The figure below shows an accelerometer, a device for measuring the horizontal acceleration of airplanes and cars. A ball is free to roll on a parabolic track described by the equation y=x^2 , where both x and are in meters. A scale along the bottom is used to measure the ball's horizontal position x. Please Register or Log in to view the hidden image! (i) Find an expression that allows you to use a measured position x (in m) to compute the acceleration a_sub x (in m/s^2). (For example, a_sub x=3x is a possible expression.) (ii) What is the acceleration if x=20cm? Now I am totally puzzled. y=x^2 is not the position function because it's 2D. How can I find the acceleration (in the x direction) in terms of x? Is it true that tan theta= y/x = x^2/x= x? Can someone please help me? I am so confused... I have never done a CURVED incline plane problem before...Please Register or Log in to view the hidden image!
You'll probably need to know that the slope of the curve at any point is 2x. Knowing that bit is easy if you've done any calculus. I don't know how you'd do it without that bit of info. Use that to determine the components of the normal force, equate the vertical component to gravity, and solve for the horizontal component to give acceleration.
But y=x^2 describes the shape of the "path" followed, not the position function x(t) and/or y(t). I know that v(t)=x'(t) and a(t)=v'(t). But the trouble here is I am not given the x(t) function in which I can find the derivatives.
I have done a bit of calculus and I know how to find the derivative of y=x^2. But how does finding the derivative of y=x^2 help me? And how can I use it to find the components of the normal force? One more thing that puzzles me: the question says y=x^2, but how come the graph shown in the figure is vertically shifted upward from the origin? And also, how can I find the acceleration in the vertical direction? (just wondering...) Thanks for your help
For vertical acceleration, I think you'd need a different sort of apparatus. The vertical origin isn't specified in the graph, but if y = x² then y=0 must be at the bottom of the curve, not the bottom of the graph. The slope of the curve is equal to tan(theta), where theta is the angle above the horizontal. Don't worry about how the ball's position changes with respect to time. For this device to work, you have to have constant acceleration, wait for the ball to stabilize, and then take a reading.
"Use that to determine the components of the normal force, equate the vertical component to gravity" <----how can I equate these if there is vertical acceleration? (i.e. Fnet.vertical not =0) "The slope of the curve is equal to tan(theta), where theta is the angle above the horizontal."<---why? and how can this fact help me? I still don't get how to calculate (i) and (ii). Sorry if this seems obvious to you, but this problem is a non-typical, strange problem to me. I haven't done anything similar in my physcis course.
You can't. The problem requires that you assume vertical acceleration is zero. You can see why by drawing a diagram. Draw a triangle with a horizontal side, a vertical side, and the third side having a gradient of (say) 3. Mark on the diagram the angle of the third side above the horizontal. What in the tan of that angle? It helps you because you need theta to break the normal force into components. Vertical component = F sin(theta), horizontal = F cos(theta), right? Don't worry, it's not obvious. It's actually a pretty cool problem... that's why I don't want to spell it out too much. That would detract from your satisfaction at solving it on your own Please Register or Log in to view the hidden image!.
you can do fancy math derivations or you can equalize the kinetic energy with potential energy i guess.
1) It's not stated anywhere in the question that the vertical acceleration is zero. How do you know that and how can I possibly know that? 2) tan theta=y/x = x^2/x= x <------tan theta =rise/run=y/x, why is this wrong?
We can't know it. The vertical acceleration might in fact be non-zero. If it is, then this device will give us the incorrect result, unless we know what the vertical acceleration is and correct for it. The device as shown can't measure vertical acceleration (the problem states that it's for measuring horizontal acceleration), so we have to know vertical acceleration through some other means. Since it's a textbook exercise, the natural thing to do is to assume it is zero. I know... you could leave g in the equation as a variable. The actual value of g to be plugged in will be the local acceleration due to gravity less any actual vertical acceleration. In practice, you could measure g directly, say with a spring balance with a known mass attached. It is good that you've picked up on this difficulty. You're thinking like a real engineer or experimental physicist. That gives you the gradient of a line from the origin to (x,y). You actually want the gradient of the tangent to the curve at (x,y). The tangent line clearly won't pass through the origin.
1) How does the accelerometer work? Is the balling rolling up and down? How can the vertical net force be zero if it is rolling down due to gravity?
Hi kingwinner (are you a chess player?), Imagine the device accelerates to the right at 5 m/s². When it begins accelerating, the ball will roll up to the left. If it keeps accelerating steadily, the ball will settle at some point on the curve and stay there. At that point: The ball has a net horizontal force acting on it to make it accelerate at 5m/s². This is the horizontal component of the normal force of the track on the ball. The ball has a zero net vertical force acting on it. This means that the vertical component of the normal force of the track on the ball must balance the gravitational force on the ball. So, now we know that for this particular case, the vertical component of the normal force must be twice the horizontal component of the normal force, right? (because g = twice 5m/s²Please Register or Log in to view the hidden image!... so what does that tell us about the gradient of the track at that point? Pete
No, I am not a chess player. Your explanation is very good, but I wonder where you get all this detail? I am reading the same question as you, but I can't figure out where in the question it says the whole thing is accelerating horizontally and the ball is remaining stationary... seeing such a curvature on the diagram, I am interpreting the ball as rolling up and down along the curve periodically...
I don't know... it just seemed to be the only way that makes sense. Maybe it's justs experience with textbook type questions... they're always much simpler than real-lifef situations, so you get used to making simplifying assumptions.
Kingwinner, Pete is right in his assessment. Part of being good with these types of simple Newtonian problems is to get a good intuition about them. Not about the numbers necessarily, but about motion itself. Like Pete, as soon as I saw your diagram and said it was being accelerated in the X axis I knew the motion would leave the ball at some height on the curve. This makes sense if you consider the device inside a closed box. In this case the acceleration of the box is indistinquishable from a force in the X axis. If you then look at the 2 force vectors on the ball, the horizontal one from the acceleration and the vertical one from gravity you get a resulting force vector that will be somewhere between pointing down (no acceleration all gravity) and pointing completely left and right (note that this is a limit, it can never ACTUALLY be completely vertical due to the gravity component). At some point on the curve, this resulting force vector will be normal to the curve. THIS is the point where the ball will sit. Granted none of this solves your problem, but it gives you an intuitive sense of what the motion is so that you can then start forming the equations needed to solve for the acceleration. Also note that you may want to consider the diagram differently since this is obviously a simplified model. Consider it instead as a TUBE whose center line is the equation y = x^2. This way you don't even THINK about any motion away from the wall of the ramp. -AntonK EDIT: Changed 'tangent' to 'normal' as Pete pointed out my typo
You are EXACTLY correct. And it was indeed a typo, in my head I said normal. Thanks for the correct. If it were tangent to the curve that ball would be accelerating forever Please Register or Log in to view the hidden image! -AntonK
