Quick question, acceptable way to solve for acceleration?

[phantomcow2]

Hi there, I have a pretty quick and straight forward question here.

Say a ball was dropped from a ramp with a hypotenuse length of 1m. The ball is allowed to roll from exactly 1m down to the ramp. I need to calculate the velocity by the time it just comes off the ramp. The ball was just dropped on, so initial velocity is 0m/s. It takes 2.04 seconds to reach the end.

So my thoughts were to simply use the X1=X0+V0+1/2at^2 equation.
Solve for acceleration, since I already know every other variable. Then, just plug it into the v1=v0+at equation to find final velocity.

The question..

Is it safe to assume that I need to solve for both vectors? Get a final V for both vectors, and use the pythagorean theorem to find the magnitude of velocity?

Thanks very much!

 

[berkeman]

Two problems. Your equation for X1 is missing something from the Vo term, but since Vo = 0, that missing term won't enter in.

The second problem is that you are not including the energy stored in the ball's rotation. Use the moment of inertia and velocity as a function of time to account for it. If they don't give you the mass and diameter of the ball, I guess you could assume it's massless, in which case the rotational energy won't be a factor.

As for the two vector part of your question, yes, you have forces in both the x and y directions, and both contribute to the motion of the ball.

 

[chroot]

Is it safe to assume that I need to solve for both vectors?

I don't see any vectors in your question, so I have no idea what you're asking.

- Warren

 

[Doc Al]

Since the ball only moves parallel to the plane, if you use your method to solve for the acceleration and velocity parallel to the plane you won't have to worry about adding components.

(I also assume you are to ignore the rotational energy of the rolling ball.)

 

[Chi Meson]

Hi there, I have a pretty quick and straight forward question here.

Say a ball was dropped from a ramp with a hypotenuse length of 1m. The ball is allowed to roll from exactly 1m down to the ramp. I need to calculate the velocity by the time it just comes off the ramp. The ball was just dropped on, so initial velocity is 0m/s. It takes 2.04 seconds to reach the end.

So my thoughts were to simply use the X1=Xo+Vot+1/2at^2 equation.
Solve for acceleration, since I already know every other variable. Then, just plug it into the v1=v0+at equation to find final velocity.

so far so good, (although the equation contains an error,I have corrected it above.

Is it safe to assume that I need to solve for both vectors?

OK, what are you talking about when you say "both vectors"?

Get a final V for both vectors, and use the pythagorean theorem to find the magnitude of velocity?

you will find the magnitude of velocity by using the equations you have identified.
Did you mean something else?

 

[phantomcow2]

Sorry for missing information.

I am omiting all stored energy, forces, air pressure, air resistance, all that. We are just finishing up our 2 dimensional motion unit. Forces is the next step, so in a month I will concern myself with that :).

Vectors were just the horizontal and vertical. My ramp is 1m long in the horizontal (actually a tiny bit less because the ramp itself is 1m, not the horizontal dist).

 

[Doc Al]

As mentioned earlier, you need not worry about horizontal and vertical components. The easy way is to analyze the motion (distance, speed, acceleration) parallel to the ramp.

 

[phantomcow2]

I think I understand what you mean by this. BUt, I am not sure how to do it.

So, are you saying that by utilizing the method I have chosen, I do not need to calculate for X and Y vectors at all? I simply solve for A as I have shown?

 

[chroot]

Indeed. Solve for acceleration, and, using that acceleration, solve for velocity at the bottom of the ramp.

- Warren

 

[phantomcow2]

Alright great! Would you mind checking my #'s for me? My answer just seems wrong.

Using the X1= equation...
1=0+0(2.04)+.5a(2.04)^2

a is equal to .481 w/ three significant figures.
Then V1= 0+(.481)(.204)

that means v1= .098m/s That can't be right.

 

[Doc Al]

that means v1= .098m/s That can't be right.

Why can't it? (I assume that the angle that the ramp made with the horizontal was quite small.)

 

[phantomcow2]

Why can't it? (I assume that the angle that the ramp made with the horizontal was quite small.)

Well, the angle was 4.47degrees with the horizontal. Still, it took 2.04 seconds for the ball to go from start to end of ramp.

 

[Doc Al]

If this is just a "textbook" problem, then your numbers are fine. If it's a lab experiment, are the data really accurate to 3 sig figures? (Considering that you are ignoring the fact that the ball rolls, which reduces the linear acceleration.)

 

[phantomcow2]

It is a lab experiment. I know it is not accurate to 3 sig figs, we used meter sticks :). THe most accurate device used was a digital stop watch, accurate to 1/100 of a second.
So you would agree then that the velocity at the end of the ramp is .098m/s?
The angles were calculated using trig though, that's the reason it was more sig figs.