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.