Question

CAN anyone figure this out?!

Answer 1

A ball with uniform mass density rolls without slipping on a turntable. Show that
the ball moves in a circle (as viewed from the inertial lab frame), with a frequency
equal to 2/7 times the frequency of the turntable.

Answer 2

this is what i keep getting..................

Answer 3

https://www.physics.harvard.edu/uploads/files/undergrad/probweek/prob21.pdf
:)

Answer 4

ya its an assignment i have to answer the questions of the week

Answer 5

@BioEpic

Answer 6

Since the ball rolls without slipping, we can consider the
point q as an instantaneous axis of rotation of the ball.
Starting from rest with zero angular speed to a final constant angular speed
ωtb, the turntable acquires the angular acceleration α = Ωtb/Δt where Δt is the
elapsed time. This gives rise to the tangential acceleration At = α*Ro = Ωtb*Ro/Δt
at a point a distnace Ro from the center of the turntable where its axis of rotation
is. Assuming that the ball on the turntable is located at Ro when it begins to roll
without slipping, then the fictitious force
(1) ..... F = ΔP/Δt = M*Vcm/Δt = M*At = M* Ωtb*Ro/Δt
causes the center of mass (CM) of the ball to move with velocity Vcm = Ωtb*Ro
The force in (1) also produces the torque N about the instantaneous axis of
rotation q that causes the ball with radius Rb to have a change in angular
momentum:
.......... N = Rb*F = Rb*M*Ωtb*Ro/Δt = Iq*ωb/Δt
(2) ..... Rb*M*Ωtb*Ro = Iq*ωb
where Iq is the moment of inertia of the ball about its instantaneous axis at q
and ωb is the angular velocity of the ball about q.
Using the parallel axis theorem, we find that
(3) ..... Iq = Icm + M*Rb^2 = (2/5)*M*Rb^2 + M*Rb^2 = (7/5)*M*Rb^2
under the assumption that the ball is a solid sphere.
Substituting (3) in (2), we get
.......... (Rb*M)*Ωtb*Ro = (7/5)*M*(Rb^2)*ωb = (7/5)*(M*Rb)*(Rb*ωb)
.......... Ωtb*Ro = (7/5)*(Rb*ωb) = (7/5)*Vcm
(4) ..... Vcm = (5/7)* Ωtb*Ro
Denote by Tb the period of rotation of the ball around the axis of rotation of the
turntable. Then
.......... Vcm*Tb = (5/7)* Ωtb*Ro*Tb = 2*π*Ro
.......... (5/7)* Ωtb*Tb = 2*π
(5) ..... 2*π/Tb = (5/7)* Ωtb
But that’s a (5/7) and not a (2/7)!

Answer 7

Let us denote by q the point of contact between the ball and the surface
of the turntable. Since the ball rolls without slipping, we can consider the
point q as an instantaneous axis of rotation of the ball.
Starting from rest with zero angular speed to a final constant angular speed
ωtb, the turntable acquires the angular acceleration α = Ωtb/Δt where Δt is the
elapsed time. This gives rise to the tangential acceleration At = α*Ro = Ωtb*Ro/Δt
at a point a distnace Ro from the center of the turntable where its axis of rotation
is. Assuming that the ball on the turntable is located at Ro when it begins to roll
without slipping, then the fictitious force
(1) ..... F = ΔP/Δt = M*Vcm/Δt = M*At = M* Ωtb*Ro/Δt
causes the center of mass (CM) of the ball to move with velocity Vcm = Ωtb*Ro
The force in (1) also produces the torque N about the instantaneous axis of
rotation q that causes the ball with radius Rb to have a change in angular
momentum:
.......... N = Rb*F = Rb*M*Ωtb*Ro/Δt = Iq*ωb/Δt
(2) ..... Rb*M*Ωtb*Ro = Iq*ωb
where Iq is the moment of inertia of the ball about its instantaneous axis at q
and ωb is the angular velocity of the ball about q.
Using the parallel axis theorem, we find that
(3) ..... Iq = Icm + M*Rb^2 = (2/5)*M*Rb^2 + M*Rb^2 = (7/5)*M*Rb^2
under the assumption that the ball is a solid sphere.
Substituting (3) in (2), we get
.......... (Rb*M)*Ωtb*Ro = (7/5)*M*(Rb^2)*ωb = (7/5)*(M*Rb)*(Rb*ωb)
.......... Ωtb*Ro = (7/5)*(Rb*ωb) = (7/5)*Vcm
(4) ..... Vcm = (5/7)* Ωtb*Ro
Denote by Tb the period of rotation of the ball around the axis of rotation of the
turntable. Then
.......... Vcm*Tb = (5/7)* Ωtb*Ro*Tb = 2*π*Ro
.......... (5/7)* Ωtb*Tb = 2*π
(5) ..... 2*π/Tb = (5/7)* Ωtb
But that’s a (5/7) and not a (2/7)!
Are you sure it should be (2/7)?

Answer 8

Are you the person who asked it on yahoo answers?

Answer 9

yup

Answer 10

no im the one who replied

Answer 11

http://www.personal.kent.edu/~fwilliam/Chapter%208%20Angular%20Momentum,%20Part%20II%20(General%20L).pdf

Question number 28

Answer 12

i replied to it on yahoo and later came across it while doing the harvard questions

Answer 13

thx @BioEpic

Answer 14

No problem.