Lω/ω2

E = 1/2 Lω

Hence, we have two different equations to calculate the rotational kinetic energy in terms of the angular momentum.

Frequently Asked Questions

What is the velocity of the ball rolling down from the slide at a height of 8 meters from the ground?

The object at a height above the ground has the gravitational potential energy equal to mgh.

This gravitational potential energy is converted into the translational kinetic energy and the rotational kinetic energy too as the ball is round in shape.

PE = KEtrans + KErot

mgh = 1/2 mv2 + 1/2ω2

For spherical object I = 2/5 mr2

And ω=v/r

Therefore,

mgh = 1/2 mv2 + 1/2 * 2/5 * (v2/r2)

Eliminating ‘m’ and rearranging the equation we get

gh = v2/2 + v2/5

gh = v2 ( 1/2 + 1/5 ) = 7/10 v2

v2 = 10/7 gh

Hence the velocity of the object is

v= √ 10/7 gh

= √ 10/7 * 9.8 * 8

= √ 112 = 10.6 m/s

The velocity of ball is 10.6 m/s.

What is the rotational energy of a ring rotating at a constant axis running from its center if the mass of the ring is 556 grams and the radius is 12cm with an angular velocity is 1.8rad/s?

Given: m=556grams=0.556kg

r=12cm=0.12m

ω = 1.8 rad/s

The density of the matter in the ring is the total mass in the volume of the ring.

The linear mass density with respect to the angle of rotation

dm/dθ is equal to the ration of

M/2π

Hence, dm =M/2π dθ

The moment of inertia is the integral of the total mass of the ring from the center of mass

I = ∫r2 dm

I = Mr2/ 2πθ

I = Mr2/ 2π * 2π

I = Mr2

The rotational kinetic energy of the ring is

E = 1/2 Iω2

= 1/2 Mr2ω2

= 1/2 * 0.556 * 0.122* 1.82

= 0.013 Joules

The rotational kinetic energy of a ring at a point is 0.013 Joules.

AKSHITA MAPARI

Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess. Connect me on LinkedIn - linkedin.com/in/akshita-mapari-b38a68122

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