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The angular acceleration is 0.19 and the number of turns is 41.07 revolutions.
m = 87 kg be the mass of the grindstone,
r = 0.35 m be the radius of the grindstone,
I = mr²/2 be the grindstone's moment of inertia,
ω = 95 revolution/min = 95 2π/60s = 9.9 s-1 be initial angular velocity,
F = 15 N be the force pressing radially,
f = 0.2 be the coefficient of friction,
α (unknown) be the grindstones angular acceleration.
a)The tangential force slowing the grindstone is
fF
Its torque is
rfF
By Newton's Second Law
rfF = Iα
Therefore
α = rfF/I = 2rfF/mr² = 2fF/mr
Substituting actual numbers
α = 2×0.2×15/(87×0.35) = 0.19 s-2
Note that the above is the absolute value of the acceleration.
Depending on which direction is considered positive or negative,
the answer may need to be negated.
b)
Let
t be the time it takes the grindstone to come to rest,
θ be the total angle by which the grindstone will turn before coming to rest.We have the following identities, which follow from the definition of acceleration:
ω = αt
θ = αt²/2
From the first
t = ω/α
Plug into the second
θ = α(ω/α)²/2 = ω²/2α
Substitute actual numbers
θ = 9.9²/2×0.19 = 257.92
The above result is in radians.
Since the result is supposed to be given in revolutions,
we need to divide by the number of radians per revolutions:
257.92/2π = 41.07 revolutions
Therefore, the angular acceleration is 0.19 and the number of turns is 41.07 revolutions.
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