Force Homework: Mechanics Theorem - Find the Force in terms of Theta

[lifeonfire]

Homework Statement

Let a mass move in an orbit given by r = a*(theta)
(a) If theta is a linear function of time t, what is the Force in terms of theta?
(b) How should theta depend on t so that the force is central?

Homework Equations

The Attempt at a Solution

Given that theta is a "linear function of time". I can write: theta = A t + B
Thus, r = a*(At +B)
and, v = dr/dt = a*(t)
and a = dv/dt = a

Therefore Force = m*a ? Where m = mass.
The force doesn't depend on theta?
Also, I have no clue what the second question is asking for?

 

[CompuChip]

What do you get when you differentiate a*A*t? It's not a*t!
You should be more careful anyway, the "r" here is just the radial component (i.e. the distance to the origin) in polar coordinates.
Of course, these coordinates are defined by
$$x(t) = r \cos\theta, y(t) = r \sin\theta,$$
where the usual notation $\vec r(t) = (x(t), y(t))$ is causing some confusion here.

Once you solved a, I suggest repeating it for a more general $\theta(t)$. If you equate the force you then calculate to the standard expression for central force, you will get a differential equation for $\theta$.

 

[collinsmark]

Hello lifeonfire,

I had posted something yesterday, but I quickly deleted it after recognizing a misleading mistake I made. I think I have it right this time.

Given that theta is a "linear function of time". I can write: theta = A t + B
Thus, r = a*(At +B)

Okay, so far so good. Here r is the magnitude of the position vector r.

and, v = dr/dt = a*(t)

This is where things seem to be going wrong. Given the way the problem was phrased, it seems r is a vector. It has a magnitude r and a direction that points θ radians away from the x-axis.

Let's take a step back for a moment. Imagine that we have such a vector r and we want to "wiggle" both r and θ by small (infinitesimal) amounts, and calculate the resulting differential vector (the difference between the vector r before and after the wiggling). In polar coordinates, this vector is

$${\bold d} {\bold r} = dr \hat r + rd \theta \hat \theta$$

$$dr \hat r$$ is the radial component extending away from the center. Perpendicular to that is tangential component $$rd \theta \hat \theta$$. Note that the tangential component contains an r in it. That's because as we wiggle θ by a given amount, the effect that it has on the differential vector is proportional to the radius.

Now divide everything by dt and you have the velocity.

$$\dot {\bold r} = \dot r \hat r + r \dot \theta \hat \theta$$

Differentiate that with respect to time, and you have the acceleration vector. (Don't forget to use the chain rule on the $$\hat \theta$$ component.)

Now you can make your substitutions, differentiating terms as appropriate.

Also, I have no clue what the second question is asking for?

A central force is a conservative force in the direction of $$\hat r$$. Gravity and the coulomb forces are examples. But what it really boils down to is that a central force will be a function of r and will be completely in the $$\hat r$$ direction.

That last part is pretty important. Essentially what the second part of this problem is asking is "how does θ have to vary with time such that the $$\hat \theta$$ component of the force is zero?"

I'm going to give you a big hint. When you set the $$\hat \theta$$ component of the acceleration vector equal to zero, and solve for θ, you end up solving for a second-order, nonlinear, ordinary differential equation. Yuck. My hint is as follows: You've already done half the work in your previous steps! Saying that the $$\hat \theta$$ component of the acceleration must be equal to zero is the same thing as saying that the $$\hat \theta$$ component of the velocity must be equal to a constant. An arbitrary constant. By recognizing this, you'll end up only needing to solve a 1st order diffy-Q instead of a 2nd order.

 

[lifeonfire]

Thanks :)