Category: 06 Circular Motion

1002 Gravitron

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. Those people are moving in circles, which required a centripetal force. The normal contact force that the wall exerted on the bodies provided this required centripetal force.

The motion is fully explained by the presence of a centripetal force, not a centrifugal force.

When we see the people struggling to move towards the centre of the spinning drum, we are again tempted to think that some centrifugal forces are flinging them outward towards the wall. Wrong again. What’s happening is that when those people leave the wall, they can only count on the friction on their shoes to provide the required centripetal force. Their shoes obviously did not provide enough friction, hence they were not able to keep up with the circular motion of the spinning drum. They thus moved in a somewhat tangential direction, and ended up looking as if they were moving along the radius of the spinning drum.

Again, the motion is fully explained by the insufficient or absence of a centripetal force. Let’s banish the centrifugal force forever.

(Worked Example explained by xmtutor)

1001 Wall of Death

If done on a horizontal surface, it would be the normal contact force holding up the weight of the bike, and the frictional force providing the required centripetal force.

When done on a vertical surface (as in this video), it is the frictional force holding up the weight of the bike, and the normal contact force providing the required centripetal force.

The key to pulling off this stunt successfully is to ride at a high speed. This forces the bike and wall to be pressed hard into each other, resulting in a large normal contact force. A large normal contact force is crucial because the amount of friction depends on it.



Surely you noticed that the track was slanted at the bends? These are called banked curves.

If you had watched F1 racing before, you should know that F1 race tracks are flat. The race cars have to slow down (F1 fans love to watch and hear the acceleration) when they go into the curves . If they do not, the required centripetal force (mv2/r) will be too large. The tires just do not have enough grip to produce the required amount of friction.

NASCAR race cars do not slow down when they turn. Banking allows the cars to utilize the normal contact force that the track surface exerts on the cars to provide the required centripetal force. In fact, at particular speed and angles, the cars can negotiate the bends without the tires providing any sideway frictional force at all. This allows NASCAR to turn at full speed, which NASCAR fans love.

(Worked example explained at xmtutor)

136 Ball tied to String in Circular Motion

The tension in the string is changing during the circular motion. Why?

Firstly, note that the vertical circular motion of a ball swung on a string is not uniform circular motion where the speed is constant.Gravity causes the ball’s speed to be fastest at the bottom and slowest at the top. (vB>vT) As such the required centripetal force is maximum at the top of the circle, and minimum at the bottom of the circle. (mv2/r)


There are thus two reasons why the string is most taut at the bottom.

  1. The required centripetal force at the bottom is maximum.(mvB2/r)
  2. Weight of the ball is in opposite direction to the centripetal direction. The tension in the string must overcome the weight before contributing to the required centripetal force. (T = mvB2/r + mg)


There are also two reasons why the string is least taut at the top.

  1. The required centripetal force at the top is minimum. (mvT2/r)
  2. Weight of the ball is in the centripetal direction. The weight assists the string to provide the required centripetal force. (T = mvT2/r  mg)

In fact, if the ball is swung too slowly, gravity may pull the ball down so quickly that the string becomes slack and the ball becomes a projectile motion instead. The minimum launch speed in order to complete the circular motion successfully is presented in this xmtutor video.


125 Water in Pail Swung in a Vertical Circle

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.

124 Ball Bearing in Circles

The wall of the container pushes the ball bearing in the centripetal direction, causing the ball bearing’s velocity to keep changing its direction. When the centripetal force dissappears (when container is lifted), the ball bearing continues at its current velocity (as according to Newton’s 1st Law), which is in the tangential direction.

123 Circling Buoys



When the bottles are going round circles, the water level in the bottle is not level anymore. Instead, water level decreases from the outside to the inside.


The ping pong ball that is submerged in water experience pressure forces due to water. Since the pressure is always larger on the outside compared to inside, the ping pong ball experiences a net pressure force towards the inside.


In other words, because the pressure of the water increases not just with depth, but also towards the outside, the ping pong experiences an additional “insidethrust” on top of the usual upthrust. It is this “insidethrust” (minus the horizontal component of tension) that provides the required for the ping pong ball to undergo circular motion.



118 Circular Motion: Loop the Loop

(The dynamics for this demo is almost identical to that of a roller coaster’s loop-the-loop. Just replace the string’s tension with normal contact force of the coaster’s ramp, and everything is identical.)

Many students have the misconception that as long as the pendulum does not come to complete rest in the midst of motion, it can successfully loop-the-loop. That is wrong. The pendulum moves in a circular path only if the string is taut. However, when the pendulum is moving through the upper half of the circle, gravitational pull is actually tending to pull the pendulum inside the circle. If the pendulum is not moving fast enough, the component of gravitational pull in the centripetal direction will be larger than the required centripetal acceleration for circular motion at that speed and radius. This means the pendulum is being pulled by gravity towards the centre of the circle too quickly. The string becomes slack. Once that happens, the pendulum is no longer constrained to a circular path but goes into a parabolic path instead.

There is a minimum drop height in order for the pendulum to be moving fast enough to loop-to-loop successfully. The solution for the drop height is presented in this xmtutor video. So in theory, the limiting point for the experiment set up in this video is roughly when r = 25.0 cm.

096 Circular Motion: Circling Nuts

So, is the required centripetal force larger for large radius or small radius?

Careful, you better not use F=mv2/r, since the tangential speed is different for nuts at different radius for this scenario. So it is not clear whether smaller r (and smaller v) requires larger (or smaller) F.

Use F=mrω2  instead, since in this scenario, all the nuts have the same angular velocity ω. Then it becomes clear that at the same ω, the outer nuts doing circular motion with a larger r require a larger centripetal force compared to the inner nuts. Outer nuts skid first.

In fact, if one is situated right at the middle, no centripetal force is required at all. This explains why the bean at the middle had the last laugh.

048 Circular Motion (Bike leaning into the turn)

Let’s consider the system of the bike and the rider as one rigid body.


Since the system is undergoing circular motion, the resultant force acting on it must be in the centripetal direction with magnitude mv2/r. So we can write down the two equations:

(vertically):         N = mg

(horizontally):    f = mv2/r

Next, since the system is not rotating (it maintains the lean angle θ), the resultant moment acting about the centre of mass of the system must be zero*. This implies that the contact force (the resultant of N and f) must be directed towards the centre of mass.

(taking moments about the C.M.):           f/N = tanθ

Putting the three equations together, we get

v2/rg = tanθ

So centripetal acceleration v2/r = gtanθ. So if we see a bike leaning at 45°, the biker is experiencing about 1g of acceleration in the centripetal direction. It will be 2g at 63°, and 3g at 72°.

*This is a simplified analysis. People familiar with precession will know more. However, for A-level purpose, this simplified analysis is more than sufficient.