Question

A ring of diameter $0.4m$and of mass $10kg$ is rotating about its axis at the rate of $2100rpm$. Find kinetic energy of the system.
(A) $9662.408J$
(B) $8450.784J$
(C) $9984.89J$
(D) $9635.408J$

Answer 1

Hint: In order to solve this question, we will first find the moment of inertia of the ring about its axis and then angular frequency from the given information and then we will solve for the total rotational kinetic energy of the system.

Formula used:
If m is the mass, r is the radius of the ring, and its rotating about its axis then the moment of inertia I is given by $I = m{r^2}$
The rotational kinetic energy of the body is given by $K.E = \dfrac{1}{2}I{\omega ^2}$ where $\omega $ is the angular frequency of the body.

Complete answer:
We have given that, ring has a diameter of
$
  d = 0.4m = 2r \\
  r = 0.2m \\
 $
r is the radius of the ring, and its mass is $m = 10kg$ then its moment of inertia about its axes is given by
$
  I = m{r^2} \\
  I = 10{(0.2)^2} \\
  I = 0.4kg{m^2} \\
 $
Now, frequency of the ring is given to us in revolution per minute which is $f = 2100rpm$ converting this into revolution per second we get, $f = 35rps$ and angular frequency is given by
$
  \omega = 2\pi f \\
  \omega = 2 \times 3.14 \times 35 \\
  \omega = 219.8{s^{ - 1}} \\
 $
Now, the total rotational kinetic energy of the ring is given by $K.E = \dfrac{1}{2}I{\omega ^2}$ on putting the values of parameters we get,
$
  K.E = \dfrac{1}{2}(0.4){(219.8)^2} \\
  K.E = 0.2 \times 48312.04 \\
  K.E = 9662.408J \\
 $
So, the Kinetic energy of the ring will be $K.E = 9662.408J$

Hence, the correct answer is option (A) 9662.408 J.

Note: It should be remembered that revolution per minute is converted into revolution per second by dividing the value by $1\min = 60\sec $ and when a body rotates only, there is no translational motion, the body will have pure rotational energy only.