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
![\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5CDelta%20%5Comega%7D%7B%5CDelta%20t%7D%20%3D%20%5Cfrac%7B21%20-%2018.85%7D%7B10%7D%20%3D%200.21%20rad%2Fs%5E2)
b) Assume the grind stone is a solid disk, its moment of inertia is
![I = mR^2/2](https://tex.z-dn.net/?f=I%20%3D%20mR%5E2%2F2)
Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.
![I = 28*0.15^2/2 = 0.315 kgm^2](https://tex.z-dn.net/?f=%20I%20%3D%2028%2A0.15%5E2%2F2%20%3D%200.315%20kgm%5E2)
So the friction torque is
![T_f = I\alpha = 0.315*0.21 = 0.06615 Nm](https://tex.z-dn.net/?f=T_f%20%3D%20I%5Calpha%20%3D%200.315%2A0.21%20%3D%200.06615%20Nm)
The friction force is
![F_f = T_f/R = 0.06615 / 0.15 = 0.441 N](https://tex.z-dn.net/?f=F_f%20%3D%20T_f%2FR%20%3D%200.06615%20%2F%200.15%20%3D%200.441%20N)
Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone
![N = F_f/\mu = 0.441/0.2 = 2.205 N](https://tex.z-dn.net/?f=N%20%3D%20F_f%2F%5Cmu%20%3D%200.441%2F0.2%20%3D%202.205%20N)
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wtjfavyw
Sorry to say but I know that t(e introduction is first and the coda is last