What is the acceleration of the bowling ball in a subway car?

[StephenPrivitera]

A bowling ball sits on a level floor of a subway car. If the car has a horizontal acceleartion a, what is the acceleration of the ball wrt the ground? Ball rolls w/o slipping.
The forces that act on the ball are its weight, a normal force, and static friction. The weight and normal offset, so Friction = MA.
But if we choose the point of contact with the ball as an axis, the net torque is zero?! So since torque is zero there is no angular acceleration and thus A=0?
Something's wrong here.

 

[Norman]

So it is good that you have an intuition about your answer being zero. Why not think of the reference frame as that of the center of mass of the bowling ball. Then you will have a torque acting at the radius of the bowling ball which will be prependicular to the "lever arm." So basically don't use the point of contact as your axis of rotation- it doesn't make sense to because the ball is not going to spin about that point- it will spin about it's center of mass. Hope this helps.
Cheers,
Norm

 

[StephenPrivitera]

But if it spins about an axis, it must spin about any parallel axis. That's what's throwing me off. The angular acceleration should be the same about any parallel axis. If I use the CM as a axis, then I get F=ma, FR=Ia/R, eliminating F, I get maR=Ia/R, but then a drops out!

 

[Doc Al]

Originally posted by StephenPrivitera
If I use the CM as a axis, then I get F=ma, FR=Ia/R, eliminating F, I get maR=Ia/R, but then a drops out!

The accelerations are different and don't drop out:

[tex]F=ma_{cm}[/tex]

[tex]FR=I\frac{a_{car}}{R}[/tex]

 

[harsh]

There is one thing you forgot. Since the frame is accelerating, the only time that Torque = (Moment of Inertia) (alpha) is when the axis is through its center of mass. The parallel-axis theorem will not be valid in an accelerating frame of reference.

- Harsh

 

[StephenPrivitera]

Fr=I(a_1-a_2)/r
F=ma_2
(ma_2)r^2=I(a_1-a_2)
a_2(m+I)r^2=Ia_1
a_2=Ia_1/(m+I)
I believe this to be the correct solution, and this is in the rest frame of the ground.
I'm not quite sure what to put in for I, though.

 

[Doc Al]

Originally posted by Doc Al
The accelerations are different and don't drop out:

[tex]F=ma_{cm}[/tex]

[tex]FR=I\frac{a_{car}}{R}[/tex]

I think I messed up. I believe the following is true, as the condition for rolling without slipping:
[tex]a_{cm} + \alpha R = a_{car}[/tex]

The force equations should be:
[tex]F=ma_{cm}[/tex]

[tex]FR=I \alpha=I\frac{a_{car}-a_{cm}}{R}[/tex]

With [tex]I=\frac{2}{5}mR^2[/tex]