Question

Suppose that the speed of a ball moving in a horizontal circle is increasing at a steady rate. Is this increase in speed produced by the centripetal acceleration? Explain why or why not.

Answer 1

Answer:

The acceleration of this ball is not centripetal. In other words, the speed of the ball isn't always pointing toward the center of the circle.

Explanation:

The word centripetal means pointing towards the center. If the acceleration of an object is "centripetal", that acceleration must point towards the center of the circular path at all times. The proposition is that if a ball is

• moving in a circular path, and
• speeding up at a constant rate,

then the acceleration of the ball will be centripetal.

To prove this proposition, one needs to show that the acceleration of the ball is pointing towards the center no matter how the velocity and position of the ball changes. However, to disprove this proposition, giving a counter-example will be sufficient. The challenge is to construct a case where

• The speed of the ball is increasing at a constant rate, and
• The path of the ball is a circle. In other words, the distance between the ball and a certain point is constant.

Start by considering a circular motion of constant speed.

Consider an object in a typical circular motion where speed stays the same.  Assume that the origin is the center of this circular path, and that the radius of this circle is $1$. What will be the 2D vector equations for the position, velocity, and acceleration of this object at time $t$?

Start with the position of this object. Keep in mind that this object is supposed to move around the origin in a circle of radius $1$. Apply the Pythagorean Theorem. The square of the $x$ and $y$ coordinates (components) of this object should add up to $1^{2} = 1$. The trigonometric functions suits this purpose:  

• Position: $\vec{x}(t) = (\cos(t), \sin(t))$, where

$\cos^{2}(t) + \sin^{2}(t) = 1$.

Velocity is the first derivative of displacement $\vec{x}(t)$with respect to time. To find the derivatives of a vector equation, simply find the derivative of each component.

• Velocity: $\displaystyle \vec{v}(t) =\frac{d}{dx}\vec{x}(t)= \left(\frac{d}{dx}\cos(t), \frac{d}{dx}\sin(t)\right) = (-\sin(t), \cos(t))$.

Similarly, take the first derivative of velocity with respect to time to find acceleration.

• Acceleration: $\vec{a}(t) =\displaystyle \frac{d}{dx} \vec{v}(t) = (-\cos(t), -\sin(t))$.

Note how acceleration and position differ only by a negative constant $-1$. In other words, the position vector and the acceleration vector of this object are parallel but in opposite directions at all time.

Construct a circular motion where speed increases constantly.

Again, the trigonometric functions and the Pythagorean Theorem could help ensure that the object is indeed moving in a circle. The task is to find a suitable function of $t$, $f(t)$, such that

• $\vec{x}(t) = (\cos f(t), \sin f(t))$, and
• $\vec{v}(t) = \left(-t\sin f(t), \quad t\cos f(t)\right)$.

Speed is the magnitude of velocity:  

$v(t) = ||\vec{v}(t)|| = \sqrt{t^{2}(\cos^{2}t + \sin^{2}t)} = \sqrt{t^{2}} = t$.

Indeed this expression increases at a constant rate relative to time $t$.

The trick is to take the first derivative of $\vec{x}(t)$ using the chain rule to obtain

$\begin{aligned} \vec{v}(t) &= \frac{d}{dt} \vec{x}(t) \\&= \left(-\frac{d}{dt}\sin f(t),\quad \frac{d}{dt}\cos f(t)\right) \\&= (-f^{\prime}(t) \cos f(t), \quad f^{\prime}(t) \sin f(t))\end{aligned}$.

Compare the two expressions for $\vec{v}(t)$. Apparently $-f^{\prime}(t)$, the first derivative of this unknown function, should be equal to $t$. The general form of such functions can be found through integration:

$\displaystyle \int{t}\; dt = \frac{t^{2}}{2} + C$,

where $C$ can be any real number. To keep things simple, take $C = 0$. The function $\displaystyle f(t) = \frac{t^{2}}{2}$ is still going to meet the requirements.

Write the vector equation for position, velocity, and acceleration using the chain rule and the product rule:

• Position: $\vec{x}(t) = (\displaystyle \cos\frac{t^{2}}{2}, \sin\frac{t^2}{2})$,
• Velocity: $\displaystyle \vec{v}(t) =\frac{d}{dx}\vec{x}(t)= \left(\frac{d}{dx}\cos\frac{t^2}{2}, \frac{d}{dx}\sin\frac{t^2}{2}\right) = (-t\sin\frac{t^2}{2}, t\cos\frac{t^2}{2})$.
• Acceleration: $\begin{aligned}\vec{a}(t) &= \frac{d}{dx} \vec{v}(t) \\&= \left(\frac{d}{dt}-t\sin\frac{t^2}{2}, \quad \frac{d}{dt}\cos\frac{t^2}{2}\right) \\&= \left(-\sin\frac{t^2}{2} - t^{2}\cos\frac{t^2}{2}, \quad \cos \frac{t^2}{2} - t^2 \sin \frac{t^2}{2}\right)\end{aligned}$.

Note that most of the time, acceleration and position are no longer in opposite directions. For example, at time $t = 1$, refer to the diagram attached. Acceleration is not pointing to the center of the circular path. This acceleration is therefore not centripetal.

Answer 2

Answer:

No.

Explanation:

Centripetal acceleration would make the ball move towards the center of the circle. Centri- means "center", -petal means "towards". But since the ball has its own velocity that is perpendicular to the centripetal force vector, the centripetal acceleration can only change the direction of the velocity. Which means that centripetal acceleration cannot increase the ball's speed.