Angular Velocity
Just above a merry-go-round, a 35-kg child rests on a low hanging tree limb. The merry-go-round is a solid disk of radius 2.0 m and a mass of 160 kg. It is spinning at 3.0 rpms. The child gently drops onto the merry-go-round at a point 1.0 m from the center.
Spinning together, what is the new angular velocity?
(okay ... Knightmare talked me into it ... maybe one good puzzle a day ... and a genuine thanks to everyone who expressed appreciation for my puzzle posting efforts)
Spinning together, what is the new angular velocity?
(okay ... Knightmare talked me into it ... maybe one good puzzle a day ... and a genuine thanks to everyone who expressed appreciation for my puzzle posting efforts)
posted by Zaux at
9:42 AM
10 Comments:
Chris ... thanks for your nice comment ... I appreciate it.
2.844... rpm
... and you're welcome.
Angular Velocity
Angular momentum , H, is momentum about a radius
w is angular velocity
m is point mass at radius r
H=m*v*r=m*w*r*r=m*w*r^2
So to find moment um of a solid disc
p=density
imagine small strips of width dr about centre
H=integral 0 to r [m_strip*r^2 *w*dr]
H=integral o to r [(p*2*Pi*r)*r^2*w*dr]
H=1/2*p*Pi*w*r^4
but mass of disk, md = Pi*r^2*p
H=1/2*md*w*r^2
Conservation of momentum
Hdisk1=Hdisk2+Hchild
1/2*md*r^2*w1=1/2*md*r^2*w2+mcc*rc^2*w2
w1/w2=(1/2*md*r^2+mc*rc^2)/( 1/2*md*r^2)
w1/w2=1+ (mc*rc^2)/( 1/2*md*r^2)
w1/w2=1+(35*1^2)/(0.5*160*2^2)
w1/w2=1.109375
w2=w1/1.109375=3/1.109375=2.704225 rpm
Answer:
The new angular velocity is 2.704225 rpm
Cam
Zaux,
Can't thank you enough for all the puzzle posts. You've been awesome.
Cam
Hmmm... my answer is a bit different from Chris's. I'm scanning for mistakes now.....
Cam
Hi Cam ... thanks for the nice words ... I really appreciate it. I'm not nearly the mathematician you and Chris are ... but, I just love puzzles, games, on-line mmorpgs, and competition of any nature.
The published answer is 2.7 rpms, and knowing your skills, I suspect 2.704225 is dead on.
Didn't know that you did physics ones Cam.
I simply remembered (but verified) that I = ½MR² for a disc.
For the kid (as radius = 1) moment of inertia = mr²
J = angular momentum = Iω
J = ½MR²ω = (½MR² + mr²)w, where w is final angular speed
M = 160, R = 2, m =35 r = 1 => ½MR² = 320 and mr² = 35.
final rpm = 3*320/(320+35) = 2.7042.. (same as Cam)
How embarrassing. I did the right physics when I posted, but must
have miskeyed in the calculator.
I goofed 2²/2 so did 3*640/(640+35).
Cam. I'm sorry that I caused you extra work.
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