Answer:
a. 0.21 rad/s2
b. 2.205 N
Explanation:
We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds
200 rpm = 200 * 2π / 60 = 21 rad/s
180 rpm = 180 * 2π / 60 = 18.85 rad/s
r = d/2 = 30cm / 2 = 15 cm = 0.15 m
a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is
[tex]\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2[/tex]
b) Assume the grind stone is a solid disk, its moment of inertia is
[tex]I = mR^2/2[/tex]
Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.
[tex] I = 28*0.15^2/2 = 0.315 kgm^2[/tex]
So the friction torque is
[tex]T_f = I\alpha = 0.315*0.21 = 0.06615 Nm[/tex]
The friction force is
[tex]F_f = T_f/R = 0.06615 / 0.15 = 0.441 N[/tex]
Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone
[tex]N = F_f/\mu = 0.441/0.2 = 2.205 N[/tex]
