Question

A car has wheels of radius $0.30m$ and is travelling at $36m/s$. Calculate:
(a) the angular speed of the wheel
(b) If the wheel describes 40 revolutions before coming to rest with a uniform acceleration.
(i) find its angular acceleration and
(ii) the distance covered

Answer 1

A car’s wheel is circular in shape and so will move with an angular velocity. Use the mathematical expression for the angular velocity and the equation of rotational kinematics to find the required answers.

Formula used:
$v = r\omega$
${\omega _f}^2 - {\omega _i}^2 = 2\alpha \theta$

Complete answer:
It is given that the,
Radius of the wheel, r$= 0.3m$
Speed of the wheel, v$= 36m/s$
(a) Angular velocity is defined as the angular position of a rotating body. It is the rate at which the particle rotates around a centre point. Linear velocity when an object is moving along a straight path is defined as the rate of change of displacement with respect to time.
The relation between the angular velocity and the linear velocity is given as,
$v = r\omega$
Where ‘v’ is the linear velocity, ‘r’ is the radius of the wheel and ‘$\omega$’ is the angular velocity.
$\Rightarrow \omega = \dfrac{v}{r}$
$\Rightarrow \omega = \dfrac{{36}}{{0.3}}$
$\Rightarrow \omega = 120rad/s$

(b) the wheel describes 40 revolutions before coming to rest with a uniform acceleration, it stops at ${\omega _f} = 0$
(i) Let ‘$\alpha$’ be the angular acceleration of the wheel. According to the equation of rotational kinematics, we get
$\Rightarrow {\omega _f}^2 - {\omega _i}^2 = 2\alpha \theta$
Since, ${\omega _f} = 0$
So, $- {\omega _i}^2 = 2\alpha \theta$
$\Rightarrow \alpha = - \dfrac{{{\omega _i}^2}}{{2\theta }}$
$\Rightarrow \alpha = - \dfrac{{{{\left( {120} \right)}^2}}}{{2 \times 40 \times 2\pi }}$
$\Rightarrow \alpha = - 28.64rad/{s^2}$
(ii) The distance covered by the wheel will be,
$d = 40 \times 2\pi r$
\eqalign{ & \Rightarrow d = 40 \times 2\pi \times 0.3 \cr & \Rightarrow d = 75.39m \cr}
Hence, (a) The angular speed of the wheel is 120rad/s.
(b) (i) If the wheel describes 40 revolutions before coming to rest with a uniform acceleration its angular acceleration is $\alpha = - 28.64rad/{s^2}$.
(ii) The distance covered is $d = 75.39m$.

Note:
Angular acceleration of an object in motion is defined as the rate of change of angular velocity with time. It is also known as rotational acceleration. Its unit is $rad/{s^2}$. Average angular is defined as the ratio of the change in angular position to the change in time.