A solid, uniform disk of radius 0.250 m and mass 48.5 kg rolls down a ramp of le

Question

A solid, uniform disk of radius 0.250 m and mass 48.5 kg rolls down a ramp of length 5.20 m that makes an angle of 12° with the horizontal. The disk starts from rest from the top of the ramp.

(a) Find the speed of the disk's center of mass when it reaches the bottom of the ramp. ________m/s

(b) Find the angular speed of the disk at the bottom of the ramp. ________rad/s

A playground merry-go-round of radius 2.00 m has a moment of inertia I = 335 kg·m2 and is rotating about a frictionless vertical axle. As a child of mass 25.0 kg stands at a distance of 1.00 m from the axle, the system (merry-go-round and child) rotates at the rate of 11.0 rev/min. The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge?
________rad/s

Explanation / Answer

Q1.

let speed at the bottom of the ramp is v m/s.

then angular speed at the bottom of the ramp=w=linear speed/radius=v/0.25=4*v rad/s

initial height=5.2*sin(12)=1.08114 m

final height=0 m

momet of inertia of the uniform disk=0.5*mass*radius^2=0.5*48.5*0.25^2=1.515625 kg.m^2

as initial speed is 0, initial kinetic energy=0

final kinetic energy=final linear kinetic energy+final rotational kinetic energy

using energy conservation principle:

initial potential energy+initial kinetic energy=final potential energy+final kinetic energy

==>mass*g*initial height+0=0+0.5*mass*speed^2+0.5*moment of inertia*angular speed^2

==>48.5*9.8*1.08114=0.5*48.5*v^2+0.5*1.515625*(4*v)^2

==>24.25*v^2+12.125*v^2=513.8658

==>v=3.7586 m/s

so speed of disk's center of mass when it reaches bottom is 3.7586 m/s

part b:

angular speed of the disk at the bottom of the ramp=4*v=15.0344 rad/s

Q2.

initial moment of inertia of the system=moment of inertia of the merry go round + moment of inertia of the child

=335+mass of the child*distance from the center^2

=335+25*1^2=360 kg.m^2

initial angular speed=11 rev/min=11*2*pi rad/(60 second)=1.152 rad/s

when the child reaches the edge, final moment of inertia=335+25*2^2=435 kg.m^2

using conservation of angular momentum:

initial moment of the system*initial angular speed=final angular moment of the system*final angular speed

==>final angular speed=360*1.152/435=0.95338 rad/s

so angular speed of the system when the child reaches the edge is 0.95338 rad/s

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