How Much Force is Needed to Stop a Grinding Wheel in 20 Seconds?

[clockworks204]

1. A piece of metal is pressed against the rim of a 2.8- kg, 20- cm-diameter grinding wheel that is turning at 2000 rpm. The metal has a coefficient of friction of 0.75 with respect to the wheel. When the motor is cut off, with how much force must you press to stop the wheel in 20.0 s?

2. angular velocity w= (rpm*2*pi)/60
angular accel a= w/time
inertia I=mr^2
Torque T=Ia
F=T/r



3. Here is what I have...I'm not sure how to work in the coefficient of friction... I assumed that it might be T(mu)/r, but it doesn't seem to be right..
w=(2000*2*3.14)/60 =209.44
a=209.44/20 =10.47
I= 2.8*.1m =.028
T= 10.47*.028 =.293
F=.29/.1 =2.2 N
Again, I'm assuming all of the above is right, but I'm unsure of how to work in the coef of friction.

 

[collinsmark]

2. angular velocity w= (rpm*2*pi)/60
angular accel a= w/time
inertia I=mr^2
Torque T=Ia
F=T/r[/b]

3. Here is what I have...I'm not sure how to work in the coefficient of friction... I assumed that it might be T(mu)/r, but it doesn't seem to be right..
w=(2000*2*3.14)/60 =209.44
a=209.44/20 =10.47
I= 2.8*.1m =.028
T= 10.47*.028 =.293
F=.29/.1 =2.2 N

I like your general idea so far. But you might want to rethink your equation for the moment of inertia. Most grinding stones are more-or-less solid disk shaped (not hoop shaped). Of course this isn't universal, and the truth is they come in different shapes and configurations (there are fancy, diamond, grinding wheels that are hoop shaped, for example). But if you walk to the corner hardware store and purchase a traditional, generic, run-of-the-mill grinding stone, it's probably solid disk shaped. I'm guessing that you should assume a solid disk shape. (I could be totally wrong here, maybe you are supposed to assume a hoop. I don't know.)

[Edit: Oh, by the way, there is some sort of error in your final F calculation. .29/.1 is definitely not 2.2. Typo?]

Again, I'm assuming all of the above is right, but I'm unsure of how to work in the coef of friction.

You've calculated 'a' force above. But which force is it? (Hint: when contemplating this question, think about the direction of the force you calculated).

The coefficient of friction relates the frictional force to the normal reaction force. Which force is the problem statement asking for? And which force did you calculate above? (Hint: consider the directions of each force. )

 

[clockworks204]

Yes, I meant to put 2.93 instead of 2.22, but either way both are incorrect. I don't understand what is wrong with the inertia formula.. I've looked again, and it seems like mr^2 is correct.. and mr^2(a) is Torque also correct so I'm still stuck.

 

[inky]

Yes, I meant to put 2.93 instead of 2.22, but either way both are incorrect. I don't understand what is wrong with the inertia formula.. I've looked again, and it seems like mr^2 is correct.. and mr^2(a) is Torque also correct so I'm still stuck.

wheel is a solid disc. I=0.5 mr²
How many forces acting on the wheel?
When you consider torque, you may consider applied force and frictional force.

 

[clockworks204]

I do recognize the equation .5mr^2, but I was unsure of it. So if I add in the coefficient of friction I get: I=.5(2.8)(1.1^2) = .014; Torque= I(a)(.75 fric coef)= .014(10.47)(.75)= .1099; F=T/r =.1099/.1 =1.099

I don't know how else to do it... Everybody in the class is stumped on this one, but I'm sure it's easier than I'm making it out to be..

 

[inky]

I do recognize the equation .5mr^2, but I was unsure of it. So if I add in the coefficient of friction I get: I=.5(2.8)(1.1^2) = .014; Torque= I(a)(.75 fric coef)= .014(10.47)(.75)= .1099; F=T/r =.1099/.1 =1.099

I don't know how else to do it... Everybody in the class is stumped on this one, but I'm sure it's easier than I'm making it out to be..

Use summation torque= I(alpha)

For summation torque=F(r)-F_fr(r)

 

[collinsmark]

Use summation torque= I(alpha)

For summation torque=F(r)-F_fr(r)

Eeek. Be careful! Forces are vectors! Furthermore, the frictional force and normal force are perpendicular to each other!

Here is a big hint:
(i) The question is asking you for the normal force, for the final answer. [Edit: I'm assuming here that there is some sort of attached stopping mechanism here that pushes against the wheel surface. If my assumption is incorrect, and there is no such mechanism, see below. (I can't tell which without a figure or more details of the setup.)] That's the force used to press against the wheel (the pressing force is normal [i.e. perpendicular] to the wheel's tangential surface).
(i) If you know the frictional force, you can calculate the normal force easily enough since you know the coefficient of friction. (In many physics problems one starts with the normal force and later calculates the frictional force. This time it's the other way around.)

Consider the directions of the normal force and the frictional force. Only one of them will cause torque. Which one?

You should be able to look this up in your textbook. But the relevant moments of inertia are:
I = mr² for a hoop, spun around its axis.
I = (1/2)mr² for a solid disk.
I'm not sure which one you are supposed to use. But it makes a difference. Perhaps you should work the problem both ways if you are unsure.

[Edit: On second thought, the problem might not be asking you for the normal force alone (but I suppose that also depends on the whether the grinding wheel has some sort of dedicated, attached stopping plunger thing or not). If there is some sort of dedicated plunger thing (attached), then yes, it's asking for the normal force (i.e. the force used to press against the attached plunger thing).

On on other hand, if one stops the wheel by simply holding a block of metal or something against the wheel freely with one's hand or something, then the pressing force is the vector sum of both the normal and the frictional force. If so, find the normal and frictional forces just like before. The you can use the Pythagorean theorem to find the magnitude of this free-pressing force. ]

 

[clockworks204]

I was able to figure it out finally by using I=.5mr^2 instead of mr^2 and dividing the final answer by the coef of friction to get the correct force. Thanks to all!