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Torque and Forces of grindstone

  • Thread starter NinjaIQ
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Homework Statement



Young University Physics 12e: Mastering Physics Problem 10.53

A 70.0-kg grindstone is a solid disk 0.540 m in diameter. You press an ax down on the rim with a normal force of 150 N. The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50 Nm between the axle of the stone and its bearings.

Part A
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 120 rev/min in 9.00 s ?

Part B
After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min?

Part C
How much time does it take the grindstone to come from 120 rev/min to rest if it is acted on by the axle friction alone?



Homework Equations



Torque = Radius(R) x Force(F) or Moment of Inertia(I) x Angular Acceleration(alpha)


The Attempt at a Solution



Not even sure how to do part A

Do I need to include the negative torque done by the ax along with the friction on the bearings in my calculation for part A?
 
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Answers and Replies

  • #2
tiny-tim
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Part A
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 120 rev/min in 9.00 s ?

Do I need to include the negative torque done by the ax along with the friction on the bearings in my calculation for part A?
Hi NinjaIQ!

You do know what a crank handle is, don't you?

You push tangentially on one end of the handle, the other end is a pivot (so the handle moves in a circle), and so you produce a torque at the pivot. That torque is transferred directly to the grindstone.

In other words: torque produced by crank handle equals torque received by grindstone. :smile:

oh, and yes … the whole purpose is to sharpen the axe … so include axe and bearings! :smile:
 

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