Question

Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.
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Answer 1

The

The only difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. ... The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy.

Answer 2

☆Translational motion:

》When we say that object is doing pure translatory motion, that means every particle on that rigid system will have the same translatory parameters:

• distance covered(s),velocity(v),linear acceleration(a)

》The velocity by which the centre of mass is moving, will be the velocities of all particles present inside.

☆Rotational motion:

》When we say that an object is doing a pure rotatatory motion, that means :

• every particle on the object is doing circular motion around the axis in which the object is rotational motion.
• every particle will have same rotational parameters,i.e, angular displacement($\theta$), angular velocity($\omega$), angular acceleration($\alpha$).

》An object always do pure rotational motion about an axis, by which it is hinged.

☆Rolling motion:

》Also defined as transrotational motion,i.e, translational + rotational motion.

》In this case object is doing both motion, it is moving forwards with 'v' and also rotating on its axis.

》Here, the object will do rotational motion around the CM.

》so CM will only have the linear velocity, and all the particles around it will have the combined velocity(trans. + rotational)

☆Case of pure rolling:

》It is that state of rolling motion, when there is no energy loss or friction acting etc. takes place.

》If this kind of motion happens in space, it will go forever and ever.

》During that motion the bottomost point is at rest with respect to the ground, which results no relative motion between the surfaces, means no friction involvement.

》object will possess tranlational kinectic energy and rotational kinectic energy.

☆ Relations:

see the solution with refer to above image!

》The bottommost point will be at rest.

• $\sf{ V– \omega R=0}$

• $\bf{V= \omega R}$

》Kinetic energy = 1/2mv²

》Rotational energy = $\dfrac{1}{2}I \omega²$

》Total kinetic energy = 1/2mv²+$\dfrac{1}{2}I \omega²$

Putting $\ V = \omega R$

• TE = $\sf \dfrac{1}{2}m( \omega R)² + \dfrac{1}{2}I \omega²$

•TE = $\bf{ \dfrac{1}{2} \omega²(I+mR²)}$

OR

• TE = $\dfrac{1}{2}mk² (\dfrac{v}{R})² + \dfrac{1}{2}mv²$

• TE = 1/2mv²($\dfrac{k²}{R²} + 1$)

{I = mk²}

☆so this in the required relation in object doing pure rolling.

》 we can see here that, this relation will only be obtained when those quantities are equal which is only in the case of pure rolling.

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