torque in physics

Question

You have a horizontal grindstone (a disk) that is 93 kg, has a 0.33 m radius, is turning at 86 rpm (in the positive direction), and you press a steel axe against the edge with a force of 21 N in the radial direction.

a) assuming that the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone in rad/s².

b) what is the number of turns, N, that the stone will make before coming to rest?

Answer 1

Let

m = 93 kg be the mass of the grindstone,

r = 0.33 m be the radius of the grindstone,

I = mr²/2 be the grindstone's moment of inertia,

ω = 86 revolution/min = 86 2π/60s = 9 s^-1 be initial angular velocity,

F = 21 N be the force pressing radially,

f = 0.2 be the coefficient of friction,

α (unknown) be the grindstones angular acceleration.

a)

The

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 = 2×0.2×21/(93×0.33) = 0.27 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 = 9²/2×0.27 = 150

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:

150/2π = 23.9 revolutions