How Do You Calculate Angular and Linear Velocities of a Rotating Rod?

[wooram83]

Homework Statement

A uniform rod of length L (2.6 m) and mass M (10.2 kg) is free to rotate on a frictionless pin
through one end. The rod is released from rest in the horizontal position, at t = 0.

(a) Use energy considerations to find the angular velocity of the rod when it has rotated an angle theta from the horizontal?

(b) Similarly, determine the linear velocity of the center of mass as a function of theta.

(c) Since you have an expression for ω as a function of theta, write down a differential equation for theta as a function of time, of the form: ω=d(theta)/dt = f (theta)

(d) By trial and error, find an expression for  as a function of t that satisfies the equation in part c.

Homework Equations

ω=d(theta)/dt
v=ωxr

The Attempt at a Solution

a) w=sqrt(3gcos(theta)/L)
b) v=sqrt(gLcos(theta))
c) ??
d)??

 

[tiny-tim]

welcome to pf!

hi wooram83! welcome to pf!

(have a theta: θ and an omega: ω and a square-root: √ )

a) w=sqrt(3gcos(theta)/L)

(it would help if you showed your full calculations)

that looks ok, except shouldn't ω increase with θ ?

b) v=sqrt(gLcos(theta))

how did you get that?

 

[Sniperman724]

I would think that you can use I=(1/3)mr^2 to find the moment of inertia, then use the rotational kinematics equations to find the rest. I could be wrong though

 

[Paulie323]

I would think that you can use I=(1/3)mr^2 to find the moment of inertia, then use the rotational kinematics equations to find the rest. I could be wrong though

I would agree here i feel taking this approach could make it a little easier in finding your solution.

 

[Apphysicist]

I=(1/3)*M*L^2 for a uniformly dense rod rotating about an endpoint (whether you do the integration or look it up in a table).

From there, for the energy considerations, I would model the rod as a point mass at the center of mass on a massless string, swinging as a pendulum. Set your zero-potential-energy point as the lowest point that the pendulum will swing through so that it's easy to determine the change in potential energy. Beware, there will be some trig creeping in there (I measured theta as the angle from the horizontal position to the pendulum as the terminal side of the angle), and then use, ∆U=(1/2)*I*ω^2, conservation of energy to solve for ω.

Then using v=ω*r, take note of what you should use as your r (you found ω for the center of mass and thus the whole rigid body...now you want, r, the distance from the axis of rotation to the center of mass for v of the center of mass).

For (c), I'm not sure what exactly to say.

For (d), to check your answer, you need to take the necessary number of derivatives of the the fcn from (c) and plug them back into the differential equation in order to find an identity (always true).

[e.g. f '(x) + f(x) = 0. If I guess f(x)=x^2, then f '(x)=2*x and 2*x + x^2 =0 is false for some x, so f(x)=x^2 is not a solution for the DE.
Now I guess e^-x, then f '(x)=-e^-x and -e^-x + e^-x =0 is 0=0 always true, so f(x)=e^-x is a solution to that particular DE. ]

If you know separation of variables or other techniques, it may be useful to you. Without having solved for (c), I'm inclined to say that the DE you come up with will be first-order, though that's based on nothing in particular.