Question

A uniform circular disc of radius 50cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0rad{s^{ - 2}}$. Its net acceleration in $m{s^{ - 2}}$ at the end of $2.0s$ is approximately:
A. 8.0
B. 7.0
C. 6.0
D. 3.0

Answer 1

It has some angular acceleration so with time it will acquire some angular velocity, then there will centripetal acceleration corresponding to angular velocity and net acceleration will be vector sum of angular acceleration and centripetal acceleration.

Complete step by step solution:
Angular acceleration is given as $\alpha = 2.0rad{s^{ - 2}}$ and
Radius is given as $r = 50cm = \dfrac{{50}}{{100}} m = 0.5m$
At the end of $2.0s$ it will acquire some angular velocity
And that angular velocity will be
$\omega = \alpha t$
$\Rightarrow \omega = 2 \times 2 = 4rad{s^{ - 1}}$
Corresponding to this angular velocity there will be linear velocity and centripetal acceleration.
Linear velocity $v = \omega \times r = 4 \times \dfrac{{50}}{{100}} = 2m/s$
And
Centripetal acceleration ${a_c} = \dfrac{{{v^2}}}{r} = \dfrac{{{2^2}}}{{0.5}} = 8m{s^{ - 2}}$
Now we know that angular acceleration and centripetal acceleration are always perpendicular to each other so total acceleration will be the square root of the sum of squares of both the acceleration.
i.e; ${a_{net}} = \sqrt {{a_c}^2 + {\alpha ^2}}$
putting values we have ,
${a_{net}} = \sqrt {{8^2} + {2^2}} = \sqrt {68} = 8.246 \approx 8m{s^{ - 2}}$
Approximately net acceleration is $8m{s^{ - 2}}$

So, the correct answer is “Option A”.

Note:
As time passes disc acquires angular velocity, there will be associated centripetal acceleration with it and with time the magnitude of angular velocity increase so does the magnitude of centripetal acceleration will increase after some time value of angular acceleration will be insignificant to centripetal acceleration.