Question

A carousel has a radius of  R=3.00m and a moment of inertia of I= 6250 kgâ‹…m2 for rotation about axis perpendicular to the its center. The carousel is rotating unpowered and without friction with an angular velocity of 1.25 rad/s. An 85.0 kg man runs with a velocity of v=8.00m/s , on a line tangent to the rim of the carousel, overtaking it. The man runs onto the carousel and grabs hold of a pole on the rim.

(Figure 1)

a moment of inertia of I= 6250 kgâ‹…m2 for rotation about axis perpendicular to the its center. The carousel is rotating unpowered and without friction with an angular velocity of 1.25 rad/s. An 85.0 kg man runs with a velocity of v=8.00m/s , on a line tangent to the rim of the carousel, overtaking it. The man runs onto the carousel and grabs hold of a pole on the rim. (Figure 1)

Part A Part complete What is the angular velocity of the carousel and the person combined? …….rad/s Part B Part complete What is the change in total kinetic energy in the process? Kf-Ki =….. (note: part a I got 1.4 is correct, but part b someone said 847 J and another said 688.77 J. hope you help me to get correct answer. Thanks

Answer 1

here,

the moment of inertia of carousel , I = 6250 kg.m^2

rqadius , R = 3 m

initial angular velocity , w0 = 1.25 rad/s

mass of Man , m = 85 kg

v = 8 m/s

a)

let the final angular speed be w

using conservation of angular momentum

Li = Lf

(I * w0 + m * v * R) = (I + m * R^2) * w

( 6250 * 1.25 + 85 * 8 * 3) = ( 6250 + 85 * 3^2) * w

solving for w

w = 1.4 rad/s

the final angular speed is 1.4 rad/s

b)

the change in kinetic energy of the system , dKe = KEf - KEi

dKe = 0.5 * (I + m * R^2) * w^2 - ((0.5 * I * w0^2 + 0.5 * m * v^2 ) )

dKE = 0.5 * ( ( 6250 + 85 * 3^2) * 1.4^2 ) - (0.5 * 6250 * 1.25^2 + 0.5 * 85 * 8^2)

dKE = - 728 J