Asked by amy
A grindstone in the shape of a solid disk with diameter 0.520 and a mass of = 50.0 is rotating at = 860 . You press an ax against the rim with a normal force of = 250 (Figure 1) , and the grindstone comes to rest in 7.40 .
Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.
Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.
Answers
Answered by
Damon
angular momentum initial = I omega
I = (1/2) m r^2 = (1/2)(50).26^2
860 what? rpm?
860 rev/min * 1/60 min/sec * 2 pi radians/rev = 90 rad/sec
so Iw= 152
torque * time = change of angular momentum
250 N * mu * 7.4 sec = 152
I am just guessing at your units for everything
I = (1/2) m r^2 = (1/2)(50).26^2
860 what? rpm?
860 rev/min * 1/60 min/sec * 2 pi radians/rev = 90 rad/sec
so Iw= 152
torque * time = change of angular momentum
250 N * mu * 7.4 sec = 152
I am just guessing at your units for everything
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