Question

A wheel starting from rest is uniformly accelerated at $2rad{{s}^{-2}}$ for 20 seconds. It is allowed to rotate uniformly for the next 10 seconds and is finally brought to rest in the next 20 seconds. The total angle rotated by the wheel (in radian) is:
$\begin{aligned} & (a)600 \\\ & (b)1200 \\\ & (c)1800 \\\ & (d)300 \\\ \end{aligned}$

Answer 1

The total angle rotated by the wheel will be a sum of angle rotated in the first 20 seconds of uniform acceleration, then secondly in the 10 seconds of constant uniform motion and lastly the 20 seconds of motion under constant deceleration, that is after which the wheel stops to move.

Complete step-by-step answer:
Let’s calculate the angle rotated by the wheel in the first part.
Let the angle rotated be given by ${{\theta }_{1}}$
Now, it has been given to us:
Initially the wheel was at rest, so initial angular velocity is equal to zero.
The initial angular acceleration (say $\alpha$) is give to us as $2rad{{s}^{-1}}$
And, the total time duration of motion in the first part is 20 seconds.
Then, using the equation of straight line motion under constant acceleration for purely rolling objects, we have:
$\Rightarrow \theta ={{\omega }_{0}}t+\dfrac{1}{2}{{\alpha }_{0}}{{t}^{2}}$
Where,
$\theta$ is the angle rotated (in radians)
${{\omega }_{0}}$is the initial angular velocity
${{\alpha }_{0}}$is the angular acceleration
$t$ is the total time duration of motion
Putting the values of these terms in the above equation, we get:
$\begin{aligned} & \Rightarrow {{\theta }_{1}}=0+\dfrac{1}{2}\times 2\times {{\left( 20 \right)}^{2}} \\\ & \Rightarrow {{\theta }_{1}}=400radians \\\ \end{aligned}$
Now, let the angular velocity of the wheel in second part be ${{\omega }_{2}}$
Thus, after 20 seconds the angular velocity of the wheel is:
$\begin{aligned} & \Rightarrow {{\omega }_{2}}=0+2(20) \\\ & \Rightarrow {{\omega }_{2}}=40rad{{s}^{-1}} \\\ \end{aligned}$
Thus, the angle rotated in the second part of the motion (say ${{\theta }_{2}}$) will be equal to:
Given, the time duration of the second part being 10 seconds.
$\begin{aligned} & \Rightarrow {{\theta }_{2}}=40\times 10 \\\ & \Rightarrow {{\theta }_{2}}=400radians \\\ \end{aligned}$
Now, in the third and last part. Since, after 20 seconds the wheel comes to rest under a constant deceleration. The deceleration then must be equal to the initial acceleration as it took 20 seconds for the wheel to achieve the same angular velocity.
Thus, the angle rotated in the last 20 seconds (say ${{\theta }_{3}}$) must be equal to the first part, that is:
$\begin{aligned} & \Rightarrow {{\theta }_{3}}={{\theta }_{1}} \\\ & \Rightarrow {{\theta }_{3}}=400radians \\\ \end{aligned}$
Therefore, the total angle rotated (say ${{\theta }_{T}}$) is equal to:
$\begin{aligned} & \Rightarrow {{\theta }_{T}}={{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}} \\\ & \Rightarrow {{\theta }_{T}}=400+400+400 \\\ & \Rightarrow {{\theta }_{T}}=1200radians \\\ \end{aligned}$
Hence, the total angle rotated by the wheel is $1200radians$ .

So, the correct answer is “Option (b)”.

Note: It should be noted that, in the last part of the question, we avoided unnecessary calculations by using simple analogy that if the angular velocity is same and time of acceleration is equal to time of deceleration, then they must have rotated the same angle.