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A ring and a disc roll on the horizontal surface without slipping with the same linear velocity. If both have same mass and total kinetic energy of the ring is $4J$ then total kinetic energy of the disc is:
A) $3J$
B) $4J$
C) $5J$
D) $6J$

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: Recall that kinetic energy is the energy of a body by virtue of its motion. When an object whose shape is like a sphere, if it is travelling along an axis, then it will have rotational kinetic and translational kinetic energy.

Complete step by step solution:
Step I: The total energy of the system is the sum of rotational and translational kinetic energy. Translational kinetic energy is the product of mass of the object and square of the linear velocity of the object. The result is then divided by $2$. It is written as $\dfrac{1}{2}m{v^2}$

Step II: The rotational kinetic energy is the product of rotational inertia and square of angular velocity. The result is then divided by $2$. It is written as $\dfrac{1}{2}I{\omega ^2}$
Where $\omega $is the angular frequency and
$I$is the moment of inertia

Step III: Also the relation between the tangential and rotational velocity is
$v = \omega r$
$\omega = \dfrac{v}{r}$
Where $v$is the tangential velocity
$\omega $is the rotational velocity
$r$is the radius vector

Step IV: The total energy of the ring is given by
$K.E{._{ring}} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}$
Also $I = m{r^2}$
$K.E{._{ring}} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{r^2}.\dfrac{{{v^2}}}{{{r^2}}}$
$K.E{._{ring}} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{v^2}$
$K.E{._{ring}} = m{v^2}$
Given that kinetic energy of the ring is $4J$,
Therefore $K.E{._{ring}} = m{v^2} = 4J$---(i)

Step V: The disc is rotating about the centre of mass, so the inertia of the disc will be $\dfrac{{m{r^2}}}{2}$. The mass of both the ring and the disc is the same. Therefore kinetic energy of the disc is:
$K.E{._{disc}} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}.\dfrac{{m{r^2}}}{2}.\dfrac{{{v^2}}}{{{r^2}}}$
$K.E{._{disc}} = \dfrac{1}{2}m{v^2} + \dfrac{1}{4}m{v^2}$
$K.E{._{disc}} = \dfrac{3}{4}m{v^2}$
From equation (i), substitute the value of kinetic energy
$K.E{._{disc}} = \dfrac{3}{4} \times 4$
$K.E{._{disc}} = 3J$

Step VI: So the kinetic energy of the disc is $3J$

Hence Option (A) is the right answer.

Note: It is to be noted that the rotational kinetic energy is due to the rotation of an object. If there is any body that is rotating about a fixed axis with an angular velocity $\omega $, then the work done by the torque and the change in kinetic energy can be known by using Work Energy Theorem.