Question

A L shaped rod of mass M is free to rotate in a vertical plane about axis A A' as shown in figure. Maximum angular acceleration of rod is
ans should be 3g/(√10L)​

Answer 1

Given that,

Mass = M

Shaped of rod is L.

Let the length of rod is L.

We need to calculate the torque

Using formula of torque

$\tau=r\times F$

Where, r = length of rod  from A to mass

F = force

Put the value into the formula

$\tau=\dfrac{L}{2}\times mg$

$\tau=\dfrac{mg L}{2}$

We need to calculate the moment of inertia of rod

Using formula of moment of inertia

$I=\dfrac{mL^2}{3}+\dfrac{mL^2}{12}+m(\dfrac{5l^2}{4})$

$I=\dfrac{20mL^2}{12}$

$I=\dfrac{5mL^2}{3}$

We need to calculate the maximum angular acceleration

Using formula of torque of mg

$\tau=I\alpha$

$r\times F=I\alpha$

Where, r = distance from A to mass

F = force

I = moment of inertia

$\alpha$ = angular acceleration

Put the value into the formula

$\dfrac{L}{2}\times mg=\dfrac{5mL^2}{3}\alpha$

$\alpha=\dfrac{3g}{10L}$

Hence, The maximum angular acceleration is $\dfrac{3g}{10L}$