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Suppose there is a car and a ball inside it can roll. Now the car accelerates with acceleration a.

Now what would be the the equations of motion for the ball in the car from both the frames of the car and ground ( both from inertial and non inertial).

I am having difficulty in analyzing the situation in both the frames because I think that when the car starts acceleration the ball will move back but then because of friction the ball will move ahead. In doing so it will accelerate forward .

Please see if how I have approached this simplified version is correct or not. if not please correct.

Please imagine a moving ball or box ( so that we can ignore rolling) inside an accelerating car with the same acceleration as the car , a.

From the ground $$F=uN=ma$$ ( since it has acceleration a and I am supposing it doesn't roll to simplify and understand my doubt).

From the car frame It is at rest in car frame. So the equation must to be $$F=uN-ma=0$$

( 0 because it has 0 acceleration in car frame and ma is pseudo force).

Is that right?

2) If the ball, had an acceleration A with respect to the car Then considering the car as the reference frame $$F=uN-ma=mA$$

And from the ground frame $$F=uN=m(a+A)$$

Are the above equations correct or is there a fault ?

I have tried to solve the question and it would be beneficial for me if I can know whether I am going on the right track.

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  • $\begingroup$ @sammygerbil Have edited the question with an attempt . Please help $\endgroup$ – E2n Mar 13 '17 at 17:44
  • $\begingroup$ Friction is $\mu_s N$ if the ball is on the point of slipping (sliding), and $\mu_k N$ if it is slipping. If the ball is rolling then $f\le \mu_s N$. $\endgroup$ – sammy gerbil Mar 13 '17 at 17:52
  • $\begingroup$ Remember that friction always opposes motion. You have to be moving backwards to have forward friction. Friction will never make the ball move forward because that would add net energy to the ball. $\endgroup$ – ja72 Mar 13 '17 at 19:41
  • $\begingroup$ @sammygerbil Please imagine a moving ball or box ( so that we can ignore rolling) inside ac accelerating car with the same acceleration as the car , a. Now from the ground uN=ma is right ( since it has acceleration a and I am supposinf it doesn't roll to simplify and understand my doubt). Now from the car frame it is at rest is car frame. So will the equation be uN-ma=0 ( 0 because it has 0 acceleration in car frame and ma is pseudo force). Is that right. 2) if the ball had an acceleration A in the car frame too then in car frame -uN-ma+mA. And from the ground frame uN=m(a+A). Where am i wrong $\endgroup$ – E2n Mar 14 '17 at 7:46
  • $\begingroup$ @ja72 Could you please just tell whether what i am saying in the following comment is right or not $\endgroup$ – E2n Mar 17 '17 at 10:14
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The friction only cares about the surfaces in contact; in the case of a sphere, it is just a point. It does not care if the sphere rotates or translates; all it wants is to avoid any relative motion at the point of contact.

With the above in mind, the question can be solved easily.


Frame of reference: car

enter image description here

$a$ or $a_{car}$ is the acceleration of the car with respect to the ground

$a_{sphere}$ is the acceleration of the sphere with respect to the car

$m$ is the mass of the sphere

$f$ is the force of friction

The frictional force will be due to the static friction if the pseudo-force is less than $\mu_sN$. If the pseudo-force is larger than the maximum static friction, then the sphere will begin to slide. The static friction adjusts itself so that the surfaces in contact tend to remain at relative rest.

This reduces to an elementary rotational motion problem where you have a sphere upon which a constant external force is applied.

The static friction tries its best to ensure that the surfaces in contact are at relative rest (condition for pure rolling). For that, the acceleration of the sphere must be such that the following condition is satisfied:

$$a_{sphere} = \alpha R$$

The only source of torque is friction.

$$\tau = fR$$

The net force on the sphere can be written as:

$$ma_{sphere} = f - ma_{car}$$

You have three equations and three unknowns.


Frame of reference: ground

The floor is accelerating; therefore, to ensure that the sphere rolls, the lowermost point of the sphere must accelerate such that the surfaces in contact are at relative rest.

enter image description here

The frictional force will cause a torque and will cause the sphere to roll as well as move translationally. Just like in the previous case; well, the answers must be the same in whichever frame of reference you solve from.

The friction will try its best to ensure that there is no slipping. For that, the following condition must be satisfied:

$$a_{sphere} = \alpha R$$

The net acceleration of the sphere can be written as the sum of the acceleration of the car and the acceleration of the sphere with respect to the car.

$$a_{net-sphere} = a_{car} + a_{sphere}$$

The only force in this frame of reference is the friction. Therefore, the net force equals friction.

$$m\left(a_{sphere} + a_{car}\right) = f$$

The only source of torque is friction.

$$\tau = fR$$

You have the same three equations with the same three unknowns (good that they agree).

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  • $\begingroup$ Thanks for the answer.I understand your point . But please have a look at the question once again . I said that ball accelerates ( but i said there is no rolling acceleration ). I was at fault if there is friction there has to be rolling. Sorry for that . But please consider all the situation for a box which accelerating with the box at a. Obviously friction causes acceleration. Consider also the case when the box has an acceleration A with r.pt to the car and somehow friction only causes this acceleration. Then are my equations correct ? $\endgroup$ – E2n Mar 27 '17 at 16:51
  • $\begingroup$ I just want to see whether i have used the equation using pseudo force correctly. Have I applied the equations correctly just for the translation of the box . Both when it has 0 acceleration and acceleration A with respect to car , both form car frame and ground ? Whether the quations in this scenario are correct or not ? $\endgroup$ – E2n Mar 27 '17 at 16:54
  • $\begingroup$ I added the equations. $\endgroup$ – Yashas Mar 27 '17 at 17:14

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