Question

Q.3 A uniform disc of mass M and radius Ris mounted on
a fixed horizontal axis. A block of mass m hangs from
a massless string that is wroapped around the rim of
the disc. The magnitude of the acceleration of the falling
block (m) is :
2M
M+2m
M+2m
2m
-9
M+2m
2m
(d) -
2M +
2M​

Answer 1

Answer:

you can get a good night's sleep and then I will send you the link to the video of the day and I will be there at this point I don't know if you have any questions please feel free to

Answer 2

$a=r\times \alpha$ is the magnitude of acceleration of the falling block.

Explanation:

Moment of inertia of the disc:

$I=\frac{1}{2} M.R^2$

Torque generated due to the tangential force of weight:

$\tau=m.g\times R$

But since the mass drives the disc so it won't fall with the acceleration g.

Now, angular acceleration of the disc:

$\tau=I\times \alpha$

$\alpha=\frac{\tau}{I}$

As we know the relation:

$a=r\times \alpha$

TOPIC: angular acceleration, moment of inertia, torque.

#LearnMore

https://brainly.in/question/4525671