Relative Acceleration of a Bolt on a Train

[optoracko]

Homework Statement

A bolt drops from the ceiling of a traincar which is acceleration northward at a rate of 2.5 m/s². What is the acceleration of the bolt with respect a stationary train station.

Homework Equations

Seems more qualitative to me, but d = V₁t + (1/2)at² could be relevant.

The Attempt at a Solution

This question is just puzzling me. I'm assuming that if you were in the train station and saw the train, it'd be passing left to right. Or should I be thinking that the train has already passed the train station?

This is a two part question, the first asking what the acceleration of the bolt is wrt the train car. That part is simple, I just get the resultant acceleration, by solving the vectors with Pythagorean theorem.

I'm thinking that the acceleration from the stationary train station is the same. It's based on inductive reasoning, because if someone in a bus had an object in their hand and the bus passed stationary pedestrian (and they saw it the bus pass coming from the left of their vision, to the right) it'd be whatever velocity the bus was traveling at.

So for the stationary train station, I'd think that the magnitudes would be the same, but the direction would be...different?

 

[shallgren]

My thinking might be wrong here, but your solution for the bolt wrt the train car seems like what it would be for the train station.

If you are riding in a train car, and you drop a bolt, it will fall straight down with an acceleration of g.

If you were at the train station, it would fall down and also appear to go forwards (north), requiring a resultant vector to determine the acceleration.

 

[optoracko]

That sounds right to me as well...still relative motion has to take into account both things' motion, and the bolt would be in free fall after the initial fall. I'm clueless.

 

[PhanthomJay]

Hint: When the bolt is falling, in the air, what forces act on it in the horizontal direction ? What would its acceleration then be in the horizonatl direction with respect to the stationary station?? Velocity and acceleration are 2 different things.

 

[optoracko]

There would be no forces acting on it, except maybe air resistance.

But still, that would mean that wrt to the train station, it is 9.8 down?
How does velocity and acceleration differ when talking about relative motion?

 

[ideasrule]

But still, that would mean that wrt to the train station, it is 9.8 down?

Yes. Gravity is the only force on it, so its acceleration must be 9.8 m/s^2 down.

How does velocity and acceleration differ when talking about relative motion?

They don't, in the sense that both velocities and accelerations can be changed from frame to frame by adding/subtracting the velocity/acceleration of the frame itself. PhantomJay meant that the bolt could have a non-zero velocity in the horizontal direction while still having a zero velocity.

 

[PhanthomJay]

A person standing at the station platform sees the bolt moving forward from left to right at a constant speed and downwardly accelerating at 9.8m/s/s (parabolic projectile motion curve). It's velocity in the x direction is non-zero, but it's acceleration in the x direction is zero. Now what's the acceleration of the bolt with respect to a person in the train?

 

[optoracko]

A person standing at the station platform sees the bolt moving forward from left to right at a constant speed and downwardly accelerating at 9.8m/s/s (parabolic projectile motion curve). It's velocity in the x direction is non-zero, but it's acceleration in the x direction is zero. Now what's the acceleration of the bolt with respect to a person in the train?

It would be the resultant vectors of the acceleration sideways and up and down? The x vector would be 2.5 m/s^2 [Back] and 9.8 m/s^2 [down]. The resultant would be 10.1 m/s^2.

 

[PhanthomJay]

It would be the resultant vectors of the acceleration sideways and up and down? The x vector would be 2.5 m/s^2 [Back] and 9.8 m/s^2 [down]. The resultant would be 10.1 m/s^2.

Yes, that would be the magnitude. It is sometimes easier and more clear to leave the vectors in their x and y component form.

 

[optoracko]

If this because of the use of the phrase with respect to? Because the bolt in the first situation is relative to the train, the acceleration of the train must be taken in consideration, but with a bystander at a train station, since the train has no acceleration, it does not need to be considered? Would the same apply if the word was relative?

 

[PhanthomJay]

If this because of the use of the phrase with respect to? Because the bolt in the first situation is relative to the train, the acceleration of the train must be taken in consideration, but with a bystander at a train station, since the train has no acceleration, it does not need to be considered? Would the same apply if the word was relative?

The terms relative' and 'with respect to' are synonymous in this context. But you are confusing terms. A bystander at the train station sees the train pass by accelerating at 2.5m/s/s with respect to him, and sees the bolt accelerating in the horizontal direction at 0 (constant speed) with respect to him. Is this clear??

 

[optoracko]

The terms relative' and 'with respect to' are synonymous in this context. But you are confusing terms. A bystander at the train station sees the train pass by accelerating at 2.5m/s/s with respect to him, and sees the bolt accelerating in the horizontal direction at 0 (constant speed) with respect to him. Is this clear??

But since the bolt is on the train, then why are the magnitudes for both answers not the same?

 

[PhanthomJay]

But since the bolt is on the train, then why are the magnitudes for both answers not the same?

The train is accelerating with respect to the rails or an observer at the station at 2.5 m/s/s (the engine provides the force through the force of the rails on its wheels). A person sittting on the train is accelerating at 2.5m/s/s with respect to the rails (the floor and seat provide the friction force necessary to accelerate him at the same rate as the train). The bolt, while still fastened to the ceiling, is accelerating at 2.5 m/s/s with respect to the rails (the ceiling to which the bolt is fastened provides the force). Now with respect to each other, they all have no acceleration (see ideasrules post).

Now the bolt breaks loose and falls. There is now no more horizontal force on it, and thus, its acceleration with respect to the rails must be zero, per Newton 1 or 2 . But the train and person on the train are still accelerating at 2.5 m/s/s with respect to the rails, and thus, the acc of the bolt relative to the person on the train is 0 -2.5 = 2.5 m/s/s to the left. And of course, to repeat, the acc of the bolt relative to a person on the station platform is zero.