A disk rotates through 10 radians in 4 second. The disc experienced uniform acceleration. If the disc is resting, what is the angular velocity after four seconds?
A) 2.5 radian/sec
B) 5 radian/sec
C) 7.5 radian/sec
D) 10 radians/sec

Hint:Use the equation of motion in circular motion, \[\theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2}\] and \[{\omega _f} = {\omega _i} + \alpha t\].

Complete solution:
We know from the question that the angular displacement of the disc is $\theta = 10\;{\rm{rad}}$ and the time required to displace is $t = 4\;{\rm{s}}$. Since the disc started from rest, we know that the initial angular velocity is ${\omega _i} = 0$.

Now we use the equation of motion in circular motion to calculate the angular acceleration of the particle,
\[\theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2}\]
Here, $\alpha $ is angular acceleration of the particle.

Now we substitute \[10\;{\rm{rad}}\] as $\theta $ , 0 as ${\omega _i}$ and $4\;{\rm{s}}$ as $t$ in the above expression, we have,
$
10 = 0 + \dfrac{1}{2} \times \alpha \times {4^2}\\
\alpha = \dfrac{{2 \times 10}}{{16}}\\
 = \dfrac{5}{4}\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}
$

Now we use another equation of motion in circular motion to calculate the final angular velocity of the particle,
\[{\omega _f} = {\omega _i} + \alpha t\]
Here, $\alpha $ is angular acceleration of the particle.

Now we substitute $\dfrac{5}{4}\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}$ as $\alpha $ , 0 as ${\omega _i}$ and $4\;{\rm{s}}$ as $t$ in the above expression, we have,
$
{\omega _f} = 0 + \dfrac{5}{4} \times 4\\
 = 5\;{\rm{rad/s}}
$

Hence, the angular velocity of the disk after four seconds $5\;{\rm{rad/s}}$ and option (B) is correct.

Additional information:The velocity of the body is constant so its acceleration is zero. Thus, it is uniform linear motion. But in uniform circular motion, the velocity is not constant because the direction of the body changes at every point, but its magnitude is constant.

Note:Uniform circular motion is analogous to the uniform linear motion. The equations of motion in circular motion are as follows:
\[{\omega _f} = {\omega _i} + \alpha t\]
\[\theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2}\]
$\omega _f^2 = \omega _i^2 + 2\alpha \theta $