Question

A body having a moment of inertia about its axis of rotation equal to 3 kg-m² is rotating with the angular velocity of 3 rad s⁻¹. The kinetic energy of this rotating body is the same as that of a body of mass 27 kg moving with velocity v. The value of v is-

• A. 1 m/s
• B. 1.5 m/s
• C. 2.5 m/s
• D. 2 m/s

Answer 1

The correct option is A

1 m/s

Correct option is (A) 1 m/s

Step 1 Given data

Moment of inertia of body(I) = $3kg-m^2$

Angular velocity of the body$\left(\omega \right)$ = $3rad/s$

Mass of the second body(M) = 27 kg

The velocity of the second body = v

Step 2 Finding the value of v

We know,

The kinetic energy of a body performing rotational motion is given as

Kinetic energy = $\frac{1}{2}I{\omega }^2$

Where,

I = Moment of inertia

$\omega$ = Angular velocity

The kinetic energy of a body performing translational motion is given as

Kinetic energy = $\frac{1}{2}m{v}^2$

Where,

m = Mass of the body

v = Velocity of the body

Now, according to the question

$\frac{1}{2}I{\omega }^2$ = $\frac{1}{2}m{v}^2$

Putting all the values

$\frac{1}{2}\times 3kg-m^2\times {\left(3rad/s\right)}^2=\frac{1}{2}\times 27kg\times v^2$

After solving

$v^2=1$

Or, v = 1 m/s

Hence, value of v is 1 m/s