Question

Derive an expression for torque in terms of the moment of inertia.

Answer 1

Step 1: Consideration

Consider a rigid body rotating about a given axis with a uniform angular acceleration α, under the action of torque.

Let the body consist of particles of masses m₁, m₂, m₃, . . … , m_n at a distance r₁, r₂, r₃, . . … r_n respectively from the axis of rotation.

Step 2: Formula used

$$\begin{gathered} a=r\alpha (wherea=linearaccelerationaand\alpha =angularacceleration \\ andf=ma(fisforceandmismass) \end{gathered}$$

Step 3: Diagram

Step 4: Derivation of expression for torque

If a₁, a₂, a₃, . . … , a_n are the respective linear acceleration of the particles, then,

$a_1=r_1\alpha ,a_2=r_2\alpha ,a_3=r_3\alpha \dots ..$

Force on particles of mass m is

$f_1=m_1{a}_1=m_1{r}_1\alpha$

Moment of this force about the axis of rotation $f_1\times r_1=\left(m_1{r}_1\alpha \right)\times r_1=m_1{r_1}^2\alpha$

Similarly, the moment of forces on other particles about the axis of rotation is

$m_2{r_2}^2\alpha ,m_3{r_3}^2\alpha ———-m_n{r_n}^{2^{}}\alpha$

The torque acting on the body

$\tau =m_1{r_1}^2\alpha +m_2{r_2}^2\alpha +m_3{r_3}^2\alpha ———-m_n{r_n}^2\alpha$

So,$\tau =\sum _{i=1}^n\left(m_i{r_1}^2\right)\alpha$

or $\tau =I\alpha$

$where\sum _{i=1}^n{m}_i{r_i}^2=I=momentofinertiaofthebodyaboutthegivenaxisofrotation$