 Posers and Puzzles

1. 13 Oct '08 22:18
A mass (m) is hanging on a massless, unstretchable rod of length l, and it swings like a pendulum. There is a frictinal force proportional to the velocity of the mass.

a) What is the equation of motion for the mass?
b) In the approximation of small ampiltudes (sin x = x), show that the natural frequency is (g/l)^(1/2), where g is the gravitational field strength.
c) Neglecting friction, but considering large amplitudes ( sin x =/= x), determine the period T.
2. 15 Oct '08 15:341 edit
Originally posted by Jirakon
A mass (m) is hanging on a massless, unstretchable rod of length l, and it swings like a pendulum. There is a frictinal force proportional to the velocity of the mass.

a) What is the equation of motion for the mass?
b) In the approximation of small ampiltudes (sin x = x), show that the natural frequency is (g/l)^(1/2), where g is the ...[text shortened]... ecting friction, but considering large amplitudes ( sin x =/= x), determine the period T.
Missed this one! Now let's see...

(a) What is the equation of motion for the mass?

The arc length "s" swept out through an angle "theta" by a pendulum of length "L" is given by:

s = L * theta

To find the velocity, we differentiate with respect to time:

v = ds/dt = L * dtheta/dt

The friction force is proportional to the velocity, so we have:

Ff = kL * dtheta/dt

Differentiating "v" with respect to time we get:

a = dv/dt = L * d^2theta/dt^2

Multiplying this by the mass gives us the net force:

F = ma = mL * d^2theta/dt^2

Also, a force balance on a free-swing mass attached to the pendulum that is constrained to swing on a circular arc reveals that the resultant force is:

F = -mg * sin(theta)

This value is negative, because a positive displacement results in a force back along the arc. Summing this force and the friction force gives us:

F(total) = kL * dtheta/dt - mg * sin(theta)

Equating this expression with our other force expression, we get:

mL * d^2theta/dt^2 = kL * dtheta/dt - mg * sin(theta)

Simplifying, we have:

d^2theta/dt^2 - (k/m) * dtheta/dt + (g/L) * sin(theta) = 0

This is the expression for the motion of the pendulum when the friction force is proportional to the velocity. I think it can be solved using Laplace transforms, but I haven't done them in a while so I'll leave them for another day. 😉

(b) In the approximation of small ampiltudes (sin x = x), show that the natural frequency is (g/l)^(1/2), where g is the gravitational field strength.

In approximating small amplitudes, I'll make the assumption that the velocity is small as well and so we can neglect friction. Therefore, our expression above reduces to:

d^2theta/dt^2 + (g/L) * theta = 0

Using the D-operator method to solve this, we get:

(D^2 + (g/L))(theta) = 0

D = +/- i*(g/L)^0.5

We could plug this value into the characteristic expression, but all we're really interested in is the magnitude of the constant, which denotes the frequency (g/L)^0.5 as required.

(c) Neglecting friction, but considering large amplitudes ( sin x =/= x), determine the period T.

For large amplitudes with no friction, our expression reduces to:

d^2theta/dt^2 + (g/L) * sin(theta) = 0

According to Wikipedia, after solving this expression for dtheta/dt and then inverting, the result can't be integrated so you can only approximate the answer using a power series.

http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Arbitrary-amplitude_period
3. 05 Nov '08 11:18
Originally posted by PBE6
Missed this one! Now let's see...

[b](a) What is the equation of motion for the mass?

The arc length "s" swept out through an angle "theta" by a pendulum of length "L" is given by:

s = L * theta

To find the velocity, we differentiate with respect to time:

v = ds/dt = L * dtheta/dt

The friction force is proportional to the velocity, so we ...[text shortened]...

http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Arbitrary-amplitude_period[/b]
PBE6 i always love reading your posts. even though i've never taken any serious physics, the running commentary always carries the reader right through the steps in a logical and concise manner. just wanted to thank you on behalf of the math/physics nerd community 🙂
4. 05 Nov '08 15:44
Originally posted by Aetherael
PBE6 i always love reading your posts. even though i've never taken any serious physics, the running commentary always carries the reader right through the steps in a logical and concise manner. just wanted to thank you on behalf of the math/physics nerd community 🙂
It all comes from old engineering test habits - going for part marks!! 😀

Thanks for the kind words, though. I always strive for clarity and concision when writing a solution, for my own benefit as much as anyone else's. I find that if I can't write a solution out in words, then I haven't understood the ideas. I'm glad to hear this ensures my ideas are clearly translated to the reader (right or wrong 😉).
5. 07 Nov '08 05:18
Originally posted by PBE6
It all comes from old engineering test habits - going for part marks!! 😀

Thanks for the kind words, though. I always strive for clarity and concision when writing a solution, for my own benefit as much as anyone else's. I find that if I can't write a solution out in words, then I haven't understood the ideas. I'm glad to hear this ensures my ideas are clearly translated to the reader (right or wrong 😉).