Question

A 10.0 kg wheel has a diameter of 50.0 cm. How much is the moment of inertia of the wheel? If the wheel is pressed by an axe with a normal force of 10.0 N, the coefficient of kinetic friction between the axe and the wheel is 0.200. How much torque is the wheel experiencing from the axe (in Nm)? If the wheel was spinning with 300 rpm before the axe touches it, how long does it take for the axe to stop it? How many revolutions does the wheel do before coming to rest?

Answer 1

part a:

wheel can be approximated as a ring.

so its moment of inertia is given by mass*radius^2

=10*0.5^2

=2.5 kg.m^2

part b:

normal force=10 N

friction force=friction coefficient*normal force=0.2*10=2 N

torque=radial force*radius

=2*0.5=1 N.m

part c:

initial angular velocity=w0=300 rpm=31.416 rad/s

angular deceleration=torque applied/momentum of inertia

=1/2.5=0.4 rad/s^2

final angular velocity =w1=0

then time taken=(final angular velocity - initial angular velocity)/deceleration

=(0-31.416)/(-0.4)

=78.54 seconds

part d:

angle moved=initial angular speed*time+0.5*angular acceleration*time^2

=31.416*78.54-0.5*0.4*78.54^2

=1233.7 rad

as 1 rev=2*pi rad

total revolution=1233.7/(2*pi)

=196.35

so total 196 revolutions are completed before wheel comes to rest.