Question

A wheel starting from rest is uniformly accelerated at $4rad/{{s}^{2}}$ for 10seconds. It is allowed to rotate uniformly for the next 10 seconds and is finally brought to rest in the next 10seconds. Then the total angle rotated by the wheel is
A. 600rad
B. 800rad
C. 400rad
D. 200rad

Answer 1

As a first step, you could read the question carefully and note down all the given values in the question. Then, you could divide the whole motion into three parts and do the calculations accordingly. So, you could find the angle rotation that took place in each part and then take their sum to get the net rotation.
Formula used:
Angular acceleration,
$\alpha =\dfrac{\Delta \omega }{\Delta t}$
Angle rotated,
$\theta =\dfrac{{{\omega }_{f}}^{2}-{{\omega }_{i}}^{2}}{2\alpha }$

Complete step-by-step solution:
In the question we are actually given details about three sets of motion which are: uniformly accelerated motion, constant angular velocity and then uniform deceleration.
(A) Uniformly accelerated motion,
${{\omega }_{i}}=0$
$t=10s$
$\alpha =4rad/{{s}^{2}}$
${{\omega }_{f}}=\alpha t=40rad/s$
Now for finding the angle rotated by the wheel in the part of the motion,
${{\omega }_{f}}^{2}-{{\omega }_{i}}^{2}=2\alpha {{\theta }_{1}}$
$\Rightarrow {{40}^{2}}-{{0}^{2}}=2\times 4\times {{\theta }_{1}}$
$\therefore {{\theta }_{1}}=\dfrac{1600}{8}=200rad$
So, we found the angular rotated in this part of the motion to be 200radian.
(B) Constant angular velocity,
$\omega =40rad/s$
$\omega =\dfrac{\Delta \theta }{\Delta t}=\dfrac{\Delta \theta }{10}=40rad/s$
$\therefore \Delta {{\theta }_{2}}=400rad$
We found the angle rotated in this part of the motion to be 400radian.
(C) Uniformly decelerated motion,
${{\omega }_{i}}=40rad/s$
${{\omega }_{f}}=0$
$t=10s$
Now we have angular acceleration as,
$\alpha =\dfrac{\Delta \omega }{\Delta t}=\dfrac{0-40}{10}=-4$
Now for finding the angle rotated here,
${{\omega }_{f}}^{2}-{{\omega }_{i}}^{2}=2\alpha \Delta \theta$
$\Rightarrow {{0}^{2}}-{{40}^{2}}=2\left( -4 \right)\Delta {{\theta }_{3}}$
$\therefore {{\theta }_{3}}=\dfrac{1600}{8}=200rad$
Angle rotated here is found to be 200radian.
So, the total rotation that is caused would be,
${{\theta }_{total}}={{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}}=200+400+200$
$\therefore {{\theta }_{total}}=800radian$
Hence, the total angle rotated by the wheel is 800radian. Option B is correct.

Note: In case if you wonder how we divided the whole motion into three parts you could read the question again. The question clearly indicates that the wheel starts rotating from rest, and then undergoes uniform rotation and finally the wheel goes to rest each part happens in an interval of 10s.