Question

Billy (mass = 50 kg) stands on the edge of a merry-go-round, which is a disc with a radius R = 2.0 m and a mass M = 100 kg. Annie begins spinning the merry-go-round (and Billy) by pushing on it with a force of 20 N tangent to its edge for 20 s.

(a) What are (i) the torque Annie imparts on the merry-go-round (with Billy on it) and (ii) the resulting angular acceleration during the 20 second interval? Note, the total momenet of intertia of the system is the sum of the merry-go-round and Billy's moments where Billy can be treated like a point object.

(b) What is the final angular speed of the merry-go-round after it accelerates over this 20-second interval?

(c) As the merry-go-round continues spinning at this angular speed, what is the minimum value of the coefficient of static friction (causing Billy's centripetal acceleration) between Billy's shoes and the merry-go-round so tat he doesnt slide off the merry-go-round as it spins? Again, Billy is standing at the edge.

Answer 1

Moment of inertia of billy = mr² = 50*2² = 200 Kg.m²

Moment of inertia of merry go round = 1/2mr² = 1/2*100*2² = 200 Kg.m²

Now, torque = force * moment arm

torque = 20*2 = 40 N.m

Also

torque = moment of inertia * angular acceleration

angular acceleration = torque / moment of inertia

angular acceleration = 40 N.m / 400 kg.m²

angular acceleration = 0.1 rad/sec²

Let the initial angular speed be zero

then we can use rotational kinematics

= _o + t

=  t

= 2 rad/sec

To find coefficient of friction, we can equate

Force of friction = centripetal force

umg = mv²/r

ug = v²/r

here v = r = 2*2 = 4 m/s

u*9.8 = 4² / 2

u = 0.816