Question

A playground ride consists of a disk of mass M=41 kg and radius R=1.6 m mounted on a low-friction axle (see figure below). A child of mass m=24 kg runs at speed v=2.5 m/s on a line tangential to the disk and jumps onto the outer edge of the disk.

(b) What is the change in the kinetic energy of the child plus the disk?

Where has most of this kinetic energy gone?

(d) 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. (Assume the child is initially moving in the positive direction. Indicate the direction with the sign of your answer.)

What caused this change in the linear momentum?

(f) What is the change in the kinetic energy of the child plus the disk, from the beginning to the end of the walk on the disk?

Answer 1

Given

Mass of disc (m1) = 41 kg

Radius of disc (r) = 1.6m

Mass of child (m2) = 24 kg

Speed of child (v) = 2.5 m/s

Moment of inertia of disc (I) = 1/2m1r^2 = ½*41*1.6^2 = 52.48 kgm^2

Moment of inertia of the child (I2) = m2r^2 = 24*1.6^2 = 61.44 kgm^2

Initial angular velocity of disc (w1) = 0

Initial angular velocity of child (w2) = v/r = 2.5/ 1.6 = 1.56 rad/s

Initial angular momentum = final angular momentum

I1w1 + I2w2 = (I1+I2)w

52.48*0 + 61.44*1.56 = (52.48+61.44)w

=> w = (52.48*0 + 61.44*1.56)/ (52.48+61.44) = 0.84 rad /s

b) Initial KE = ½ * I2 * w2^2 = ½*61.44*1.56^2 = 74.76 J

Final KE = ½* (52.48+61.44)* 0.84^2 = 40.19 J

Change in KE = 74.76 – 40.19 = 34.57 J

C) Most of KE is lost as heat due to friction