Answer to Question #186450 in Mechanical Engineering for Deepanshi singh

Question

A flywheel rotates with a constant retardation due to breaking from t=0to t=10 seconds. it made 400 revolutions . At time t=8 sec , it's angular velocity was 45πrad/sec. Determine

1 . Value of constant ratardation.

2. Total time taken to come to rest.

3. Total revolution made till it comes to rest.

Answer 1

$t_o = 0s\\ \omega_o= ?\\ t_1 = 8s \\ \omega_1 = 40π rad/s\\ t_2 = 10s\\ \omega_2 = ?$

Retardation is constant throughout,

$\therefore \alpha = \dfrac{\omega_1- \omega_o}{t_1- t_o}=\dfrac{\omega_2- \omega _1}{t_2- t_1}$

$\dfrac{40π-\omega_o}{8} = \dfrac{\omega_2 - 40π}{2}$

$\dfrac{40π-\omega_o}{4} = {\omega_2 - 40π}$

$40π-\omega_o= 4{\omega_2 - 160π}\\ 4\omega_2+ \omega_o=200π---(i)$

$\theta =\dfrac{\omega_o +\omega_2}2× t\\$

$400(2π) =\dfrac{\omega_o +\omega_2}2× 10$

$800π = \dfrac{\omega_o +\omega_2}2× 10$

$160π= \omega_o +\omega_2---(ii)$

Comparing equation (i) with equation (ii),

$\omega_o = 461\ rad/s, \quad \omega_2= 52\ rad/s$

$\theta= \omega_o t + 0.5\alpha t²\\ 800π =461(10) + 0.5\alpha(10)^ 2$

$\alpha = -42\ rad/s²$