Question

A wheel has a speed of 1200 revolutions per minute and is made to slow down at a rate of $4{\text{rad}}{{\text{s}}^{ - 2}}$. Find the number of revolutions it makes before coming to rest.
A) 143
B) 272
C) 314
D) 722

Answer 1

Hint: The given speed of the wheel is actually the frequency of the rotation and the given rate at which the wheel slows down is its angular retardation. As the wheel comes to rest, we can take the final angular velocity of the wheel to be zero. The number of rotations made by the wheel can be calculated once the angular displacement of the wheel is known.

Formulas used:
The final angular velocity of a rotating body is given by, ${\omega _f}^2 = {\omega _i}^2 + 2\alpha \theta $, where ${\omega _i}$ is the initial angular velocity of the body, $\alpha $ is its angular acceleration and $\theta $ is its angular displacement.
The number of rotation of a body is given by, $n = \dfrac{{{\theta _T}}}{{2\pi }}$, where ${\theta _T}$ is the total angular displacement of the body.

Complete step by step answer:
Step 1: List the parameters given in the question.
The problem at hand involves a rotating wheel which slows down at a given rate and eventually comes to rest. We have to determine the number of rotations made by the wheel before it comes to a halt.
The frequency of the wheel is given to be $f = 1200{\text{rpm}}$.
The angular retardation of the wheel is given to be $\alpha = - 4{\text{rad}}{{\text{s}}^{ - 2}}$.
Let ${\omega _i}$ be the initial angular velocity ${\omega _f} = 0$ be the final angular velocity of the wheel.
Then we have ${\omega _i} = 2\pi f = 2400\pi {\text{radmi}}{{\text{n}}^{ - 1}}$.
Let ${\theta _T}$ be the total angular displacement of the wheel.
Step 2: Express the relation for the final angular velocity to obtain the angular displacement of the wheel.
The final angular velocity of the rotating wheel can be expressed as ${\omega _f}^2 = {\omega _i}^2 + 2\alpha \theta $ .
$ \Rightarrow {\theta _T} = \dfrac{{{\omega _f}^2 - {\omega _i}^2}}{{2\alpha }}$ --------- (1)
Substituting for ${\omega _i} = 40\pi {\text{rad}}{{\text{s}}^{ - 1}}$, ${\omega _f} = 0$ and $\alpha = - 4{\text{rad}}{{\text{s}}^{ - 2}}$ in equation (1) we get,
$\Rightarrow {\theta _T} = \dfrac{{0 - {{\left( {40\pi } \right)}^2}}}{{2 \times \left( { - 4} \right)}} = 200{\pi ^2}{\text{m}}$
Thus the angular displacement of the wheel is ${\theta _T} = 200{\pi ^2}{\text{m}}$.
Step 3: Using the obtained angular displacement of the wheel, obtain the number of rotations performed by the wheel before stopping.
Let $n$ be the number of rotations performed by the wheel before coming to rest.
One rotation $\left( {n = 1} \right)$ corresponds to an angle $\theta = 2\pi $.
So for the obtained angular displacement of the wheel, the number of rotations performed will be
$n = \dfrac{{{\theta _T}}}{{2\pi }}$ ------- (2)
Substituting for ${\theta _T} = 200{\pi ^2}{\text{m}}$ and $\pi = 3 \cdot 14$ in equation (2) we get,
$\Rightarrow n = \dfrac{{200{{\left( {3 \cdot 14} \right)}^2}}}{{2 \times 3 \cdot 14}} = 314$
Thus the number of rotations is $n = 314$.

Hence the correct option is C.

Note:
While substituting values of different physical quantities in an equation, all the quantities should be expressed in their respective S.I units. Here the initial velocity is converted into its S.I unit of radian per second as ${\omega _i} = \dfrac{{2400\pi }}{{60}} = 40\pi {\text{rad}}{{\text{s}}^{ - 1}}$ before substituting in equation (1). Angular retardation is the negative of angular acceleration and so we substitute $\alpha = - 4{\text{rad}}{{\text{s}}^{ - 2}}$ in equation (1).