Answer to Question #105566 in Mechanics | Relativity for Shambhawi singh

Question

A thin rectangular plate ( length=b and breath =a)is hanging with the help of two identical massless springs as shown .just after one of the spring is cut, angular acceleration of the plate is

Answer 1

As per the given question,

length and breadth of the rectangular sheet = a and b

Let the mass of the sheet is m,

Initially it was hanged with the help of two string,

so, 2T= mg

Where T is the tension in the string and g is the gravitational acceleration, α is the angular acceleration after the string breaks and I is the moment of inertia of the rectangular sheet.

Hence $T=\dfrac{mg}{2}$

When strings breaks,

Now applying the torque equation,

$mg \dfrac{b}{2}=I\alpha$

$\Rightarrow \alpha=\dfrac{mgb}{2I}----------------(i)$

Moment of inertia of the rectangular sheet =$\dfrac{m(a^2+b^2)}{12}+\dfrac{mb^2}{4}=\dfrac{m(a^2+4b^2)}{12}$

Now substituting the of I in the equation,

$\alpha=\dfrac{12mgb}{2m(a^2+4b^2)}=\dfrac{6gb}{a^2+4b^2}$