Question

A torque of $2{\text{ }}Newton - m$ produced an angular acceleration of $2rad/{\sec ^2}$
a body. If its radius of gyration is$2m$, mass will be-
A. $2kg$
B. $4kg$
C. $1/2kg$
D. $1/4kg$

Answer 1

The torque is always related to change of angular momentum which is defined as moment of inertia times angular velocity. Angular acceleration can be obtained by differentiating the angular velocity with respect to time.

Formula Used: The torque is related to moment of inertia as: $\vec \tau = I\overrightarrow \alpha$

The formula for radius of gyration is: $K = \sqrt {\dfrac{I}{M}}$

Complete step by step solution: Moment of inertia of a rigid body is the sum of the product of masses of different particles and the square of their radial distances from the axis of rotation. The torque is measured as the product of the force and the perpendicular distance of the line of action of the force from the axis of rotation. Therefore, torque can also be defined as the rate of change of angular momentum $\vec L$.Thus,
$\vec \tau = \dfrac{{d\vec L}}{{dt}}$ $\to (1)$
But $\overrightarrow L = I\overrightarrow \omega$ $\to (2)$
where, $I$ is the moment of inertia and $\overrightarrow \omega$ is the angular velocity.
Substituting equation (2) in equation (1)
$\vec \tau = I\dfrac{{d\overrightarrow \omega }}{{dt}} = I\overrightarrow \alpha$
where, $\overrightarrow \alpha$ becomes the angular acceleration.
Therefore, $I = \dfrac{{\vec \tau }}{{\overrightarrow \alpha }}$ $\to (3)$
Radius of gyration $K$ is the square root of the ratio of the moment of inertia $I$ of a rigid body and
its mass $M$
$K = \sqrt {\dfrac{I}{M}}$
Therefore, $I = M{K^2}$ $\to (4)$
Equating equation (3) and equation (4)

}}{{\overrightarrow \alpha \times {K^2}}}$$ $$ \to (5)$$ It is given that $$\vec \tau = 2{\text{ }}Newton - m$$, $$\overrightarrow \alpha = 2rad/{\sec ^2}$$ and $$K = 2m$$. Substituting in equation (5) $$M = \dfrac{{\vec \tau }}{{\overrightarrow \alpha \times {K^2}}} = \dfrac{2}{{2 \times {2^2}}} = \dfrac{1}{4}kg$$ **Hence, option (D) is the correct answer.** **Note:** The angular momentum is $$\overrightarrow L = I\overrightarrow \omega $$. In absence of external torque, $$I\overrightarrow \omega $$ is always a constant vector. Therefore, one cannot further derive the angular velocity. This is the law of conservation of angular momentum. It states that, in the absence of external torque, the angular momentum of a body remains constant.