Question

Obtain an expression relating the torque

with angular acceleration for a rigid

body​

Answer 1

Answer:

$\bigstar\:\underline{\boxed{\bf{\red{\tau=I\alpha}}}}$

↗ Consider a rigid body rotating about a fixed axis AB.Consider a particle P of mass m rotating in a circle of radius r.

❄ The radial acceleration of the particle

$\dashrightarrow\bf\:a_r=\dfrac{v^2}{r}={\omega}^2r$

❄ Thus, the radial force on it

$\dashrightarrow\bf\:F_r=m{\omega}^2r$

❄ The tangential acceleration of the particle

$\dashrightarrow\bf\:F_t=\dfrac{dv}{dr}$

❄ Thus, the tangential force on it

$\dashrightarrow\bf\:F_t=m\dfrac{dv}{dt}=mr\dfrac{d\omega}{dt}=mr\alpha$

↗ The torque of mω²r about AB is zero as it intersects the axis and that of m,α is mr²α as the force and the axis are skew and perpendicular. Thus, the torque of the resultant force acting on P is mr²α. Summing over all the particles, the total torque of all the forces acting on all the particles of the body is

$:\implies\sf\:\tau_{total}=\sum{m_i{r_i}^2\alpha}\\ \\ \dag\sf\:I=\sum{m_i{r_i}^2}\\ \\ :\implies\underline{\boxed{\bf{\gray{\tau_{total}=I\alpha}}}}$