An interesting physics question!! but it is like very hard..

A student of mass 36.9 kg wants to measure the mass of a playground merry-go-round, which consists of a solid metal disk of radius R = 1.50 m that is mounted in a horizontal position on a low-friction axle. She tries an experiment: She runs with speed v = 5.99 m/s toward the outer rim of the merry-go-round and jumps on to the outer rim, as shown in the figure. The merry-go-round is initially at rest before the student jumps on and rotates at 1.30 rad/s immediately after she jumps on. You may assume that the student’s mass is concentrated at a point.

a) What is the mass of the merry-go-round?
b) If it takes 46.9 s for the merry-go-round to come to a stop after the student has jumped on, what is the absolute value of the average torque due to friction in the axle?
c) How many times does the merry-go-round rotate before it stops, assuming that the torque due to friction is constant?

http://i1104.hizliresim.com/2011/4/22/12550.png

(a) Mom. of Inertia for disc roundabout .. Ir = ½MR² .. (M = roundabout mass)
For girl on the roundabout .. Ig = mR² .. (m = girl's mass)
Total Mom. of Inertia .. It = R²(M/2 + m)
Total angular mom. (It.ω) gained from girl's angular momentum as she lands on roundabout (mvR)

mvR = R²(M/2 + m).ω
M = 2([mv/Rω] - m) = 2([36.9*5.99 / 1.50*1.30] - 36.9) .. .. ►M = 153 kg

(b) Rotational distance to stop given by .. θ(rad) = (Av. rot. vel. in rad/s) * time (s)
θ = 1.30/2 * 46.9 = 30.49 rad
Frictional torque = T (Nm)
Work done against friction = Tθ = KE lost = ½ (It)ω²
Tθ = ½ (It)ω²
T = ½ ([R²[M/2 + m])(1.30)² = ..... 215.6 / 30.49 .. .. ►T = 7.07 Nm


(c) No. of revs = θ/2π = 30.49 / 2π ..... ►= 4.85 revs.

I'm pleased my answer was so helpful to you.
Thanks for your comments - regards

RB

Report Abuse