Question

A playground ride consists of a disk of mass M = 43 kg and radius R = 2.2 m mounted on a low-friction axle. A child of mass m = 29 kg runs at speed v = 2.1 m/s on a line tangential to the disk and jumps onto the outer edge of the disk.
a) Calculate the change in linear momentum of the system consisting of the child plus the disk (but not including the axle), from just before to just after impact, due to the impulse applied by the axle. Take the x axis to be in the direction of the initial velocity of the child. (pf-pi)
b) The child on the disk walks inward on the disk and ends up standing at a new location a distance R/2 = 1.1 m from the axle. Now what is the angular speed?

Answer 1

Given,

v = 2.1 m/s

w = v/R = 2.1/2.2 = 0.955 rad/s

a)The intial moment of inertia of the system was

I1 = 1/2 m R^2 = 0.5 x 43 x 2.2^2 = 104.06 kg-m^2

intial ang momentum

L1 = I1 omega1 = 104.06 x 0.955 rad/s = 99.38 J-s

Final moment of inertia is:

I2 = 1/2 mR^2 + MR^2 = 104.06 + 29 x 2.2^2 = 244.42 kg-m^2

L2 = 244.42 x 0.955 = 233.42 Js

Delta L = L2 - L1 = 233.42 - 99.38 = 134.04 Js

Hence, delta L = 134.04 Js

b)The moment of inertia of sustem now become:

I2 = 1/2 mR^2 + MR^2 = 104.06 + 29 x 1.1^2 = 139.15

We know from conservation of angular momentum that

Lf = Li

139.15 wf = 99.38

wf = 0.71 rad/s

Hence, wf = 0.71 rad/s