Question

A wheel has an angular acceleration of $3.0rad{\sec ^{ - 2}}$ and an initial angular speed of $2.00 rad{\sec ^{ - 1}}$. In a time of $2 sec$, it has rotated through an angle (in radian) of:
A) 6 radians
B) 10 radians
C) 12 radians
D) 4 radians

Answer 1

Hint: Newton's laws of motion and Newton's equation hold for every case of motion exists. When the type of motion I changed the variables of that motion will also satisfy the newton`s equation. When there is rotational motion the angle by which the object rotated will be the same as the distance travelled in the transverse motion.

Complete step by step solution:
Here we need to understand the analogies for the transverse and rotational motions as we generally remember the formulae of motion for transverse, motion only.
Following are the three formulae on transverse motion
$s = ut + \dfrac{1}{2}a{t^2}$
$v = u + at$
And, \[{v^2} = {u^2} + 2as\]
Where
$s$ is the distance traversed by the object in transverse motion
$u$ is the initial speed of the object before the motion
$v$ is the final velocity of the object after the motion
$a$ is the acceleration of the object
$t$ is the time duration of the motion
Now as we can the are the basic 5 dimensions which define the transverse motion.
Now, as we know that in rotational motion the object rotates about the fixed axis and does not traverse but this is also a case of motion so it should also obey formulae of motion.
In this case, the angle by which the object rotates on application force is the same as the distance travelled by the object in transverse motion.
Now, as the object rotates by an angle so the speed of rotation will also be called angular speed.
And similarly, the acceleration here will be called angular acceleration.
Now, we can summarize the above analogies as
$s \to \theta $
Where \[\theta \] is the angle turned by the object
$\Rightarrow a \to \alpha $
Where $\alpha $is the angular acceleration
$\Rightarrow u,v \to \omega $
Where \[u,v\] are the angular velocities.
Now our formulae for motion looks like
$\Rightarrow \theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2}$
$\Rightarrow {\omega _f} = {\omega _i} + \alpha t$
And, \[{\omega _f}^2 = {\omega _i}^2 + 2\alpha \theta \]
Now here in this question, we are given that
A wheel has an angular acceleration of 3.0$rad{\sec ^{ - 2}}$ and an initial angular speed of 2.00$rad{\sec ^{ - 1}}$ and we have to find out that in 2 secs how much angle it has rotated by:
Now given values are
$\Rightarrow \alpha = 3.0rad{\sec ^{ - 2}}$
$\Rightarrow {\omega _i} = 2.00rad{\sec ^{ - 1}}$
$\Rightarrow t = 2\sec $we need to find out $\theta $
The best-suited formula for this situation is the one which shows the relation between$\theta ,{\omega _i},t,\alpha $ and not uses ${\omega _f}$
Where ${\omega _i}$is initial angular velocity and ${\omega _f}$is the final angular velocity
So the best formula is
$\theta = {\omega _i}t + \dfrac{1}{2}\alpha {t^2}$
Substituting values we get
$\Rightarrow \theta = 2 \times 2 + \dfrac{1}{2} \times 3 \times {2^2}$
$\Rightarrow \theta = 4 + 6$
$\Rightarrow \theta = 10$ radians

Final answer is (B), the wheel will rotate by 10 radians.

Note: In transverse motion, the body traverses a path but does not rotate like sliding. In rotational motion, the body rotates about a fixed axis and does not have a traversed path like a fan. In general motion, the body rotates as well as transverse a path like a ring rolling down a hill. Here in this question, only the rotational motion of the wheel is considered, though generally the wheel of an automobile is in transverse motion.