Question

You have a horizontal grindstone (a disk) that is 85 kg, has a 0.31 m radius, is turning at 88 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 the kinetic coefficient of friction between steel and stone is .2, calculate the angular acceleration of the grindstone in rad/s^2



b) how many turns will the stone make before coming to rest

Answer 1

Step-by-step explanation:

moment of inertia of wheel

I = 1/2 mr² , m is mass and r is radius of disk .

= .5 x 85 x .31²

= 4.08425 kg m²

88rpm

= 1.467 x 2π rad /s

= 9.21 rad / s

force of friction acting = 21 x .2

= 4.2 N

torque produced by friction to stop it

= friction x radius

= 4.2 x .31

= 1.302 Nm

angular deceleration = Torque / moment of inertia

= 1.302 / 4.08425

= .31878 rad /s²

angular deceleration = - .31878 rad /s²

b )

initial angular speed ω₁ = 9.21

Final angular speed ω₂ = 0

angle of turn = θ radian

ω₂² = ω₁ ² - 2 αθ , α is angular deceleration

0 = 9.21² - 2 x .31878 θ

θ = 133.04 radian

no of turns = 133.04 / 2π

= 21.18 turns .