Non-Uniform Circular Motion

38 Non-Uniform Circular Motion

I want to explore circular motion a bit more. To do so, let’s start by warming up with an exercise.

Exercise 38.1: The Sphere of Death

You may have seen videos of the “motorcycle sphere of death.” It’s a spherical cage in which one or more motorcycles go in, and drive around, doing loop-the-loops and other acrobatic stunts. Here’s a small snippet of a video showing one such show.

Let’s focus on what’s going on at the end, when the motorcycles are driving along the horizontal plane going through the middle of the sphere.

From the video, I estimate the height of the sphere to be $\approx 5\ {\rm m}$ . Assuming the static frictional coefficient of the tires is $\mu=0.7$ , what is the minimum speed which the drivers must maintain in order to be able to drive in a horizontal circle around the middle of the sphere?

If you are struggling, check that your picture matches mine by using the slider below.

Now that we are warmed up, let’s do a quick review of a piece of math we will need for the next bit: the binomial approximation.

Math review: the binomial approximation.

The binomial approximation states that

$(1+x)^\alpha \approx 1+\alpha x$

if $|x|\ll 1$ . You can readily show this is true using a Taylor Series expansion. Given any function $f(x)$ , the derivative of $f$ is

$f'(x)\approx \frac{f(x+\Delta x) - f(x)}{\Delta x}.$

This approximation becomes exact if we take the limit $\Delta x \rightarrow 0$ . But let’s not do that, and use the approximation. We can solve for $f(x+\Delta x)$ to find

$f(x+\Delta x)\approx f(x)+f'(x)\Delta x$ .

Consider now the function $f(x)=x^\alpha$ , for $x=1$ . We have then:

$(1+\Delta x)^\alpha = f(1+\Delta x) \approx f(1)+f'(1)\Delta x = 1 + \alpha \Delta x$ .

This is what we wanted to show, i.e. the binomial approximation: $(1+x)^\alpha \approx 1+\alpha x$ . In practice, the approximation usually comes up when dealing with expressions of the form $(a+\epsilon)^\alpha$ where $a\neq 1$ . However, we can factor out the $a$ insider the parenthesis, so that

$(a+\epsilon)^\alpha = a^\alpha (1+ \epsilon/a )^\alpha \approx a^\alpha (1+\alpha\epsilon/a)$ .

Bottom line: You should know and remember the binomial approximation. It comes up all the time.

We are now in a position to answer the question I wanted to address: how does the speed of an object change when it is subject to a force that is always perpendicular to the direction of motion?

Exercise 38.2: Perpendicular Forces and Speed

Let’s start by thinking again about the impact of forces on a moving object.

The last answer may sound problematic, but hear me out. Say the object moves along the x-axis, and you push it in the y-direction for a small amount of time $\Delta t$ .

A. Find $v_y$ after this tiny push.

B. Find the corresponding change in the speed $\Delta v$ , and simplify using the binomial approximation.

C. Divide $\Delta v$ by the amount of time $\Delta t$ , and take the limit $\Delta t \rightarrow 0$ to find $dv/dt$ . What does your answer tell you about how $v$ changes with time?

Bottom line: forces perpendicular to the direction of motion change only the direction of the velocity vector, not its magnitude.

This all gives rise to an obvious question:

D. If forces perpendicular to the direction of motion can only change the direction, how do big perpendicular forces differ from small ones?

Let’s make sure that the lesson from the above problem sinked in.

Exercise 38.3: Non-Uniform Circular Motion

Let’s look at two problems involving non-uniform circular motion.

Exercise 38.4: Vertical Ball on a Rope

Imagine you have a tennis ball tied to the end of a rope. You grab the other end of the rope, which has length $L$ , and spin it around so that the ball traces a vertical circle.

Hint: the following questions may be multiple choice, but I don’t expect you to be able to just “figure them out” without drawing anything!

As you spin the ball around the vertical circle, what is the minimum speed that the ball can have as it reaches the top of the circle such that the ball will complete a full revolution? The alternative is the object just “falls down” without completing a full revolution around the circle).

Exercise 38.5: A Race Car Taking a Turn

A race car approaches an S-shaped curve as shown at below. As the driver reaches the curve, he starts breaking, and continues to do so until he completes the first half of the S (shown in BLUE). The driver then hits the gas and starts to speed up as he pulls out of the second half of the S (shown in RED). We will assume that each half of the “S” is a semi-circle or radii $R_1$ and $R_2$ respectively.

A. The diagram below shows two yellow points in each of the two halves of the S curve. Click on the vectors that point in the direction of the car’s acceleration at that point in the S curve.

Note: the length of the vectors in the diagram is not to scale.

B. Assume the race car enters the “S” curve at some speed $v_0$ , and exits at the same speed. During which half of the S curve was the centripetal acceleration the largest?

C. During which half of the S curve was the tangential acceleration the largest?

D. Let $t_1$ be the time that it takes for the race car to complete the first half of the S turn, and $t_2$ be the time that it takes for the race car to complete the second half of the S turn. What is the ratio $t_2/t_1$ ?

E. Does the description of how the race car driver took the S turn sound reasonable to you, or do you think there are obvious “problems” with the way he took the turn, in terms of minimizing how long it takes to go through the curve? Explain your answer.

The whole discussion about how race car drivers should take corners to optimize their drivers is actually pretty interesting. If you’d like to hear more, check out this Vox video.

Key Takeaways

38 Non-Uniform Circular Motion

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