Bowling Ball rolling on a smooth floor

  • #1
Hi there! I chanced upon this problem whilst trying to brush up my Classical Mech knowledge and found it confusing. Hope someone out there can provide an insight!

Homework Statement


A bowling ball of mass m and radius R sits on the smooth floor of a subway car. If the car has a horizontal acceleration ##\mathbb{a_1}## , what is the acceleration ##\mathbb{a_2}## of the ball? Assume that the ball rolls without slipping. The gravitational acceleration is g.

Homework Equations



c is the center of mass ( just a reference point)
Rolling motion: ##a_c## = ##α_cR##
Torque : ##τ_c## = ##I_cα_c##
Newton's 2nd Law: ##F## = ##ma##
Moment of Inertia of a sphere rotating about an axis through its center: ##\frac {2MR^2} {5} ##


The Attempt at a Solution


[/B]
Some predictions I had (may or may not be right):
Smooth floor; therefore there is no static friction between the floor and the ball.

Yet the ball is rolling; therefore there must be a horizontal force F acting on the ball ( I call it a "generated" friction force produced somehow by the acceleration of the subway car floor. Not sure how this is produced, though.) The rolling motion could not have been produced by the vertical forces, for the only vertical forces are weight and the contact force between the ball and the floor. The contact force must always be acting along the same line of action as the center of mass c, through which gravity acts, so there is no possibility of either force producing a torque with respect to the center of mass.

The only possible force, is a horizontal force F that acts at the contact between the floor and the ball.

This force is the one responsible for the acceleration ##a_2## experienced by the ball, so $$F = ma_2 (1)$$
This force is also responsible for the torque about c, so $$τ_c = FR = I_cα_c (2)$$
The ball is rolling without slipping, so $$ a_2 = α_cR (3)$$

And this is where I am stuck, because manipulating equations (1) and (2) and making ##a_2## the subject does not give me equation (3). So one of my equations must be wrong, and yet they seem physically sound and valid to me.

Can anyone out there provide an insight? Really sorry if this seems trivial and the answer is actually staring at me.
 

Answers and Replies

  • #2
334
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ithout slipping, so
a2=αcR(3)​
the mistake is here. You must consider relative acceleration
 
  • #3
Doc Al
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Smooth floor; therefore there is no static friction between the floor and the ball.
As you suspect, this cannot be right. There must be static friction between the ball and the floor.

Hint: Start by viewing things from the accelerating frame of the subway car.
 
  • #4
the mistake is here. You must consider relative acceleration
Am I right to think that it's the relative acceleration that produces the horizontal force necessary for ##a_2##?

As you suspect, this cannot be right. There must be static friction between the ball and the floor.

Hint: Start by viewing things from the accelerating frame of the subway car.
Hmm. I think I get it. My equations (1),(2) and (3) were derived from the reference frame of a stationary. But my equation (3) was odd, because the direction of rolling motion, and the direction of translational motion, were different. In that reference frame I thought up a horizontal force F, acting, in the same direction as ##a_1##, on the ball. This produced the acceleration ##a_2##, acting in the same direction as ##a_1## (Let's just assume to the right). Yet, this force F acted at the contact point of the ball producing a counter-clockwise torque, which meant that the rolling motion was to the left.

The magnitude of ##a_2## is less than the magnitude of ##a_1##, simply because with no relative acceleration between the floor and the ball, there is no force between the floor and the ball, and thus there must be a difference.

In the reference frame of the accelerating subway car though, with ##a_2## less than ##a_1##, the ball is moving to the left with an acceleration (##a_1 - a_2##) , exactly what the rolling motion is all about. Is that what the question meant when it said that the ball rolls without slipping (that it was doing so in the reference frame of the accelerating subway car)?
This would imply $$a_1-a_2 = α_cR$$ which is enough to allow me to solve the problem.

##a_2 = \frac {2a_1} {7}##

Interesting though, does this mean that, from the reference point of a stationary, the ball is moving to the right, whilst rotating ccw?
 
  • #5
334
61
Am I right to think that it's the relative acceleration that produces the horizontal force necessary for a2a_2?
do not put everything in the same pile;
horizontal force is produced by friction; formula we discuss is a pure kinematic thing it has no relation to the forces it just condition of non slipping
 
  • #6
Doc Al
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Is that what the question meant when it said that the ball rolls without slipping (that it was doing so in the reference frame of the accelerating subway car)?
Rolling without slipping means that the point of contact of the ball is stationary with respect to the car. That's true in any frame.

which is enough to allow me to solve the problem.

##a_2 = \frac {2a_1} {7}##
Redo that.

Interesting though, does this mean that, from the reference point of a stationary, the ball is moving to the right, whilst rotating ccw?
Yes.
 
  • #7
Redo that.
Here's what I have done:
I have $$F=ma_2 -(1)$$ $$τ_c =FR = I_cα_c -(2)$$ and $$a_1 - a_2 = α_cR -(3)$$
Substituting (1) into (2) gives me $$α_c = mRa_2/I_c$$
We know that the moment of inertia of a sphere (I assumed that the bowling ball was a uniform sphere) about a line of axis going through its center is ##I_c = \frac{2}{5} mR^2##, which would make ##α_c = \frac {a_2}{ \frac{2}{5} R} = \frac{5}{2R} a_2##

Bringing that into equation (3) gives me: $$a_1 - a_2 = \frac{5}{2}a_2$$, and thus $$a_2 = \frac{2}{7}a_1$$
 
  • #8
haruspex
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Here's what I have done:
I have $$F=ma_2 -(1)$$ $$τ_c =FR = I_cα_c -(2)$$ and $$a_1 - a_2 = α_cR -(3)$$
Substituting (1) into (2) gives me $$α_c = mRa_2/I_c$$
We know that the moment of inertia of a sphere (I assumed that the bowling ball was a uniform sphere) about a line of axis going through its center is ##I_c = \frac{2}{5} mR^2##, which would make ##α_c = \frac {a_2}{ \frac{2}{5} R} = \frac{5}{2R} a_2##

Bringing that into equation (3) gives me: $$a_1 - a_2 = \frac{5}{2}a_2$$, and thus $$a_2 = \frac{2}{7}a_1$$
Looks good to me.
 
  • #9
Great! Thanks everyone for helping!
 
  • #11
Doc Al
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Looks good to me.
Looks good to me too.

Well, let's see if Doc Al still disagrees.
I had made a sign error in my own solution. o0) D'oh!
 

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