Category: 06 Circular Motion

Got to be Fast

As the ball loses speed, it eventually becomes too slow to complete the loop successfully.

Why? The next video explains.

Why doesn’t the water fall out of the bottle?

Let’s consider the water when the bottle is at the top most position. At this instant, the water has a horizontal velocity. The earth’s gravitational pull would have caused the water to continue along a parabolic arc. The bottle, however, intends to follow a circular path. Relative to the circular path of the bottle, the parabolic path of the water would have taken the water out of the bottle through the TOP, not the BOTTOM, thus puncturing the base of the bottle.

Of course, the bottle would not allow itself to be punctured by water. So what it does is to exert a downward normal contact force on the water, just enough to push the water into following the same circular path as the bottle.

The water falls out of the bottle only if it falls faster than the pail. As long as the bottle is swung at a high enough speed, this will not happen. In fact, the bottle must press down on the water to make the water fall faster, as fast as the bottle.

How do the passengers not drop out of the car when they are inverted at the top of the loop?

The following video explains why.

Centrifugal Force?

When we see the bodies pinned against the wall, it is tempting to imagine that there are some centrifugal forces pushing those people outward against the wall. But there is none. The motion is fully explained by the presence of a centripetal force, not a centrifugal force. The following video will explain.

In the next video, you can see the same principle being applied by these stuntmen. The key to pulling off this stunt successfully is to ride at a high speed. The inertia of the bike (tendency to keep moving straight) causes it to pressed hard against the wall, resulting in a large normal contact force. A large normal contact force is crucial because the amount of friction depends on it. It is the frictional force that holds up the weight of the car.

Similarly, in the video below, the nuts start to slip not when the “centrifugal force” overcomes the frictional force, but when the required centripetal force is too large and cannot be provided by the frictional force.

Explanation at xmdemo.wordpress.com/096

Conical Circular Motion

All the circular motion featured here have very similar dynamics.

(1) Conical Pendulum


(2) Flying Chairs


(3) Airplane turning a horizontal circle


(4) Cars turning on banked curve

NASCAR race cars do not slow down when they turn. The tracks are banked (slanted) at the bends. Banking allows the cars to utilize the normal contact force that the track surface exerts on the cars to help provide the required centripetal force. This allows NASCAR to turn at full speed. In fact, at particular speed and angles, the cars can negotiate the bends without the tires providing any sideway frictional force at all. The following video explains.

Uniform Circular Motion

Qualitatively, a force in the centripetal direction is what’s needed for uniform circular motion.

Explanation at xmdemo.wordpress.com/124

Quantitatively, the magnitude of the required centripetal force is related to the speed and radius of circular motion by the formula F=mv2/r.

centripetalForce0

On the other hand, for the same v, a stronger centripetal force F would result in a tighter circular motion (smaller r). In the animation below, the objects are experiencing centripetal forces of 4F, 2F and F, resulting in circular motion of radii R, 2R and 4R.

centripetalForce1

For the same centripetal force F, a lower speed v would result in a tighter circular motion (smaller r). In the animation below, the objects have speeds of 1/√2 v, v and √2v, resulting in circular motion of radii R, 2R and 4R.

centripetalForce

Angular Velocity

How do the linear velocity v and angular velocity ω of these two circular motions compare?

rw1

v=

These two circular motions have the same ω. The blue one has twice the radius, so its v is double that of the red one.

rw2

ω=v/r

These two circular motions have the same v. The blue one has twice the radius, so its ω is half that of the red one.

812 Lissajous Figures on CRO

When the CRO is set to the xy mode, it uses the Channel 1 input as the time-base. This means the x and y position of the CRO trace is controlled by inputs from Channel 1 and 2 respectively.

If both channels are fed with sinusoids with the same frequency, then the following traces wil result, depending on the phase relationship between the two sinusoids.

Picture1

In-phase. Much like the relationship between force and acceleration.

Picture2

Anti-phase. Much like the relationship between acceleration and displacement for a SHM.

Picture3

A quarter-cycle out of phase. Much like velocity leading displacement for a SHM.

In this demo, because I was using analogue signal generators, I was unable to set the inputs to the exact frequencies I wanted. So even though I was aiming for 100 Hz for both channels, I could only get them to be close to but not exactly 100 Hz. This means there was a slight difference in the periods of the two signals. This causes their phase relationship to keep shifting, alternating between in-phase and anti-phase. This results in the lovely outcome of the tracing alternating between diagonals and flipped diagonals, including the ellipses between.

When Channels 1 and 2’s frequencies are a nice integer ratio of each other, very intricate patterns are formed. By counting the number of times the trace cuts each axis, one can figure out the exact ratios. If you are interested and want to read up more, just google “Lissajous Figures”.