Question

A grinding wheel attained a velocity of 20 rad/
sec in 5 sec starting from rest. Find the number
of revolutions made by the wheel.

Answer 1

$\huge{\mathfrak{\underline{\underline{Answer:-}}}}$

$\huge{\sf{\boxed{8 \: Revolutions}}}$

$\huge{\mathfrak{\underline{\underline{Explanation}}}}$

Wheel starts from rest :-

$\large{\sf{Angular \: acceleration \: of \: the \: wheel}}$

$\huge{\bigstar{\boxed{\sf{\alpha \: = \: (\omega - \omega0)/T}}}}$

$\large{\rightarrow{\sf{(20-0)/T}}}$

$\large{\rightarrow{\sf{4rad/sec^{2}}}}$

$\large{\rightarrow{\sf{\theta \: = \: ({\omega0 * T \: + 0.5 \: at^{2}}}}}$

$\large{\rightarrow{\sf{0 \: + (0.5 * 4) * 5}}}$

$\large{\rightarrow{\sf{50 \: rad}}}$

$\large{\rightarrow{\sf{Number \: of \: revolutions \: = \: 50/2{\pi}}}}$

$\huge{\sf{8 \: revolutions}}$

Answer 2

Answer:

8 revolutions

Explanation:

It is given that,

Initially, the wheel is at rest, $\omega_i=0\ rad/s$

Velocity of the wheel, $\omega_f=20\ rad/s$

Time taken, t = 5 s

Firstly, calculating the angular acceleration of the wheel as :

$\alpha =\dfrac{20-0}{5}=4\ rad/s^2$

Let $\theta$ is the number of revolutions made by the wheel. Using second equation of motion as :

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2$

$\theta=0+\dfrac{1}{2}\times 4\times (5)^2$

$\theta=50\ rad$

On converting radian to revolution, number of revolution made by the wheel is given by :

$\theta=\dfrac{50}{2\pi}=7.95\ rev$

or

$\theta=8\ revolution$

So, the number of revolutions made by the wheel is 8. Hence, this is the required solution.