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PHYSICS QUESTION: A grindstone in the shape of a solid disk with diameter 0.450 m and a mass of....?
m = 50.0 kg is rotating at = 880 rev/min . You press an ax against the rim with a normal force of F = 260 N , and the grindstone comes to rest in 7.00s.
Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings. (Picture Link below)
http://session.masteringphysics.com/problemAsset/1040261/2/yf_Figure_10_44.jpg
The only force acting to bring the grindstone to rest is the force of friction, since it acts opposite of motion, it is given a minus sign: -μn. We can use the fact that the net torques (∑τ) is equal to the product of moment of inertia (I) and angular acceleration (α), which in turn is equal to the product of the net force and the distance r at which the force acts, or:
∑τ = Iα = rF
The given angular speed should be in SI units:
880rpm = 880rev/min(2πrad/1.00rev)(1.00min/60.0s)
= 92.2rad/s
We are now able to find the angular acceleration:
ω = ω₀ + αt
α = (ω - ω₀) / t
= (0 - 92.2rad/s) / 7.00s
= -13.2rad/s²
Then from the torque equation, we can write:
Iα = rF
mr²α / 2 = -rμn
mrα = -2μn
μ = -mrα/2n
= -(50.0kg)(0.225m)(-13.2rad/s²) / 2(260N)
= 0.286
Hope this helps.
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Maxfli Golf balls 100pk AAAA AAA MIX Fire Rev Solid US $39.99
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100 MAXFLI Rev Solid Revolution Rev Distance More Used Golf Balls US $29.95
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24 Maxfli Rev Solid Golf Balls Nice US $.01
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