Answer:
[tex]\theta=50\ revolution[/tex]
Explanation:
It is given that,
Mass of the grindstone, m = 3 kg
Radius of the grindstone, r = 8 cm = 0.08 m
Initial speed of the grindstone, [tex]\omega_i=600\ rpm=62.83\ rad/s[/tex]
Finally it shuts off, [tex]\omega_f=0[/tex]
Time taken, t = 10 s
Let [tex]\alpha[/tex] is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]
[tex]\alpha =\dfrac{0-62.83}{10}[/tex]
[tex]\alpha =-6.283\ rad/s^2[/tex]
Let [tex]\theta[/tex] is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :
[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]
[tex]\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }[/tex]
[tex]\theta=\dfrac{-62.83^2}{2\times -6.283}[/tex]
[tex]\theta=314.15\ radian[/tex]
[tex]\theta=49.99\ revolution[/tex]
or
[tex]\theta=50\ revolution[/tex]
So, the number of revolutions of the grindstone after the power is shut off is 50.
