Loop-the-loop problem

• Joshb60796

Homework Statement

A small toy car of mass m is released from rest on an inclined track that starts as high as twice the radius of the loop the loop which is on the floor. How high is it above the floor when it leaves the track, neglect friction.

Homework Equations

The answer is 1.67R but I don't know why.

The Attempt at a Solution

I would've thought that since there is no friction all the potential energy would be converted to kinetic energy at the bottom of the incline, and start of the loop the loop, and the car would go up to a height of 2R, though it does make sense that a circle of radius 2R, flattened out so it was just 2 vertical lines infinitesimally close together, would be taller than 2R I'm not sure what to do with that idea. I know in most loop-the-loop problems you use the radial acceleration equally V-squared over R equation but in this situation I'm puzzled.

It has enough energy to reach a height of 2R, but that doesn't prove it stays on the track to do so. What are the forces on it at the top of the loop? What does the net force need to be to keep it on the track?

Well at the bottom of the loop, all the potential energy will have converted to kinetic energy. At the bottom, the normal force will equal the force of gravity, which is the weight, which is mass times gravity. As it travels up the side of the loop the normal force will always point perpendicularly from the surface of the track and the force of gravity that was holding the vehicle to the track in contrast to the normal force will give way to the radial acceleration which will push the vehicle towards the track which has to be greater than the 9.81 m/s^2 right? the normal force at the top of the loop as to be zero for a minimum speed to go around the loop but I won't ever get there because it seems the vehicle runs out of speed to do that and falls off the track prematurely but I'm not sure how to get there.

Suppose it makes it up to angle theta to the vertical, measured from the top, say. What will its speed and radial acceleration be there? What does the free body diagram look like?

I don't understand the first part of what you said exactly but I imagine as the car starts at theta = 270 degrees the kinetic energy would start to go down and normal force would start to go up. Radial acceleration would go down with the velocity from that point as well. I believe radial acceleration is velocity squared over the radius so as the velocity dropped in half, the radial acceleration would be a fourth of it's starting magnitude. At theta 2pi(or 0) all forces would be between what they would be at pi over 2 and 3pi over 2(or 270 degrees). I imagine at theta=pi over 2 that normal force would be close to zero and the car would be essentially weightless and it would follow projectile motion until normal force started to increase as the car passed from pi over two towards pi.

I don't understand the first part of what you said exactly but I imagine as the car starts at theta = 270 degrees
measured with theta increasing, yes?, so this is on the way up?
the kinetic energy would start to go down and normal force would start to go up.
KE will start to go down as soon as PE starts to rise, i.e. just after 180 degrees (the bottom of the loop). The normal force will be greatest at 180 degrees.
We need to allow theta to be any angle, not just thinking about multiples of pi/2. As function of theta, what is the PE? What does that tell you about the KE? what does that tell you about the velocity? Drawing the FBD, what do you then get for the normal force?

When you say 180 degrees do you mean pi on the unit circle? as in x=(-1)y=0? My benchmarks would be the standard unit circle where counterclockwise is increasing theta and 0 and 2pi are on the right (1,0) just as a common reference so we are both talking about the same thing. I understand PE and KE being inversely proportional to each other so the car enters the loop at the bottom with no PE and all KE and as it starts to enter the loop and go up and then counterclockwise KE is decreasing and PE is increasing. On a standard unit circle are you saying that the normal force is the greatest at the very bottom?

"As a function of theta what is PE", I know PE is mass*gravity*height and in this circle I believe height would be the only thing that changed with theta and height would be the sine of the theta multiplied by the hypotenuse but I'm not sure what the hypotenuse would be...maybe the velocity? would the height be velocity*sine of theta? so PE would be mass*gravity*velocity*sine of theta?

When you say 180 degrees do you mean pi on the unit circle? as in x=(-1)y=0?
I had previously written "measuring from the top". If you want to measure from half way the ascent, ok.
My benchmarks would be the standard unit circle where counterclockwise is increasing theta and 0 and 2pi are on the right (1,0) just as a common reference so we are both talking about the same thing. I understand PE and KE being inversely proportional to each other so the car enters the loop at the bottom with no PE and all KE and as it starts to enter the loop and go up and then counterclockwise KE is decreasing and PE is increasing. On a standard unit circle are you saying that the normal force is the greatest at the very bottom?
Yes, because a) the speed is greatest there and b) gravity is entirely in the radially outward direction.
"As a function of theta what is PE", I know PE is mass*gravity*height and in this circle I believe height would be the only thing that changed with theta and height would be the sine of the theta multiplied by the hypotenuse but I'm not sure what the hypotenuse would be...maybe the velocity? would the height be velocity*sine of theta? so PE would be mass*gravity*velocity*sine of theta?
A height cannot be a velocity. It's a circle radius R. If the height is 0 at the bottom it's 2R at the top. At θ=0 it's R. Yes, sin(theta) is highly relevant. Have another go at deciding on the hypotenuse.

Thank you so much for all your help again, ok hmm, the hypotenuse...That would have to be R right? If I drew the radius as a point in the middle of the circle pointing towards the edge of the circle that would even look like a vector pointing to the theta where the car might fall off the track. So if that was my hypotenuse I would have MgRsine(theta) for the PE and I know the PE in the beginning would be Mgh and in that situation the height was given as 2R so I'd have 2RMg=MgRsine(theta) right?

I re-read that and tried to work on finding theta from that...it turns out I'm way off haha

how bout a combination of r and theta, like r over theta. There aren't a whole lot of other dimensions in a circle that could relate to the height.

Try using a bit of geometry as a change from guesswork. Draw a circle. Mark your theta=0 position on the circumference at the right, and another generic position, P, on the circumference at an angle theta somewhere above that.
Draw a radius from the circle's centre to each of the two marked points on the circumference. Drop a perpendicular from P to the horizontal radius.
How high is the centre of the circle above the bottom of the circle?
How high is P above the centre of the circle?

The height to the center of the circle would be R and the height from the horizontal to the P mark would be Rsine(theta) so the height from ground to P would R+(Rsine(theta))?

The height to the center of the circle would be R and the height from the horizontal to the P mark would be Rsine(theta) so the height from ground to P would R+(Rsine(theta))?

Yes. So what is the KE at that position?

Since all the energy is conserved in this problem and I start out with 2RMg of potential energy I would have a KE equal to the original PE minus the PE at that point. KE=2RMg-(Mg(R+Rsin(theta)))

Good. So if the speed is v=v(θ) at that point you can write one equation involving v2.
Next, you need an equation for the normal force at that point.

v2 as in velocity squared or velocity times 2 or velocity sub 2?

Isn't the normal force proportional to the radial acceleration? So the more radial acceleration I have, the more normal force I have? Ar=(V2)/R ?...though normal force usually is the opposite of the force of gravity mg and since our radius isn't increasing or decreasing that means it has factors of m in there too right?

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v2 as in velocity squared or velocity times 2 or velocity sub 2?
I did not write v2 - I wrote v2 (v squared). Are you viewing this in some way that doesn't honour superscripts?
Isn't the normal force proportional to the radial acceleration?
So the more radial acceleration I have, the more normal force I have? Ar=(V2)/R ?
That's the equation for radial acceleration. Include the mass to get the centripetal force. But the centripetal force is not the same as the normal force. To get the object to go around in a circle, the centripetal force must be the resultant of all the forces acting on the body.
...though normal force usually is the opposite of the force of gravity mg
Only because in most problems the surface is underneath the mass.
Draw the free body diagram for the object at angle theta in the loop. What are the forces acting on it? If the centripetal force is the resultant, what equation can you write?