Question

A wheel of mass 10 kg and a radius 20cm is rotating at an angular speed of 100rev/min when the motor is turned off. Neglecting the friction at the axis, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions. (Consider wheel to be a uniform circular disc)
A) $0.86N$
B) $0.87N$
C) 0.82N
D) $0.89N$

Answer 1

Hint: The speed of the object in rotational motion is called angular speed. The unit of angular speed is radians per second. The angular speed can be calculated as distance covered by the body in the terms of revolutions to the time taken. Orbital and spin angular velocity are the two types of angular velocity.

Formula used:
To solve this type of question we use the following formula.
The way we have the equation of motion for linear motion we have the equation of motion for angular/ circular motion.
${\omega ^2} = \omega _0^2 + 2\alpha \theta $
$\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}$
$\omega = {\omega _0} + \alpha t$
Where, $\theta$ is angular displacement, $\omega$ is final angular velocity, $\omega_0$ is the initial angular velocity, $\alpha$ is angular acceleration and $t$ is the time.

Complete step by step answer:
Given,
Mass of a wheel =$ 10 kg$
Radius of wheel = $20 cm$
Angular speed ($\omega$) = $100$rev/m= $\dfrac{5}{3}$ rev/sec
The number of revolutions after the motor is off = $10$.
We have to find the tangential force which must be applied.
Let us now use the formula ${\omega ^2} = \omega _0^2 + 2\alpha \theta $ to calculate the value of $\alpha$.
We already have the following values.
$\omega=0$, total revolutions $\theta=10$ and $\omega_0=\dfrac{5}{3}$ rev/s
Let us substitute these values in the above formula.
$\Rightarrow 0 = {(5/3)^2} + 2\alpha \times 10$
Let us simplify it.
$\Rightarrow \alpha = - \dfrac{{25}}{{9 \times 20}} = \dfrac{{ - 5}}{{36}}rev/{s^2} = - \dfrac{{10\pi }}{{36}}rad/{s^2}$
Now let us calculate the moment of inertia of the wheel which is given below.
$I = \dfrac{1}{2}m{r^2}$
Let us substitute the values.
$\Rightarrow I = \dfrac{1}{2} \times 10 \times {(0.2)^2} = 0.2kg{m^2}$
Now let us calculate the torque.
$\tau = I\alpha $
Also, if $F$ is the force then torque is given as below.
$\Rightarrow \tau = F \times r = Fr$ [as $\theta=90^\circ$]
Hence, we can write the following.
$I\alpha = Fr$
Let us substitute the values.
$\Rightarrow F \times 0.2 = - 0.2 \times \dfrac{{10\pi }}{{36}}$
Let us further simplify it.
$\Rightarrow F = - \dfrac{{10\pi }}{{36}} = - 0.872N$
Here, the negative sign indicates the direction of force is opposite to the direction of motion.

$\therefore$ A force of 0.872N is to be applied to the wheel to bring it to rest in 10 revolutions. Hence, option (B) is the correct option.

Note:
- We can always solve these types of questions using the three equations of motion. However, we have to decide the equation to be used based on the variables given in the question.
- The friction force is the force that always opposes the motion of an object on another surface and its direction is always opposite to the direction of motion.