Question

a disc of mass 5kg and radius 0.25 m is capable of ratation about an axis passing through its centre and at right angles to its plane . if a constant torque of 10.25 N-m acts on it . find angular acceleration​

Answer 1

So we have,

$M=5\ kg\\\\R=0.25\ m\\\\\tau=10.25\ N\ m$

We have to find the angular acceleration, $\alpha.$

First we are going to find the moment of inertia of the body,

The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by,

$I=\dfrac {1}{2}MR^2$

So,

$I=\dfrac {1}{2}\cdot 5\cdot(0.25)^2\\\\\\I=0.15625\ kg\ m^2\\\\\\I=\dfrac {5}{32}\ kg\ m^2$

And we know that,

$\tau=I\alpha\\\\\\\alpha=\dfrac {\tau}{I}\\\\\\\alpha=\dfrac {10.25\times 32}{5}\\\\\\\boxed{\alpha=\mathbf {65.6\ rad\ s^{-2}}}$

Thus the angular acceleration is 65.6 rad s^(-2).

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Answer 2

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