A 70.0 kg grindstone is a solid disk 0.550 m in diameter. You press an ax down on the rim with a normal force of F = 180 N (Figure 1). 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. • 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? F = Value Units • Part B After the grindstone attains 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? ? F = Value Units Figure • 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? m- 50.0 ke ? Value Units

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Chapter1: Units, Trigonometry. And Vectors
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A 70.0 kg grindstone is a solid disk 0.550 m in diameter. You press an ax down on the rim with a
normal force of F = 180 N (Figure 1). 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.
• 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?
?
F =
Value
Units
• 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?
?
F2 =
Value
Units
Figure
• 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?
m- 50.0 kg
?
Value
Units
Transcribed Image Text:A 70.0 kg grindstone is a solid disk 0.550 m in diameter. You press an ax down on the rim with a normal force of F = 180 N (Figure 1). 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. • 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? ? F = Value Units • 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? ? F2 = Value Units Figure • 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? m- 50.0 kg ? Value Units
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