Bowling Ball rolling on a smooth floor

In summary, the bowling ball experiences an acceleration due to the static friction between it and the floor of the subway car.
  • #1
pancakewaffles
4
0
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.
 
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  • #2
pancakewaffles said:
ithout slipping, so
a2=αcR(3)​
the mistake is here. You must consider relative acceleration
 
  • #3
pancakewaffles said:
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
zwierz said:
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##?

Doc Al said:
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
pancakewaffles said:
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
pancakewaffles said:
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.

pancakewaffles said:
which is enough to allow me to solve the problem.

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

pancakewaffles said:
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
Doc Al said:
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
pancakewaffles said:
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!
 
  • #10
pancakewaffles said:
Great! Thanks everyone for helping!
Well, let's see if Doc Al still disagrees.
 
  • #11
haruspex said:
Looks good to me.
Looks good to me too.

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

Related to Bowling Ball rolling on a smooth floor

1. How does friction affect the motion of a bowling ball rolling on a smooth floor?

The presence of friction between the bowling ball and the smooth floor will cause the ball to slow down over time. This is because friction acts in the opposite direction of an object's motion, creating resistance and reducing the ball's velocity.

2. What factors influence the speed of a bowling ball rolling on a smooth floor?

The speed of a bowling ball rolling on a smooth floor is influenced by several factors, including the initial force applied to the ball, the ball's weight and size, the smoothness of the floor, and the presence or absence of friction. The angle and direction of the force applied to the ball can also impact its speed.

3. Can a bowling ball continue to roll on a smooth floor without any external force?

No, a bowling ball cannot continue to roll on a smooth floor without any external force. This is because of the principle of inertia, which states that an object at rest will remain at rest and an object in motion will continue in motion with a constant velocity unless acted upon by an external force.

4. How does the weight of a bowling ball affect its rolling motion on a smooth floor?

The weight of a bowling ball does not significantly affect its rolling motion on a smooth floor. This is because the force of gravity acting on the ball is equal to the force of the floor pushing back on the ball, resulting in a balanced force and allowing the ball to roll smoothly.

5. What is the difference between a bowling ball rolling on a smooth floor and a bowling ball rolling on a rough surface?

The main difference between a bowling ball rolling on a smooth floor and a rough surface is the presence of friction. On a smooth floor, the friction is minimal, allowing the ball to roll farther and with less resistance. On a rough surface, the friction is greater, causing the ball to slow down more quickly and potentially change direction as it encounters bumps or uneven areas.

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