Projectile motion

[kingwinner]

1) A table is inclined to the horizontal. An air puck is shot from an edge of the table and follows a projectile motion on the table. The horizontal acceleration is 0m/s^2 and the vertical acceleration is 0.6867m/s^2[down]. Calculate the angle at which the table was inclined to the horizontal.

 

Can someone teach me how to calculate this? Thank you! :)

[cwes99_03]

155 posts and all of them questions from your homework.

 

That being said the questions you have posted often seem lackluster at best, so might I suggest finding a new textbook.

Take this question for example. A table implies a smooth flat surface. How does a smooth flat surface imply projectile motion? The answer: the table is at an incline and the puck is being shot at an angle across the surface of the table at an upward angle. To me this implies a 3-d problem, not a 2-d problem as the question suggests with vertical and horizontal accelerations.

 

What I believe they may mean is that the vertical and horizontal with respects to the surface of the table, not with respect to the unit vector of gravity.

 

Since horizontal (i.e. across the table) has 0 m/s^2 that means that the table is not inclined in two directions (such as being tipped up on a corner of the table) but just one. Since the pull down the table (at an angle with respects to the gravitational vector) is 0.6867 m/s^2 then it becomes just a simple sliding the puck up and down an inclined plane without any projectile motion (at least as far as the equations go). 0.6867/9.81x = % of gravitational force acting on the puck which is proportional to the tangent of the incline.

[InfiniteNow]

Quick question. Do you actually have a teacher, or has hypography become your sort of free online degree in physics and math?

155 posts and all of them questions from your homework.

Well, duh, of course there's a teacher... Who do you think assigns the homework? :)

[TheBigDog]

Well, duh, of course there's a teacher... Who do you think assigns the homework? :)

It could be that kingwinner is the teacher, but woefully unqualified and trying desperately to stay one step ahead of the curriculum.

 

Bill

[InfiniteNow]

It could be that kingwinner is the teacher, but woefully unqualified and trying desperately to stay one step ahead of the curriculum.

 

Bill

Good point TBD, but I'm not too sure that's much better... :)

[cwes99_03]

Well, I can't remember but I was thinking that Kingwinner came from the land down under and down there there are quite a few regions where one has to teach themselves, or take classes litterally over the FM radio. So I figured if someone one of those kids had gotten ahold of the internet, maybe this was an effort to be self-taught. If so, I'm glad to help out as I did, but considering the staggering amount of time in which he seems to be letting others do his work for him rather than listening to a teacher or maybe giving a little bit more effort, I would not want to go about making life that much easier for him.

[CraigD]

1) A table is inclined to the horizontal. An air puck is shot from an edge of the table and follows a projectile motion on the table. The horizontal acceleration is 0m/s^2 and the vertical acceleration is 0.6867m/s^2[down]. Calculate the angle at which the table was inclined to the horizontal.

 

Can someone teach me how to calculate this? Thank you! :lol:

What is the magnitude and direction of the force exerted by a table on an object sitting on it? Clue – since objects on tables neither fall through them, not fly away from them, its related in, to use a popular phrase, an equal and opposite way to the force of gravity.

 

If a table is level, the force it exerts on an object is exactly opposite that of gravity. If the table is tilted, it only opposes the component of the force of gravity perpendicular to its surface.

A sketch like one attached can help with this kind of problem. Assume any mass for the object you like – it will cancel when you divide force by mass to get acceleration. At what angle must the tabletop be inclined for the component of the force of gravity parallel to it to produce an acceleration of 0.6867 m/s/s?

temp.bmp

[cwes99_03]

Yes, I would agree with that.

 

Hey, Craig, could you comment on my thought about the question itself. I'm just wanting to make sure that I didn't miss anything imbedded in the meaning of the question.

 

A table implies a smooth flat surface. How does a smooth flat surface imply projectile motion? The answer: the table is at an incline and the puck is being shot at an angle across the surface of the table at an upward angle. To me this implies a 3-d problem, not a 2-d problem as the question suggests with vertical and horizontal accelerations.

 

What I believe they may mean is that the vertical and horizontal with respects to the surface of the table, not with respect to the unit vector of gravity.

 

Since horizontal (i.e. across the table) has 0 m/s^2 that means that the table is not inclined in two directions (such as being tipped up on a corner of the table) but just one. Since the pull down the table (at an angle with respects to the gravitational vector) is 0.6867 m/s^2 then it becomes just a simple sliding the puck up and down an inclined plane without any projectile motion (at least as far as the equations go). 0.6867/9.81x = % of gravitational force acting on the puck which is proportional to the tangent of the incline.

[CraigD]

Hey, Craig, could you comment on my thought about the question itself. I'm just wanting to make sure that I didn't miss anything imbedded in the meaning of the question.

A table implies a smooth flat surface. How does a smooth flat surface imply projectile motion? The answer: the table is at an incline and the puck is being shot at an angle across the surface of the table at an upward angle. To me this implies a 3-d problem, not a 2-d problem as the question suggests with vertical and horizontal accelerations.

 

What I believe they may mean is that the vertical and horizontal with respects to the surface of the table, not with respect to the unit vector of gravity.

 

Since horizontal (i.e. across the table) has 0 m/s^2 that means that the table is not inclined in two directions (such as being tipped up on a corner of the table) but just one. Since the pull down the table (at an angle with respects to the gravitational vector) is 0.6867 m/s^2 then it becomes just a simple sliding the puck up and down an inclined plane without any projectile motion (at least as far as the equations go). 0.6867/9.81x = % of gravitational force acting on the puck which is proportional to the tangent of the incline.

I think whether to call this problem 2-d or 3-d is not too important. As you note, the “table” is really just a disguised inclined plane, which can be represented in 2 dimensions, inclining me (pun inadvertent) to call it 2-d.

 

The formula I get for the angle of the table (I) is

I = InverseTangent( A/((g^2 –A^2)^.5) )

Where A is the observed acceleration (0.6867), and g is the acceleration of gravity (9.81)

 

This isn’t quite the same as saying “% of gravitational force acting on the puck which is proportional to the tangent of the incline” (Tangent(A) = A/g) , since, as the incline approaches Pi/2 (90°), Tangent(I) approaches infinity, not the expected 1.

 

For small angles, such as in this problem, InverseTangent( A/((g^2 –A^2)^.5) ) and InverseTangent(A/g) don’t differ until the 3rd significant digit (about .0701 vs. about .0698 radians), so the simpler formula, or, even simpler, just A/g (.07) works well enough.

[cwes99_03]

Thanks brother, I'll take that as a yes (though you do accurately catch me in a failure to fully calculate the answer a/(sqrt(g^2-a^2)) :shrug:)

 

I couldn't quite figure out why they called the problem a projectile motion problem as projectile motion in simplest form is a 2-d problem, but by calling it projectile motion across an inclined plane, they are turning it into 3-d only to ask for a simple 2-d portion of that whole process. Why do you suppose the writer did something like that?

 

Kingwinner, was there a diagram in the book showing you this problem?

[TheBigDog]

I think the projectile stuff is just there to confuse. With the board at 90 degrees to the horizontal the vertical accelleration of the puck would be 9.8 m/s/s. With the board at 0 degrees to the horizontal the vertical accelleration would be 0 m/s/s. As the board tips further in the horizontal, the vertical accelleration of the puck increases. I think that at 4 degrees the accelleration would be .6853 m/s/s. So the answer is about 4 degrees. I did this by multiplying the tangent of the angle by 9.8 (unrestricted). I am probably wrong, but I am close.

 

Bill

[cwes99_03]

Yep, that was my thought, on the just there to confuse part.

 

So this brings up another question. Should questions like these be in a text book that is supposed to be teaching?

 

I understand the value of asking hard questions, and getting students to sort out the eroneous information, but really, this question seems to be crossing a line. In a real world situation, should I need to calculate this angle, I would take all the measurements I needed. I would not tell myself, "Now this is projectile motion because the puck is moving in an arc across the surface of the table." If I didn't need that information.

 

I recently stumbled upon this site, possibly from another thread on this very site, and I'm going to take a look into it. It is Motion Mountains online physics book. http://www.motionmountain.net They ask only two things for its free use.

 

1) that you offer this book free to anyone else

2) that you make suggestions for improvement after reading each section

 

I applaude places like this. I started my own book about 3 years ago, just after graduating college. Unfortuneately it has sat on a shelf for about 2 of those years. Maybe I shall revisit it, and participate in the motionmountain work, just because of you Kingwinner.

[TheBigDog]

The only reason I could figure they were talking about shooting the puck would be to avoid confusion with static versus moving friction. But with an air puck that is already a moot point.

 

Bill

[Qfwfq]

The inclined plane has been used since Galileo to simulate a reduced gravity.

 

As for 2D or 3D, if the puck moves on a surface and you don't consider rotations about its centre, it's a case of two degrees of freedom. You could say the same for a curved surface too, treating it with differential geometry.

 

A projectile above the ground has three degrees of freedom but, for a central force, it can be reduced to two once the initial velocity is known because angular momentum is a prime integral. Actually, if v0 = 0 it even reduces to one coordinate.

[cwes99_03]

Degrees of freedom is one way of saying that we can rotate our coordinates and reduce the number of dimensions needed. In this case we can't because we don't know the angle of the table's incline (in fact that is what we are trying to find.)

 

In the physics world we define dimension based upon the direction of the normal forces. In this case we have an initial normal force across the board, a normal force created by the board pointing 90 degrees from the surface of the board, and the gravitational vector. If we align our system with the gravitational vector being our z axis (which we must because we don't know the angle of the table), then the initial vector of the puck has an x, y, and z component and the table's normal vector has a y and a z component (if it had an x component that would be because the table was standing on a corner not layed along a flat side).

However the question is only asking about the z component of the problem. In order to figure out the z component, one only need know information about the y and the z. So why even mention anything about the initial velocity of the puck having an x component (this being the initial velocity of the puck across the surface of the table) when all I need know is the acceleration of the puck up and down the table to find the component of that vector that is in the z direction.

[Qfwfq]

Degrees of freedom in classical mechanics means that, if there are n of them, then the configuration space is an n-dimensional differentiable manifold. Now, there are no two ways about that.:steering:

 

I would talk about tangent, not normal forces. The board is a 2-dimensional manifold embedded in the 3-dimensional one, effective g for the problem is therefore the tangent component of the vertical effective g.

[kingwinner]

I figured out how to calculate the angle...is to draw a free body diagram and break down into 2 components, the angle can be found using newton's 2nd law and trig.

 

I got that the inclined angle is about 4 degrees

 

To cwes99_03:

There is no diagram in the text, but by vertical, I guess it means vertical with respect to the inclined table, not the ground...this has confused me before...