How Does Friction Affect the Stopping Time of a Rotating Grindstone?

[mizzy]

torques and angular rotation...

1. Homework Statement [/b
a grindstone used to sharpen tools is rotating at an angular speed of 8 rad/s when an axe is pressed against it. because of the torque produced on the grindstone by the force of friction between it and the axe, it comes to rest in four rotations.
a)Find the frictional force produced, assuming the grindstone is a cylinder of mass 5kg and radius 0.4m

Homework Equations

Ff = mu N

Fc = mrw^2

The Attempt at a Solution

The normal force will be equal to the centripetal force. The coefficient of friction wasn't given. How do I find that?

Can someone help me please?

 

[rock.freak667]

Well if the wheel goes from 8 rad/s to 0 rad/s in 4 revolutions, what is the angular acceleration of the wheel?

Then what is the mass moment of inertia of the wheel itself?

How does torque relate to the two above quantities?

Also, how does it relate to the frictional force? (Remember, torque is also force*distance)

 

[mizzy]

Well if the wheel goes from 8 rad/s to 0 rad/s in 4 revolutions, what is the angular acceleration of the wheel?

Then what is the mass moment of inertia of the wheel itself?

How does torque relate to the two above quantities?

Also, how does it relate to the frictional force? (Remember, torque is also force*distance)

angular acceleration = angular speed/ time (this is a dumb question, but what is the time? We are given 4 revolutions, is that frequency? therefore, Time is 1/f?)

torque is equal to the mass moment of inertia x the angular acceleration.

i'm not sure about the frictional force

 

[rock.freak667]

angular acceleration = angular speed/ time (this is a dumb question, but what is the time? We are given 4 revolutions, is that frequency? therefore, Time is 1/f?)

For constant angular acceleration, do you know the rotational kinematic equations?

torque is equal to the mass moment of inertia x the angular acceleration.

This is correct.

i'm not sure about the frictional force

We don't really need to worry about this for now.

 

[mizzy]

For constant angular acceleration, do you know the rotational kinematic equations?


This is correct.



We don't really need to worry about this for now.

omega = omegao + alpha x time

omega^2 = omegao^2 + 2alpha x angular displacement

angular displacement = omega x t + 1/2alpha t^2

 

[rock.freak667]

omega = omegao + alpha x time

omega^2 = omegao^2 + 2alpha x angular displacement

angular displacement = omega x t + 1/2alpha t^2

Right so which one will given you angular acceleration when you have initial and final angular velocities and an angular displacement?

Then what is the moment of inertia of the wheel?

 

[mizzy]

The second equation should be used:
initial angular velocity = 8rad/s
final angular velocity = 0rad/s
angular displacement = ? I'm not sure about this one. We know it went around 4 times, so therefore is it 4(2*pi*r)??

If so angular acceleration:
omega^2 - omegao^2/ 2*angular displacement
= 0^2 - 8^2/ 2(4pir)
= -64/8pir
= -6.37rad/s^2

Moment of inertia for a cylinder is I = 1/2mr^2

therefore I = 1/2 (5kg)(0.4)^2
= 0.4 kgm^2

Is that right?

 

[rock.freak667]

The second equation should be used:
initial angular velocity = 8rad/s
final angular velocity = 0rad/s
angular displacement = ? I'm not sure about this one. We know it went around 4 times, so therefore is it 4(2*pi*r)??

Nearly, 1 revolution = 2*pi radians

If so angular acceleration:
omega^2 - omegao^2/ 2*angular displacement
= 0^2 - 8^2/ 2(4pir)
= -64/8pir
= -6.37rad/s^2

Good but you need to use a different angular displacement after you read above.

Moment of inertia for a cylinder is I = 1/2mr^2

therefore I = 1/2 (5kg)(0.4)^2
= 0.4 kgm^2

Is that right?

Yes.

So now you can get the value of the torque by using T=Iα.


Now consider force wise. As the tool touches the wheel, there is only one force acting. What is this force? (Hint: They are asking you to find it).

How do you get torque, if you know the force and the distance it acts at?

 

[mizzy]

Nearly, 1 revolution = 2*pi radians




Good but you need to use a different angular displacement after you read above.



Yes.

So now you can get the value of the torque by using T=Iα.


Now consider force wise. As the tool touches the wheel, there is only one force acting. What is this force? (Hint: They are asking you to find it).

How do you get torque, if you know the force and the distance it acts at?

so 4 rev = 8*pi radians?

We are asked to find the frictional force.

 

[mizzy]

I recalculated the angular acceleration and I got -1.27rad/s^2. (It is negative because the grindstone is slowly down, correct?)

The moment of inertia stays the same at 0.4kgm^2.

The torque, therefore, is I * angular acceleration = -0.508 (Did i make a mistake? The value is negative.)

We are looking for frictional force(f).

So, f * r = I *angular acceleration/r
f = I * angular acceleration/ r
f = -0.508/ 0.4
f = -1.27N

I don't know why I get a negative value

This is difficult. Once the questions add rotational motion, i get confused!

 

[rock.freak667]

I recalculated the angular acceleration and I got -1.27rad/s^2. (It is negative because the grindstone is slowly down, correct?)

The moment of inertia stays the same at 0.4kgm^2.

The torque, therefore, is I * angular acceleration = -0.508 (Did i make a mistake? The value is negative.)

Now that is correct. The negative sign you don't really need as it signifies that the torque is in the opposite direction to the motion of the wheel.

We are looking for frictional force(f).

This is true!

So, f * r = I *angular acceleration/r
f = I * angular acceleration/ r
f = -0.508/ 0.4
f = -1.27N

I don't know why I get a negative value

This is difficult. Once the questions add rotational motion, i get confused!

Actually that should be correct. Your negative sign, as I said above, just says that your frictional force is in the opposite direction of the motion of the wheel.

 

[mizzy]

There's a second part to this question:

If the coefficient of friction between the axe and grindstone is 0.4, find the normal force with which the axe was pressed against the grindstone.

equations that can be used:

Ff = mu * N

To answer this, do we just solve for N using the above equation? Or do we have to put in account centripetal acceleration?

 

[mizzy]

Can someone confirm if the last part is right? thanks!

 

[rock.freak667]

There's a second part to this question:

If the coefficient of friction between the axe and grindstone is 0.4, find the normal force with which the axe was pressed against the grindstone.

equations that can be used:

Ff = mu * N

To answer this, do we just solve for N using the above equation? Or do we have to put in account centripetal acceleration?

Yes you can just solve for N.

Can someone confirm if the last part is right? thanks!

Yep.