Question

13. A thin rectangular plate is hanging with the help of
two identical massless springs as shown.
Just after
one of the springs is cut, angular acceleration of
the plate iis length and breath are a and b respective ly

Answer 1

Answer:

Angular acceleration of the plate just after the spring is cut is

$\alpha = \frac{3bg}{a^2 + b^2}$

Explanation:

As we know that the spring force will remain same and will not change just after one spring is cut

So here we have

$mg = 2kx$

so spring force is given as

$kx = \frac{mg}{2}$

now when one of the spring is cut

$mg - F_s = ma$

$a = \frac{mg - 0.5 mg}{m}$

$a = \frac{g}{2}$

now the plate will accelerate down with this acceleration

Now with respect to the center of mass of the plate net torque on the plate is given as

$F_s (\frac{b}{2}) = m(\frac{a^2 + b^2}{12})\alpha$

so we have

$\alpha = \frac{\frac{bg}{4}}{\frac{a^2 + b^2}{12}}$

$\alpha = \frac{3bg}{a^2 + b^2}$

#Learn

Topic : Rotational motion

https://brainly.in/question/14357351

Answer 2

Answer:

3b/a²+b²

Explanation: