Question

Please I want some to do answer it clearly and step by step

A potter works with a uniform solid grindstone wheel that has a mass of 18 kg and a radius of 25 cm. The grindstone initially rotates at 1200 rev/min. The grindstone's motor is turned off and the potter presses an instrument perpendicular against the grindstone with a force of 40 N to bring it to a stop. The coefficient of friction with the grindstone is 0.7. The grindstone can be considered a uniform solid cylinder for which I = 1/2 mr^2. (a) What is the torque on the grindstone in stopping it? (b) Find the angular deceleration of the grindstone. (c) How long does it take the grindstone to stop?

Answer 1

a)

frictional force

f =uF =0.7*40 =28 N

Torque acting on the grindstone to stop it

T=fR =28*0.25 =7 N-m

b)

Moment of inertia

I=(1/2)*18*0.25²=0.5625 Kg-m²

SInce

T=I*alpha

angular deceleration is

alpha=T/I =7/0.5625 =12.44 rad/sec²

c)

from

W=W_o+alpha*t

since grinding wheel stops ,therefore W=0 ,since it is decelerating alpha is negative

0=1200*(2pi/60)-12.44*t

t=10.1 sec