An interesting Rotational Force Problem

[kidwithshirt]

Hi all, thanks for reading. This is from my High school AP physics lab and I really couldn't figure out what to do

we attempted to solve this by actually doing the lab instead of using math (which is what you are suppose to do)

I would love to see the mathematical explanation of this. I am in Calculus so anything Calc would not daunt me >.<

Homework Statement

http://img181.imageshack.us/img181/7653/physicspd6.th.jpg

An object is tied onto the string the string with the length N. It is dropped at an angle perpendicular to something that holds up the whole system (which is perpendicular to the ground, technically the string should be parallel to the ground) at initial velocity of 0.

But when the object completes 90 degrees of rotation, after it reaches the bottom of the circle, the midsection of the string will hit a rod (that red dot in the middle of the graph) that shortens the length the string by some percent. The distance from top of the system to the rod is M.

The question is, what is the ratio of M/N that will guarantee that the string will complete the rotation (meaning that it goes over 270 degrees from the initial position) AND taut . Well there is a point that it will make the rotation complete, below that the object may rotate more than once.

Homework Equations

The Attempt at a Solution

Our lab result is somewhere around 0.6. So if the string is 30cm the rod should be around 18cm.

 

[JoAuSc]

Correct me if I'm wrong, but here's the problem as I see it:

We have a pendulum consisting of a massless string and a mass m attached to the string. Let theta be the angle of the pendulum as measured from the bottom. The pendulum is released at $$\theta = -90$$ (degrees), and if there were no friction or other complicating factors it would oscillate between theta = +90 and -90.

We have a rod that sticks out so that at $$\theta = 0$$ the string (of length N) is split into two, where the upper portion is of length M and the lower portion is of length N-M. At the moment the pendulum string hits the rod, this pendulum of length N magically transforms into a pendulum of length N-M. (At this point let's let L = M - N.) The tangential velocity is the same for the bottoms of the N-pendulum and the L-pendulum, but at this moment we change our reference point for angular momentum from the pivot of the N-pendulum to the rod, the pivot of the L-pendulum. Thus, the two angular velocities are different, since the pivot has changed and since it'd be sensible to measure angular momenta from the pivot. Specifically,

$$\omega_{N} = \frac {v_y}{N}$$
and
$$\omega_{L} = \frac {v_y}{L}$$

where $$\omega_{N}$$ is the angular velocity at $$\theta_{N} = 0$$, $$\omega_{L}$$ is the angular velocity at $$\theta_{L} = 0$$, and $$v_y$$ is the tangential velocity at $$\theta_{N} = 0$$ and at $$\theta_{L} = 0$$. (The tangential velocity is the horizontal velocity of the mass at the bottom of the pendulum, which happens to be the same for both pendulums since one pendulum ends when another begins.)

So, to summarize, a big pendulum turns into a small pendulum, and the angular velocities are related to the tangential velocity as shown above. The tangential velocity can be calculated from conservation of energy.

That was the easy part. The next part is beyond high school physics:

Let $$f(x)$$ be a function. $$\frac {df(x)}{dx}$$ is its derivative. If we know f(x), we can find df/dx, and if we know that df/dx = g(x), we can find f(x) from g(x); this is basic calculus. For example, if f(x) = x^2, we can find df/dx = 2x, and if g(x) = e^x, then we integrate both sides of
$$\frac {df(x)}{dx} = g(x)$$
to get
$$f(x) = \int g(x)dx + C = e^x + C$$
where C is a constant of integration.

What's more complicated are equations like
$$\frac {df(x)}{dx} = f(x)$$
where f(x) is unknown. Equations dealing with derivatives where we have to solve for a function are called "differential equations", and they're central to physics. If we drop a ball from height y_o, we have the following differential equation

$$\frac {d^2 y}{dt^2} = -g$$

with the following initial values

$$y(t=0) = y_o, \frac {dy(t=0)}{dt} = v_o$$

and the corresponding solution

$$y(t) = y_o + v_o t -1/2 g t^2$$

This is just about the simplest example of a differential equation you can find. For a spring, we have
$$y'' = -Cy$$
where C is some constant, and where y'' means the second derivative of y. The solutions are sinusoidal oscillations.

The reason I brought that up is because to figure out when a pendulum with an initial angular velocity is going to be exactly at the top of the pendulum (theta = 180 degrees) requires solving a differential equation that looks something like

$$\theta'' = -C sin(\theta)$$

The simplest way to solve this would be a computer simulation to find values of N-M that will result in an initial angular velocity for a pendulum that would do what you want it to do. Here's the steps I would take if I were you:
1. figure out how the angular velocity of the original pendulum at theta = 0 (at the bottom) relates to the length of the pendulum
2. derive an expression for the angular velocity of a pendulum of length L = N-M in terms of the answer to #1
3. Figure out what the angular acceleration at $$\theta = 180$$ must be for the centripetal acceleration to be greater than the acceleration due to gravity.
4. find a java applet online that simulates a pendulum. (There should be a bunch.) Make sure this is a single pendulum, that you can vary the length and initial angular velocity and angle, and that you have all other parameters (such as damping) set to zero. Set the angle so that it starts at the bottom.
5. For the applet, vary the length, then vary the initial angular velocity according to the expression you derived in #2. Run the simulation, see if the pendulum just reaches the top, then repeat this step as many times as desired until you get a fairly accurate answer. Write down N-M and see if M/N is ~ .6 .
6. If possible, when doing #5, see what the maximum N-M would be requiring that the centripetal acceleration is greater than what you got for #3.

 

[m2003]

Hi there,

First of all, it is great that you attempted to do the lab instead of just relying on math. Hands-on experimentation is an important part of the scientific process.

Now, let's take a look at the problem. This is a classic example of rotational motion, where an object is attached to a string and rotates around a fixed point. In this case, the fixed point is the red dot in the middle of the graph.

To solve this problem mathematically, we can use the equations of rotational motion, specifically the equation for centripetal force:

F = m * (v^2 / r)

where F is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circle.

In this case, we can assume that the initial velocity is 0, since the object is dropped from rest. We also know that the radius of the circle is equal to the length of the string, N. So we can rewrite the equation as:

F = m * (0^2 / N) = 0

This means that at the bottom of the circle, when the object is perpendicular to the ground, the centripetal force is 0. But as the object continues to rotate, it will experience a change in velocity due to the shortened length of the string. This change in velocity will result in a non-zero centripetal force, which will keep the object moving in a circular path.

To determine the ratio of M/N that will guarantee the completion of the rotation, we need to consider the forces acting on the object at the point of contact with the rod. At this point, the object is experiencing two forces: the tension force from the string and the force of gravity. The tension force is directed towards the center of the circle, while the force of gravity is directed downwards.

To maintain the circular motion, the centripetal force must be equal to the sum of these two forces. So we can write the equation:

F = T + mg

where T is the tension force, m is the mass of the object, and g is the acceleration due to gravity.

Now, we can substitute in the equation for centripetal force and solve for the ratio of M/N:

m * (v^2 / r) = T + mg

T = m * (v^2 / r) - mg

T = m * (v^2 / N