Question

. A car has wheels of radius 0.30 m and is travelling at 36 m/s. Calculate :-

(a) the angular speed of the wheel.

(b) If the wheels describe 40 revolutions before coming to rest with a uniform acceleration.

(i) find its angular acceleration and

(ii) the distance covered

Answer 1

do not put the value of pie it is easy to solve

Answer 2

Angular velocity, $\omega=120\ rad/s$

Angular acceleration, $\alpha =-28.64\ rad/s^2$

Distance covered, d = 75.39 m

Explanation:

It is given that,

Radius of the wheel, r = 0.3 m

Speed of the wheel, v = 36 m/s

(a) The relation between the angular velocity and linear velocity is given by :

$v=r\omega$

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{36\ m/s}{0.3\ m}$

$\omega=120\ rad/s$

(b) The wheels describe 40 revolutions before coming to rest with a uniform acceleration. Finally, it stops, $\omega_f=0$

(i) Let $\alpha$ is the angular acceleration of the wheel. Using equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$-\omega_i^2=2\alpha \theta$

$\alpha =-\dfrac{\omega_i^2}{2\theta}$

$\alpha =-\dfrac{(120)^2}{2\times 40\times 2\pi}$

$\alpha =-28.64\ rad/s^2$

(ii) The distance covered by the wheel is given by :

$d=40\times 2\pi r$

$d=40\times 2\pi \times 0.3$

d = 75.39 m

Hence, this is the required solution.

Learn more,

Rotational kinematics

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