Respuesta :
Explanation:
It is given that,
Mass of the grindstone, m = 90 kg
Radius of the disk, r = 0.34 m
Angular speed of the disk, [tex]\omega=90\ rev/min=9.42\ rad/s[/tex]
Radial force acting on the disk, F = 20 N
(a) The kinetic coefficient of friction between steel and stone is 0.2. The torque due to frictional force is given by :
[tex]\tau=f\times r[/tex]
[tex]\tau=(\mu N)\times r[/tex]
[tex]\tau=(0.2\times 20)\times 0.34[/tex]
[tex]\tau=1.36\ N.m[/tex]
The relationship between the torque and the angular acceleration is given by :
[tex]\tau=I\times \alpha[/tex]
[tex]\alpha =\dfrac{\tau}{I}[/tex], I is the moment of inertia of the disk
[tex]\alpha =\dfrac{\tau}{(1/2 mr^2)}[/tex]
[tex]\alpha =\dfrac{2\tau}{(mr^2)}[/tex]
[tex]\alpha =\dfrac{2\times 1.36}{(90\times 0.34^2)}[/tex]
[tex]\alpha =0.26\ rad/s^2[/tex]
(b) Let [tex]\theta[/tex] are the number of turns make by the stone. Using the equation of rotational kinematics to find it as :
[tex]\theta=\dfrac{\omega_f-\omega_i}{2\alpha}[/tex]
[tex]\theta=\dfrac{0-(9.42)^2}{2\times 0.26}[/tex]
[tex]\theta=-170.64\ radian[/tex]
[tex]\theta=-27.15\ revolution[/tex]
Hence, this is the required solution.
