Force, Torque and Angular Acceleration Problem

[cdbowman42]

1. A man sharpens a knife by pushing it against the rim of a grindstone. The 30-cm diameter stone is spinning at 200 rpm and has a mass of 28 kg. The coefficient of kinetic friction between the knife and the stone is 0.20. If the stone loses 10 % of its speed in 10s of grinding, what is the force with which the man presses the knife against the stone.


2. friction force=Normal Force(coefficient of static friction)
friction force=Torque/r
Torque=angular acceleration(a)*I
I=1/2Mr^2
angular acceleration(a)=(final angular velocity(w2)-initial angular velocity(w1))/t





3. The force he applies to the stone has no affect on slowing it down because it is directed toward the axis of rotation. It is the friction force that slows it down, because it is perpendicular to the moment arm.

w1=200rpm*2pi/60=20.9rad/s
w2=200rpm-.10(200rpm)=180rpm
180rpm*2pi/60=18.8rad/s

a=(18.8-20.9)/10= -.21rad/s^2

I=1/2(28)(.15)^2
I=.315kg*m^2

Torque= -.21(.315)=.06615 N*m
Friction force= -.06615/.15=-.441 N

Normal force of man=.441/.2= 2.205 N

Final answer: -2.205 N

The book says 9.9 N

Where am i messing up?

 

[ideasrule]

All your calculations seem correct to me. Perhaps the book's answer is wrong?

 

[ideasrule]

It may be better to do it using angular impulse and momentum. In this case we can assume the force is constant. We should have

$$\tau \Delta t = \Delta (I \omega)$$

then

$$r F_f \Delta t = I (\omega_{initial} - \omega_{final})$$

This gives you the frictional force of the knife on the wheel. Then you can get the compression force $$C$$ by

$$F_f = \mu C$$

Try doing it this way and see what you get.

That's exactly the way the OP used.

 

[AlexChandler]

It may be better to do it using angular impulse and momentum. In this case we can assume the force is constant. We should have

$$\tau \Delta t = \Delta (I \omega)$$

then

$$r F_f \Delta t = I (\omega_{final} - \omega_{initial})$$

This gives you the frictional force of the knife on the wheel. Then you can get the compression force $$C$$ by

$$F_f = \mu C$$

Try doing it this way and see what you get.

 

[AlexChandler]

That's exactly the way the OP used.

Yes I suppose it is really the same. But I prefer to use impulse in cases like this when you only need consider an initial and final state. And actually I get the same exact answer 2.2 N.
Sorry to delete and repost, I was having some trouble with editing the mistakes I made in LaTeX.

 

[cdbowman42]

Impulse isn't even covered that far into the book, so I don't think that's the way to go. I think we may be missing something.

 

[ideasrule]

Impulse isn't even covered that far into the book, so I don't think that's the way to go. I think we may be missing something.

It may not be the intended solution, but it gives the same answer as both your solution and mine. Unless somebody else can spot an error, I think it's safe to assume that the textbook answer is incorrect.

 

[cdbowman42]

Ok well thanks!

 

[cdbowman42]

Unless there is a way to get 9.9 N somehow?