Class for Wed 19 Feb / upcoming exam

We'll continue our discussion of momentum a little bit, time permitting, at least as far as figuring out how to handle collisions. A larger portion of the class will be related to homework problems, which are directly relevant for the exam, and another lab on programming.

For the homework, you should be a little bothered by #6, I'll outline 2 methods to solve this one. Number 7 should be quite mysterious, and that is OK - the technique you need to solve it is related to the experimental propagation of uncertainty - if you change one variable a little bit, how do the others change? This is related to how you move along surfaces in 3D, something you are learning or have learned in Cal III.

Number 8 requires some thought - the equilibrium spacing is where U(r) is minimum, or where dU/dr = -F = 0 and the net force is zero. Find this first. The breaking point of molecule is when you exceed the maximum restoring force implied by U(r). If you find F(r) = -dU/dr and look for its maximum, this will occur for a particular r, beyond which the force binding the atoms together is exceeded and the atoms will come apart. Mathematically, that means setting dF/dr = 0 to find the maximum, that's the radius beyond which you break the molecule. Using the result for the equilibrium spacing, you can write it in terms of only n, m, and the equilibrium radius. I might have asked this question before if you are willing to dig a little.

For the exam, the format is exactly like the last time. It will likely consist of 6 problems, of which you have to solve 4. The questions will only be on work, kinetic energy, potential energy, and conservation of energy - 2 chapters in the book. You'll have a formula sheet given like last time, and can bring 1 sheet of paper of your own. I will post HW3 solutions by Thursday morning, hopefully, to help you study.

Finally, the lab: we'll continue with coding. You have two basic tasks.

1) I assert that the sum of the first n cubes of integers (sum of i^3 from i=1 to i=n) is the square of the sum of the first n integers (square of the sum of i from i=1 to i=n). Write a program that can check this for specific values of n. Print out your code and results for n=10 and n=17. Basically: sum the integers, square the sum, and compare to the sum of the cubes of integers.

2) Write and evaluate a program to calculate the range of a projectile under only the influence of gravity (no drag forces). Verify that it gives the correct result (within a few percent numerical error) for a launch speed of 25 m/s and a launch angle of 45 degrees. (You already know how to calculate the range without a drag force ...) Note that links I gave previously, and specific folders here are highly useful. Print out your code and note your results for the conditions noted.

Homework 3, number 7

This is a very tough one, and I planned on going over it in class on Wednesday. I'll sketch out the approach below.

If power is constant (which we're basically told it is for the default car), then the work done is W=P*T where T is the time over which the power is being supplied. This work done must be equal to the car's change in kinetic energy. If the car starts from rest, the work just equals the final kinetic energy, 

W =PT = (1/2)mv^2

This relates P, T, and velocity, but we don't know velocity. We do know the track length though. What we want to do is solve that for v, and integrate it to get x. Since the length of the track (x) is fixed, that will let us relate power and time by themselves.

v = sqrt(2PT/m) 

x = (integral) v dt = sqrt(4PT^3 / 3m)

The question now is what happens if we vary P by some little amount dP, what happens with T? By how much does it decrease dT? The distance x is a function of the variables P and T. It is fixed, so any change in P will have to be accompanied by a change in T to keep it constant. 

The question we're really asking then is for the function x(P,T) to remain constant, what must the rates of change in P and T be? We'd need to know the slope along the "P axis" (so dx/dP) and multiply by the tiny change in P (let's call that change DP instead of dp to keep the change straight from the derivative). We'd also need to know the same along the "T axis". Basically, the change in any function is slope times displacement for each axis, all added together. This is the same way we propagate experimental uncertainties, by the way, something I hope we will cover soon.

If the function were f(x,y), we'd approximate a small change in f due to small changes DX and DY in x and y as

Df = f(x+dx) - f(x) = (df/dx)*DX + (df/dy)*DY

In the simpler case, if you just have y(x), all this says is DY = (dy/dx)*DX. Back to the problem at hand, if we have x(P,T), 

Dx = (dx/dP)*DP + (dx/dT)*DT

Since the track length is fixed, we know Dx = 0. Thus,

(dx/dP)*DP = - (dx/dT)*DT

or   DT = -DP*(dx/dP)/(dx/dT)

Given the function above, take the derivatives with respect to P and T, divide them, and that times the change in power gives you the corresponding change in time.

I'll plan on going over this on Wednesday too.

Last problem

the possible modes of the string have

f = (n/2L)sqrt(T/u)

where n is an integer, L is the length of the string, T its tension, and u the density. You know T=mg, tension is provided only by the hanging mass. You also know f is fixed by the resonator attached, f=120Hz. The only variables are m and n.

If the two given masses m1 and m2 work, but nothing in between does, then they must be adjacent harmonics - one is n and the other is (n+1).

Plug that into the above equation separately for both masses, set equal, and you can find n. Once you have n, you know everything in the equation above except u ...

16.25

You can find the velocity of a wave at any point along the string if you know the tension and linear density. At any point along the string, the tension is provided only by the rope below that point - you can't push on a rope.

Finding the time is handled just like the last problem in HW10 - t is the integral of dy/v(y), with the limits being the two ends of the rope.

16.34

The latter parts of this question will make more sense after tomorrow's lecture, if they don't from reading the chapter. To get you started:

The average power transmitted along the string is

P = \frac{1}{2}\mu v \omega^2 y_m^2

Here you don't know the velocity, but you can relate it to the tension and linear density.

If you superimpose two equal-amplitude sine waves together of the same frequency (so kx-wt is the same for both), but they differ only by a phase phi, the resulting wave is given by Eq. 16-51:

y(x,t) = \left[2y_m \cos{\left(\frac{\varphi}{2}\right)}\right]\sin{\left(kx-\omega t + \frac{\varphi}{2}\right)}

The amplitude then depends on the relative phase as well as the amplitude of each wave. If the waves are in-phase, the total amplitude is just double, but if they are 180 degrees out of phase, everything cancels.

Hint on the 2nd homework problem

When you have the torsion spring connected to the pendulum, you don't have any sort of simple equation you can use any more ... you have to just add the torques up.

First, you have a torque due to the weight of the hanging deal, something like (weight)(radial distance)(sin of inclination angle) if you're not into vectors. That pulls the thing back to equilibrium.

The torsion spring provides a little 'kick' in the opposite direction to keep it going. The torque provided is just proportional to the angle, (kappa)(theta)

In total you have then (weight)(radial distance)(sin theta) - (kappa)(theta) = net torque. The net torque must be the moment of inertia times alpha, the second derivative of angle with respect to time. If you use the low-angle approximation sin(theta)~theta, you can recover an equation for simple harmonic motion: angular acceleration is (omega)^2 times angle.

HW 11: 15.106

Problem 15.106
a) Translational - 1/16
b) Rotational - 1/32
c) it is a proof ... show that acceleration is proportional to -(const) times position.

HW 10 question 9

Updates to the hint to fix the glaring error. (with the proper link this time)

I might try to revise it a bit more to make it less terse ... I'll post here if I get that done.

Problem 9 / HW10

There is a mistake in the beginning of the hint ... the first expression for velocity v(y) is inverted.

I will fix that and clean it up a bit in the next hour. The proper expression for velocity is

v(y) = (const)*sqrt[y/(R+y)]

where "const" involves G's, M's, and R's. Possibly even a 2.

Keep an eye out here for the revision soon.

Problem set 10, number 9

Having worked it out myself last night, I realized that problem 9 is a bit more ... punishing than it really ought to be. The physics is easy, but the integral you need to solve is somewhat pathalogical.

Thus, the MASSIVE HINT, which sets up the integral for you and gives you the basic result. Note that I said "Use any means necessary to evaluate the integral required." This means you can look it up, perhaps with the Wolfram Integrator.

If you read the hint carefully, there are bonus points for solving the thing the hard way, and further bonus points for proving that it reduces to our usual expression for small heights.

Further hint: don't reinvent the wheel. What are the odds I made this problem up, and what are the odds that it comes from any number of advanced mechanics books?

One more hint

11.13

On 11.13, we managed to find alpha, but we can't find how long the ball slides because we don't know omega at that point. We know we can solve for the rest of the problem, but we're stumped right here.

Ah, but you do know omega at the point that slipping stops - at that point, it is pure rolling motion, and v = r * (omega). That's what you found in the first part.

You know that the linear velocity starts out at v_i, and ends up at v at the moment t that the sliding stops. That means

v(t) = v_i + at

where a is the acceleration you already found. For the angular part, you know that the ball starts out *without* rotation, so (omega)_i = 0. At the moment rolling without slipping starts, you know (omega) = (omega)_i + (alpha)t = (alpha)t. You also know what when rolling without slipping starts, v = r(omega). Put that together ...

v(t) = v_i + at = r(omega) = r*(alpha)*t

Now you know everything but t in the equation above ...

compiled HW hints

This is stuff that I sent to various people this afternoon ... just so you all get the same info. Forgive the lack of useful typesetting on the math, this is cut & pasted from plain-text emails.

11.13 (#3)

for number 3. i doesn't give us a mass of the ball, so are we supposed to calculate exact values or just symbolic solutions?

in the end, you shouldn't need it ... the mass should cancel everywhere.

for instance, on part b, you want the acceleration. the only force is friction, f = (mu)mg. acceleration is f/m, or (mu)g.

for part c, you want angular acceleration, which is (torque)/(moment of inertia). The force is the same f as above, acting at a distance r. the mass occurring in the torque will cancel with the factor occurring in the moment of inertia.

the other parts are similar - you only need velocities and so on, and since the accelerations are independent of mass, so are they. Let me know if you get stuck on a specific part. The answers of this one are in the back of the book too.

10.67 (#2)

We've pretty much figured out that our tangential acceleration is not
constant, which is pretty obvious since we're told to find it at 35
degrees, but we have no idea how to derive it. We managed to solve for
omega, but that really only gave us a single number, and there's not
really another way to solve for alpha, and therefore tangential
acceleration, since it isn't constant.

you are on the right track ....

the radial acceleration is just a_r = l (omega)^2, since the radial distance is l. That's half of it.

You can differentiate omega, but you want to use the chain rule when you get to the thetas - theta *is* a function of time.

actually, its easier to differentiate (omega)^2 implicitly and avoid the square root:

d(omega^2)/dt = 2*omega*(d omega/dt) = 2 * omega * alpha

Now if you know (omega)^2 = 3g(1-cos(theta))/l, then you also know

d(omega^2)/dt = (3g/l) * d(-cos(theta))/dt = (3g/l)*(sin(theta))*(d theta/dt) = (3g/l)*sin(theta)*omega

Here you have to use the chain rule when you differentiate sin(theta) with respect to time:

d(sin(theta))/dt = [d(sin(theta))/d theta] * [d theta/dt]

Thus, 2*omega*alpha = (3g/l)*sin(theta)*omega. This gives you alpha, which relates to the tangential acceleration via a_t = r*alpha = l * alpha.

Alternatively, you can find alpha by using torque (even though the problem tells you not to, bah!) ... the total torque is moment of inertia times alpha.

10.54 (#1)

For 10.54, is part (b) talking about directions in terms of x- and y-
coordinates or in terms of clockwise/counterclockwise? I'm a little
confused about that because the forces are represented as vectors in the figure.

Just clockwise or counter-clockwise is enough. It wants the direction of the angular acceleration for (b). The angular acceleration is an axial vector, so it has a direction, but also specifies the direction of rotation about an axis. You can specify either one and that is enough, but it is just simpler I think to say clockwise or counter-clockwise ...

So basically, write down which way should it rotate, that's enough.

Even-numbered problems for this week

Numerical answers for the even-numbered problems ... just so you can see if you're on the right track.

10.32

• a: 2.3e-9 rad/s^2
• b: 2600 yrs
• c: 0.024 s

10.42

• a: 8.35e-3 kg m^2
• b: 0.22%

10.66

• 1.42 m/s

Formula sheet

Well, if you aren't going to get some rest, you can peruse the formula sheet for the exam.

Subject to small changes / additions as I find typos & omissions.

Random homework hints

9.8 Of course, you know the liquid density owing to the fact that you know the mass and volume when the can is filled. This is enough to calculate the liquid mass when the can is filled to a depth x. It is advisable to plot the resulting center of mass versus x.

9.15 You can easily find the velocity of the projectile at the top of the trajectory. At that point, apply conservation of momentum. If the falling particle has no initial velocity ...

9.45 Don't overthink this one. Conservation of momentum before and after along x and y axes. Symmetry is useful.

9.48 Both energy and momentum must be conserved, right?

*cough*

Huh, I guess I did this problem before.

PS3, Number 7

Miscellaneous correspondence with one of you:

On the top block, you have the normal force up, and its weight down, so the normal force is just the mass of the block times g.

That means the friction force is (mu)(mg). For motion to occur, you need (mu static)(mg) < (pulling force). This is true in this case, so the block is moving to the left, but also sliding against the bottom slab. Even though there is no friction between the slab and the ground, the block can pull away from the slab if the pulling force is big enough.

Since the block is moving, the *kinetic* friction force is (mu kinetic)(mg). The two horizontal forces are then this one and the pulling force, their difference gives mass times acceleration for the top block.

Now, if the top block has a friction force to the right due to the interaction with the bottom slab, the slab itself must feel the same force in the opposite direction by Newton's third law. Thus, if the top block is slipping off the bottom slab, the bottom slab has to feel the *same* friction force (but in the opposite direction) that the block feels. This is the only horizontal force on the slab, since it has no friction with the floor, nor is there a pulling force directly on it. So, (m slab)g = (friction force on top block) = (mu kinetic)(m block)g

I'll draw this out in class tomorrow and hopefully it will be clearer.

Conical pendulum

Problem 6.60 involves a conical pendulum, a classic problem you can easily turn up online with even the most cursory search ...

Problem 6.34 has numerical answers of about 6 and 1 m/s^2 respectively.

More details to follow.

Hints on PS3, cont.

The hint file now contains a free-body diagram for number 6.

Remember, the centripetal force must be the result of forces in the radial direction for circular motion. It is not a separate force acting on a body, but a constraint that must be obeyed to satisfy circular motion.

Another hint on PS3, #1

See this link.

Further hints on #2 and #3 will follow later this evening.

UPDATE: the equation for the parabola had an "R squared," but that should have been just "R." The error has been corrected in the linked file.

DOUBLE SECRET UPDATE: Problem 3 was assigned in MIT's course 8.01*, F03 semester. Check out their OpenCourseWare.

* MIT's courses are all numbered. Course 8 is physics, class 1 is mechanics. Course 18 is mathematics. Everything is numbered at MIT. Some of the buildings don't even have names. I worked in building 8 for a time, and later building NW-14.