Question

13. A flywheel slows down uniformly from 1200 rpm to 600 rpm in 5 s. Find the number of revolutions made
by the wheel in 5 s. ​

Answer 1

Explanation:

If motor revolving at 1200 rpm slows down uniformly 900 rpm in 2 seconds. Calculate ... α =5πrad/s2. The angle is ... The numbers of revolutions are given as, ... Get Instant Solutions, 24x7.

Answer 2

Answer:

• Revolution made by the flywheel in 5 sec = 50

Explanation:

Given

Initial angular velocity of the Flywheel, $\omega_0$ = 1200 rpm

Final angular velocity of the Flywheel, $\omega$ = 600 rpm

Time taken for the change in velocities, t = 5 s

To find

• The Number of revolutions made by the wheel in 5 sec, n =?

Formula required

Angular acceleration 'a' is given by

$\bf{a = \dfrac{\omega - \omega_0}{t}}$

The Second equation of Angular motion  

$\bf{\Delta \theta = \omega_0 t + \dfrac{1}{2}at^2}$

Where,

a is angular acceleration, $\Delta \theta$ is angular displacement, $\omega_0$ is initial angular velocity, $\omega$ is final angular velocity and t is time taken.

Solution

Let us first convert the velocities given in rpm into rad/s

$\to \omega_0 = 1200\;rpm = 1200\times \dfrac{2\pi}{60}\;rad/s$

$\to \omega_0 = 40\pi\;\;rad/s$

and,

$\to \omega = 600 rpm = 600\times \dfrac{2\pi}{60}\;rad/s$

$\to \omega = 20\pi \;\;rad/s$

Now, calculating the angular acceleration of the flywheel

$\to a = \dfrac{\omega - \omega_0}{t} = \dfrac{ 20\pi - 40\pi}{5}$

$\to a = -4\pi\;\;rad/s^2$

   

Calculating the angular displacement of the flywheel in 5 sec

$\to \Delta \theta = \omega_0 t + \dfrac{1}{2}at^2 = 40\pi \times 5 + \dfrac{1}{2} \times (-4\pi) \times (5)^2$

$\to \Delta \theta = 100\pi\;\;rad$

Now,

since 1 revolution = $2\pi$ rad

therefore,

calculating the number of revolution 'n' made for angular displacement $100\pi$ rad

$\to n = \dfrac{\Delta \theta}{2\pi} = \dfrac{100\pi}{2\pi}$

$\to \boxed{\;\;\bf{ n = 50 }\;\;}$

Therefore,

• Revolution made by the wheel in 5 sec = 50 revolutions.