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Physics Olympiad

Period of a Planet

Try this Problem, useful for Physics Olympiad, based on the period of a planet.

The Problem:

Suppose that the gravitational force varies inversely as the \(n^{th}\) power of the distance. Then, the period of a planet in circular orbit of radius \(R\) around the sun will be proportional to

(A) \(R^{\frac{n+1}{2}}\)

(B)\(R^{\frac{n-1}{2}}\)

(C) \(R^n\)

(D) \(R^{n/2}\)

Discussion:
The gravitational force can be given as $$ \frac{GMm}{R^n}=mR\omega^2$$

Now, we know \(\omega=\frac{2\pi}{T}\),

Hence

$$\frac{GMm}{R^n}= mR(\frac{2\pi}{T})^2$$ $$ T^2= \frac{4\pi^2R^{n+1}}{GM}$$

 

Categories
Physics Olympiad

Button on a Rotating Platform

Try this beautiful problem, useful for Physics Olympiad, based on Button on a Rotating Platform.

The Problem:

A small button placed on a horizontal rotating platform with diameter (0.320m) will revolve with the plate when it is brought up to a speed of (40rev/min), provided the button is no more than (0.150m) from the axis. What is the coefficient of static friction between button and platform?

Discussion:

The button moves in a circle so it has acceleration (a_{rad}).
The coefficient of static friction $$ \mu_s=\frac{v^2}{Rg}$$ Expressing (v) in terms of the period (T), $$ v=\frac{2\pi R}{T}$$ so, $$\mu_s=\frac{4\pi^2R}{T^2g}$$.A platform speed of (40.0rev/min) corresponds to a period of (1.50s) so $$\mu_s=0.269$$

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Physics Olympiad

Minimum Speed of a Rotating Pail of Water

Try this beautiful problem, useful for the Physics Olympiad, based on Minimum Speed of a Rotating Pail of Water.

The Problem: Minimum Speed of a Rotating Pail of Water

You tie a cord to a pail of water and you swing the pail in a vertical circle of radius (0.6m). What minimum speed must you give the pail at the highest point of the circle if no water is to spill from it?

SET UP: The water moves in a vertical circle. The target variable is the speed v.

Calculate: We calculate (a) and then get v from (a=\frac{v^2}{R}). We write

the force equation as $$ mg=m\frac{v^2}{R}$$
Therefore, $$ v=\sqrt{gR}=\sqrt{(9.80)(0.600)}=2.42m/s$$

Categories
Physics Olympiad

The Conical Pendulum Problem

Let’s discuss a beautiful problem useful for Physics Olympiad based on the Conical Pendulum.

The Conical Pendulum Problem:

An inventor designs a pendulum clock using a bob with mass m at the end of a thin wire of length L. Instead of swinging back and forth, the bob is to move in a horizontal circle with constant speed v, with the wire making a fixed angle \(\beta\) with the vertical direction. This is called a conical pendulum because the suspending wire traces out a cone. Find the tension F in the wire and the period T( the time for one revolution of the bob).
Solution:

The Conical Pendulum Problem
Conical Pendulum

SET UP: To find the tension F and period R, we need two equations. These will be the horizontal and vertical components of Newton’s second law applied to the bob. We’ll find the radial acceleration of the bob using one of the circular motion equations.
The figure shows the free-body diagram and coordinate system for the bob at a particular instant. There are just two forces on the bob: the weight mg and the tension F in the wire. Note that the entre of the circular path is in the same horizontal plane as the bob, not at the top end of the wire. The horizontal component of tension is the force that produces the radial acceleration \(a_rad\).
Execution: The bob has zero vertical acceleration.

Newton’s second law says that
$$ \Sigma F_x=Fsin\beta=ma_r$$
$$ \Sigma F_y=Fcos\beta+(-mg)=0$$
These are two equations for the two unknowns F and \(\beta\). The equation for \(\Sigma F_y\) gives $$ F=mg/cos\beta$$.
That’s our target expression for F in terms of \(\beta\).
Substituting this result into the equation for \(\Sigma F_x\) and using \(sin\beta/ cos\beta =tan\beta\), we find
$$ a_r=g tan\beta$$
To relate \(\beta\) to the period T, we use the equation
$$ T=2\pi R/v$$
in terms of the period , $$ a_R=4\pi^2R/T^2$$
so, T=\(2 \pi \sqrt{\frac{R}{gtan\beta}}\)
Now, $$ R=Lsin\beta$$. We substitute this and use \(sin\beta / tan \beta=cos\beta\):

So, $$ T= 2\pi\sqrt{\frac{Lcos\beta}{g}}$$
Evaluation: For a given length L, as the angle \(\beta\) increases, \(cos\beta\) decreases. A conical pendulum will never make a very good clock because the period depends on the angle \(\beta\) in such a direct way.

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