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10.58 A 50.0 kg grindstone is a solid disk 0.520 m in diam- eter. You press an ax down on the rim with a normal force of 160 N (Fig. P10.58). 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. (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? (b) After the grind- stone 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? (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? Figure P10.58 50.0 kg m = F = 160 N

10.58 A 50.0 kg grindstone is a solid disk 0.520 m in diam- eter. You press an ax down on the rim with a normal force of 160 N (Fig. P10.58). 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. (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? (b) After the grind- stone 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? (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? Figure P10.58 50.0 kg m = F = 160 N

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10.58 A 50.0 kg grindstone is a solid disk 0.520 m in diam- eter. You press an ax down on the rim with a normal force of 160 N (Fig. P10.58). 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. (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? (b) After the grind- stone 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? (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? Figure P10.58 50.0 kg m = F 160 N
Transcribed Image Text:10.58 A 50.0 kg grindstone is a solid disk 0.520 m in diam- eter. You press an ax down on the rim with a normal force of 160 N (Fig. P10.58). 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. (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? (b) After the grind- stone 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? (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? Figure P10.58 50.0 kg m = F 160 N
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