A solid cylinder with a mass of M = 20 kg, and a radius of R = 10 cm starts rest

Question

A solid cylinder with a mass of M = 20 kg, and a radius of R = 10 cm starts rest at the top of a ramp and rolls down without slipping. The top of ramp is h = 5.00 m above its foot. Choose the foot of the ramp to be the reference point for gravitational potential energy. The moment of inertia of a solid cylinder is given by 1/2 MR^2. Find the kinetic energy and potential energy of the cylinder when it is at the top of the ramp. Find the translation speed V_t and the angular speed omega_f of the cylinder at the foot of the ramp. If the radius of the cylinder increases to 20 cm, will V_f and omega_f increase, decrease, or stay the same? Explain why.

Explanation / Answer

part a:

when it is at top of the ramp,

potential energy=mass*g*height

=2*9.8*5=98 J

as it is at rest, kinetic energy=0

part b:

let at the foot of the ramp, linear speed is v,

as the cylinder is rolling without slipping, angular speed=linear speed/radius

=v/0.1=10*v rad/s

at the foot of the ramp, potential eenrgy is zero.

so total mechanical energy of the system=linear kinetic energy + rotational kinetic energy

=0.5*mass*v^2+0.5*moment of inertia*angular speed^2

=0.5*2*v^2+0.5*0.5*2*0.1^2*(10*v)^2

=1.5*v^2

using conservation of energy principle, initial total mechanical energy

=final total mechanical energy

==>1.5*v^2=98

==>v=8.083 m/s

then angular speed=v/R=80.83 rad/s

part c:

if radius increases to 20 cm,

angular speed=v/0.2=5*v

net mechanical energy=0.5*2*v^2+0.5*0.5*2*0.2^2*(5*v)^2

=1.5*v^2

using energy conservation principle,

1.5*v^2=98

==>v=8.083 m/s

angular speed=5*v=40.415 rad/s

so Vt will stay the same.

angular speed will decrease as radius is increasing and speed is staying constant.

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