Calculating Tangential Force for Grindstone Rotation

[henry3369]

Homework Statement

A 50.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 160 N (Fig. P10.57). The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50 N*m between the axle of the stone and its bearings. How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to in 9.00 s?

Homework Equations

Net torque = Iα

The Attempt at a Solution

α = 1.396 rad/s^2
Torque from axle + Torque from friction + Torque from crank handle = Iα
6.50 + (0.6)(.26)(160) - t_handle = (1/2)(50)^(.26)2(1.396)
t_handle = 29.1 N*m = rF
F = 29.1/.26 = 111.926 N

The answer is 67.6 N though. What am I doing wrong?

 

[gneill]

You want to rethink how the various torques add up to yield the necessary total torque. Which ones add and which ones subtract from the net?

 

[henry3369]

You want to rethink how the various torques add up to yield the necessary total torque. Which ones add and which ones subtract from the net?

Aren't the signs on the individual torques correct? The two signs of torques from friction will be the opposite from the torque from the handle because friction opposes motion, which is what I have (handle is negative because it rotates it CW, while the torque from friction wants to rotate it CCW)

 

[gneill]

The torque from the handle must overcome both the axle friction and the axe friction and leave enough left over to yield the required total.

 

[henry3369]

I don't understand how this helps with the problem. I know that the net torque has to be greater than zero because the torque from the handle must over the two frictional torques. Isn't that what the equation I used means?

 

[henry3369]

I can't find net torque (the torque left over to create the acceleration), unless I find torque from the handle first, which is what I'm solving for.

 

[henry3369]

The torque from the handle must overcome both the axle friction and the axe friction and leave enough left over to yield the required total.

Sorry, I can find net torque from Iα, which I did. And I'm solving for the torque from the handle. My equation is saying that there is more torque going CW because otherwise it would be in equilibrium.

 

[gneill]

I don't understand how this helps with the problem. I know that the net torque has to be greater than zero because the torque from the handle must over the two frictional torques. Isn't that what the equation I used means?

You wrote:

6.50 + (0.6)(.26)(160) - thandle = (1/2)(50)(.26)2(1.396)

which translates to:
$$\tau_{axle} + \tau_{friction} - \tau_{handle} = \tau_{required}$$
Surely the handle torque must exceed the frictional torques, otherwise the grindstone would rotate backwards. Further, the frictional torques should subtract from the handle torque, with what remains giving the net required torque.

 

[henry3369]

You wrote:

which translates to:
$$\tau_{axle} + \tau_{friction} - \tau_{handle} = \tau_{required}$$
Surely the handle torque must exceed the frictional torques, otherwise the grindstone would rotate backwards. Further, the frictional torques should subtract from the handle torque, with what remains giving the net required torque.

The frictional force causes it to rotate in the counter-clockwise direction though, which means it should be positive right? Why would you subtract it then?

 

[henry3369]

The frictional force causes it to rotate in the counter-clockwise direction though, which means it should be positive right? Why would you subtract it then?

Also, isn't τ_required the same as net torque?

 

[gneill]

Also, isn't τ_required the same as net torque?

I wrote "required" in order to highlight the fact that it is the amount of torque needed to achieve the desired acceleration. Perhaps I should have left it as "net". Sorry for any confusion that may have caused.

 

[henry3369]

I wrote "required" in order to highlight the fact that it is the amount of torque needed to achieve the desired acceleration. Perhaps I should have left it as "net". Sorry for any confusion that may have caused.

τ_net = Iα
τ_net = (1/2)(50)(.26^2)(1.396) = 2.35924 N*m
τ_net = τ_{axle, friction} + τ_{axle, friction} - τ_handle
τ_handle = τ_{axle, friction} + τ_{axle, friction} - τ_net
τ_handle = 6.50 (given) + μnR (torque due to friction from the axe) - 2.35924 (found above)
τ_handle = 6.50 + (0.6)(160)(0.26) - 2.35924
τ_handle = 29.10 Nm
τ_handle = perpendicular force * distance from the axis = 29.10Nm
FR = 29.10
F = 29.10/0.26 = 111.80 N

What am I doing wrong? I still haven't gotten the answer... My work shows that I am solving for the torque of the handle. I've done this step by step many times and always end up with the same answer.

 

[gneill]

You can choose your CW/CCW positive/negative convention either way, but in the end you have one torque that is providing impetus in the desired direction and two that are opposed. Thus:

$$\tau_{handle} - \tau_{axle} - \tau_{axe} = \tau_{net}$$

Note that the net torque has the same sign as the handle torque.

Adjust the signs of all terms by multiplying through by -1 if you wish.

 

[henry3369]

You can choose your CW/CCW positive/negative convention either way, but in the end you have one torque that is providing impetus in the desired direction and two that are opposed. Thus:

$$\tau_{handle} - \tau_{axle} - \tau_{axe} = \tau_{net}$$

Note that the net torque has the same sign as the handle torque.

Adjust the signs of all terms by multiplying through by -1 if you wish.

This doesn't change my results though. It just changes the sign. I'll still end up with a magnitude of 110.80, which is incorrect.

 

[henry3369]

I'm trying to figure out why the magnitude of the the force is different, not the direction.

 

[TSny]

How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to in 9.00 s?

Is there some information left out here?

Does the stone have a CW or CCW angular acceleration α? Did you consider the sign of α?

Also, did you use the correct radius for the crank in your calculation?

 

[henry3369]

Is there some information left out here?

Does the stone have a CW or CCW angular acceleration α? Did you consider the sign of α?

Oh. For some reason it didn't show up when I copied and pasted it. The final angular velocity is 120 rev/min = 4pi rad/s. I still included the acceleration above though.

 

[gneill]

Let's take a look using the magnitudes of the known torques. You have the following known numbers:

T_axle = 6.50 Nm
T_axe = 24.96 Nm
T_net = 2.359 Nm

and one unknown, T_hand

You need T_hand to exceed T_axle + T_axe by the amount T_net. That is,

T_net = T_hand - (T_axle + T_axe)

T_hand = T_net + T_axle + T_axe

 

[TSny]

Oh. For some reason it didn't show up when I copied and pasted it. The final angular velocity is 120 rev/min = 4pi rad/s. I still included the acceleration above though.

Is the angular acceleration positive or negative?

 

[henry3369]

Is the angular acceleration positive or negative?

Negative. That would only increase the torque of the handle above 111 though.

 

[TSny]

Did you use the correct radius for the torque of the crank handle?

 

[henry3369]

Let's take a look using the magnitudes of the known torques. You have the following known numbers:

T_axle = 6.50 Nm
T_axe = 24.96 Nm
T_net = 2.359 Nm

and one unknown, T_hand

You need T_hand to exceed T_axle + T_axe by the amount T_net. That is,

T_net = T_hand - (T_axle + T_axe)

T_hand = T_net + T_axle + T_axe

This would result in τ_handle = 130.15 N*m, which is still incorrect.

 

[henry3369]

Did you use the correct radius for the torque of the crank handle?

Well, the crank handle is 0.25 meters away from the axis of rotation, it also gave the length of the handle to be 0.500 m. Does this contribute to the torque at all?

 

[TSny]

It would be nice to have a picture. I was assuming the crank handle is .500 m from the axis of rotation. That appears to give the correct answer. But, if it is actually .25 m, then I don't see how to get the answer.

 

[henry3369]

http://imgur.com/M2jitam

Exact problem. I'm solving for part a. The radius is 0.520/2 = 0.26

 

[gneill]

This would result in τ_handle = 130.15 N*m, which is still incorrect.

How does 2.359 + 6.50 + 24.96 equal 130.15?

 

[henry3369]

How does 2.359 + 6.50 + 24.96 equal 130.15?

Sorry I mean't that equals 33.819 N*m.
Then 33.819/.26 = 130.15 N of force.

 

[henry3369]

I figured out that using 0.500 m does solve the problem. I was confused when the problem said the handle was 0.500 meters because I thought that he meant that the part of the handle that was perpendicular to the grindstone was 0.500 meters while the length of the handle was 0.26 meters.

Thank you!