I’m going to discuss the topic of the ratio of rotational and translational kinetic energy of a sphere in detail in this article.

**The rotational kinetic energy of a solid sphere is= ½ Iw^2=⅕ mv^2……….(1)The translational kinetic energy of the solid sphere is=½ mv^2 ………(2)Therefore,the ratio of rotational and translational kinetic energy is=(⅕ mv^2)/(½ mv^2)=⅖ [by dividing equation(1) by equation (2)]**

Before starting to describe the above derivation of the ratio of rotational and translational kinetic energy of a sphere we should acknowledge a primary idea on rotational and kinetic energy. Rotational kinetic energy is the energy which a body possesses while rotating around an axis. Translational kinetic energy is the energy which a body at rest possesses while moving with a linear acceleration.

The detailed derivation of the ratio of rotational and translational kinetic energy of a sphere is given below. The expression of rotational kinetic energy of a sphere is= ½ Iw^2

Where I = the moment of inertia of a sphere

= ⅖ mr^2……………….(3) where m is the mass of the sphere and r is the distance between the axis and the sphere and w= the angular velocity of the sphere with which it is rotating around the axis

w=v/r………………….(4) where v is the linear velocity of the sphere [ from the formula v=wr]

Therefore,the rotational energy of the solid sphere= ½ Iw^2

Putting the values of I and w from equation (3) and equation (4) respectively we get,

= ½.(⅖mr^2).(v/r)^2

=⅕ mv^2………………….(5)

The expression of translational kinetic energy is= ½ mv^2……………(6)

Therefore, from equation (5) and (6)-

The ratio of rotational and translational kinetic energy of a sphere is=⅕ mv^2/½ mv^2

= 2:5

**How to find rotational kinetic energy from translational energy?**

**A rigid rotating solid sphere is a collection of several identical particles. Here I will consider the i th single particle and will find rotational energy from translational kinetic energy. We know the expression for translational kinetic energy of the particle is ½ mivi^2. Here vi is the tangential velocity of the particle and mi is the mass of the considered particle. We know vi=wi.ri where wi is the angular velocity of the particle and ri is its distance from the fixed axis.**

Translational kinetic energy of the i th particle is = ½ mivi^2

= ½ mi.(wi.ri)^2

=½ mi.wi^2.ri^2

=½ (mi.ri^2).wi^2

Here the total mass of the sphere is M= M=Σmi=m1+m2

m3+%u2026…mi

Moment of inertia of the sphere is I= Σmi.ri^{2 }and

Angular velocity is the same for all the particles of the sphere. Hence, wi=w

Therefore,total translational kinetic energy of the sphere is = ½ Iw^2

= rotational energy of the solid sphere

This is our required derivation of how to find rotational kinetic energy from transitional energy.

**Describe the relationship between rotational and translational kinetic energy**

**It has been proved earlier that these two energies are analogous. But there is a very nominal difference between these two energies. The translational kinetic energy is the energy which a rigid body possesses when it is continuing its linear motion in a straight or linear path and with linear velocity and acceleration whereas the rotational kinetic energy is the energy of that rigid body which is possessed within it while rotating around a fixed axis.**

With the help of a simple example the difference between the rotational and translational kinetic energies can be explained. When a cycle is moving on a road its wheels possess two types of kinetic energies. When the wheels are rotating on the road they possess rotational and when they are following linear motion they possess translational kinetic energy. If the front side wheel is moved above then it possesses only the rotational one.

**Are rotational and translational kinetic energy equal?**

**The concept of radius of gyration should be introduced before starting to prove whether rotational and translational kinetic energies are equal or not. The distance between the axis of rotation of a body and the point where all the masses of the body are assumed to be concentrated,is considered as the radius of gyration. The moment of inertia at this point is equal to the moment of inertia of the original mass of the body.**

Here moment of inertia of that body=I=mk^2

Or, k=(2I/m)^½……………..(7)

Here m is the mass of the body and k is the radius of gyration.

We know the general formula of moment of inertia of a body of mass m is I=mk^{2}

Now, for a circular ring r=k

where r is the distance between the axis of rotation and a given particle of that body.

Therefore, I=mk^{2}=mr^{2}

Rotational energy of a circular ring is= ½ Iw^2

= ½ (mr^2).w^2

= ½ m(w^2r^2)

= ½ m.(wr)^2

=½ mv^2 [as v=wr]

= translational kinetic energy

**Is rotational kinetic energy less than translational kinetic energy?**

**For a solid sphere,**

**the rotational energy= ½ Iw^2**

**Putting the values of I and w from equation (3) and equation (4) respectively we get,**

** =½.(⅖mr^2).(v/r)^2**

**= ⅕ mv^2= 2/10 mv^2 …………(8)**

**The expression of translational kinetic energy is=½ mv^2= 5/10 mv^2 ……………(9)**

**From the equations (9) and (10) it can be said that,**

**2/10 mv^2 < 5/10 mv^2**

Therefore, rotational K.E. < translational K.E of a solid sphere.

**What is the ratio of the translational kinetic energy to the rotational kinetic energy when it reaches the bottom of the ramp?**

**When a rolling solid sphere reaches the bottom of the ramp,**

**Its rotational K.E.=½ Iw^2**

** =½.(⅖mr^2).(v/r)^2**

**=⅕ mv^2**

**Its translational K.E.=½ mv^2**

**Therefore,the ratio of the translational kinetic energy to the rotational kinetic energy when it reaches the bottom of the ramp= ⅕ mv^2/½ mv^2**

**= 2:5**

**How much fraction of the kinetic energy of rolling body is purely translational and rotational?**

**Total K.E. of a rolling solid sphere= 7/10 mv^2**

**Rotational K.E.= ⅕ mv^2**

**Therefore, the fraction of the kinetic energy of rolling body is purely rotational=**

**= ⅕ mv^2/ 7/10 mv^2**

**=2/7**

Translational K.E.=½ mv^2

Therefore, the fraction of the kinetic energy of rolling body is purely translational=

=½ mv^2/7/10 mv^2

=5/7

**What fraction of the total kinetic energy of a rolling sphere is translational?**

**For a rolling solid sphere** **the total K.E.= K rot + K trans**

**=⅕ mv^2+½ mv^2**

**=7/10 mv^2**

**The required fraction= ½ mv^2/7/10 mv^2**

**= 5/7**

**Can an object have translational and rotational kinetic energy?**

**The movement of wheels of a bike is an example for this purpose where there are both the kinetic energies available. When the wheels are rotating they possess rotational K.E. and when they move on the straight horizontal road they acquire translational K.E. **

**the ratio of rotational and translational kinetic energy of a rolling circular disc**

**Moment of inertia of a rolling circular disc perpendicular** **to the plane of the disc about the center of the disc, I= ½ MR^2 where M is the mass and R is the distance from the axis of the disc.**

**Rotational K.E. of the disc= ½ Iw^2**

**=½ . ½ MR^2.(v/R)^2**

**= ¼ Mv^2**

Translational K.E. =½ Mv^2

the ratio of rotational and translational kinetic energy of a rolling circular disc= ¼ Mv^2/½ Mv^2

= 1:2

**What would be the ratio of the rotational kinetic energy and translational kinetic energy of the rolling solid cylinder?**

**Moment of inertia of the rolling solid cylinder,I= ½ mr^2**

** The rotational kinetic energy = ½ Iw^2= ½ .½ mr^2.(v/r)^2= ¼ mv^2**

**The translational kinetic energy = ½ mv^2**

**Therefore, the ratio of the rotational kinetic energy and translational kinetic energy of the rolling solid cylinder=¼ mv^2/½ mv^2=1:2**

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