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Question

A solid cylinder 30.0 cm in diameter at the top of an incline which is 2.00 meters high is released from rest and rolls to the bottom of the incline without the loss of energy due to friction.
a) Find the linear and angular speeds of the cylinder at the bottom of the incline.
b) After passing over a bump at the end of an incline, the cylinder rolls along now turning at a rate of 1200 rev/min as it traverses a frictionless level surface, a fire ant clinging to the side of the cylinder at a point 0.45 cm from the center (axis of rotation) of the cylinder will experience a centripetal acceleration. What is the centripetal acceleration experienced by the fire ant?

Answer 1

The moment of inertia of a solid cylinder is
I=.5*m*r^2

The total energy of the system will equal zero
so
PE gets converted to KE
PE=m*g*h

The kinetic energy will be in two forms:

Translational
.5*m*v^2

and rotational
.5*I*w^2

since w=v/R
and I=.5*m*r^2
rotational
=0.25*m*v^2

sum the energies

m*g*h=.5*m*v^2+.25*m*v^2

v=sqrt(g*h*4/3)

=5.115 m/s
w=v/r
w=5.115/.3
w=17 rad/sec

To convert rpm to rad/sec multiply the number of rpm by .10472

1200 rpm=125.7 rad/sec (was this supposed to be 120 rpm?)

a=-w^2r
=125.7^2*0.45/100
=71.1 m/s^2

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