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Section 5.6 Circular Motion Bootcamp

Subsection 5.6.1 Position and Displacement on a Circle

Subsection 5.6.2 Velocity on a Circle

Subsection 5.6.3 Uniform Circular Motion

Subsection 5.6.4 Centripetal and Tangential Accelerations

Subsection 5.6.5 Using Polar Coordinates

Subsection 5.6.6 Miscellaneous

Figure 5.6.30 shows a pebble P stuck at the outside of a tire. As the tire rolls, the pebble's position in space changes. Suppose tire does not slide but rolls smoothly. The tire has radius \(R\) and is rolling at a constant speed \(v_0 \text{.}\) At \(t=0\text{,}\) the pebble was at the origin.

Figure 5.6.30. For Problem 5.6.29.

Find the expressions for (a) position, (b) velocity, and (c) acceleration vectors (i.e., magnitudes and directions) at an arbitraru instant \(t\) with respect to the coordinate system given in the figure.


(a) Introduce another coordinate system with origin at the center of the tire at all times and \(x\) and \(y\) axes parallel to the one given. Find position in that coordinate system first. Then translate that to the given coordinate system. (b) and (c): use derivatives.


(a) \(\vec r = (-R\sin\theta + v_0 t)\; \hat i + (-R\cos\theta)\; \hat j\text{,}\) (b) \(((1 -\cos\theta)\;v_0 , (-\sin\theta)v_0)\text{,}\) (c) \(\text{.}\)

Solution 1 (a)

(a) Let us introduce a coordinate system as shown in Figure 5.6.31. For convenience, I will define \(\theta\) that measures angle from negative \(y\) axis as shown.

Figure 5.6.31.

Let \((x,y)\) and \((x',y')\) be the coordinates of P at this arbitrary instant \(t\) with respect to the two coordinate systems. It is immediately clear from the figure that

\begin{align*} \amp y = y',\ \ x = x' + v_0 t.\\ \amp x' = -R\sin\theta,\ \ y'=-R\cos\theta. \end{align*}

Therefore, position \(\vec r\) in original coordinate system will be

\begin{equation*} \vec r = (-R\sin\theta + v_0 t)\; \hat i + (-R\cos\theta)\; \hat j. \end{equation*}

It magnitude is

\begin{align*} r \amp = \sqrt{ x^2 + y^2 }\\ \amp = \sqrt{(-R\sin\theta + v_0 t)^2 + (-R\cos\theta)^2 }\\ \amp = \sqrt{R^2 - (2Rv_0\sin\theta) t + v_0^2t^2 }. \end{align*}

We can indicate direction in space by angle with respect to \(x\) axis by using arctangent.

\begin{equation*} \theta_x = \tan^{-1}(y/x) =\tan^{-1}\left( \frac{-R\cos\theta}{-R\sin\theta + v_0 t} \right). \end{equation*}
Solution 2 (b)

(b) Now, that we know the components of position, we can obtain components of velocity by taking derivatives. Thus,

\begin{align*} \amp v_x = \frac{dx}{dt} = (-R\cos\theta)\frac{d\theta}{dt} + v_0 \\ \amp v_y = \frac{dy}{dt} = (-R\sin\theta)\frac{d\theta}{dt} \end{align*}

Since the tire is not slipping, the rate at which angle \(\theta\) changes is related to the speed with which center of the tire is moving. The tire has rotated by \(R\Delta\theta\text{,}\) its center has moved \(v_0\Delta t\text{.}\) Equating them we get

\begin{equation*} R\Delta\theta = v_0\Delta t. \end{equation*}

This gives

\begin{equation*} \frac{d\theta}{dt} = \frac{v_0}{R}. \end{equation*}

Using this, we rewrite the components of velocity as

\begin{align*} \amp v_x = (1 -\cos\theta)\;v_0 \\ \amp v_y = (-\sin\theta)v_0. \end{align*}

Therefore, magnitude of velocity, i.e., speed is

\begin{equation*} v = \sqrt{v_x^2 + v_y^2} = v_0\sqrt{ 2 - 2 \cos\theta} = 2 v_0 \left| \sin(\theta/2) \right|. \end{equation*}

Note: this tells us that the speed of pebble when touching the ground is zero (since \(\theta=0\)) and when at the very top, it is \((2\;v_0)\) (since \(\theta=\pi\)). That is, speed of the pebble in \(Oxy\) coordinate system changes as it goes around.

The direction of velocity can again be given with respect to the \(x\) axis by using arctangent.

\begin{equation*} \theta_x = \tan^{-1}(v_y/v_x) = \tan^{-1}\left( \frac{-\sin\theta}{1 -\cos\theta} \right). \end{equation*}

You can further simplify this expression by using double angle formulas. Try that. You should be able to express this without arctangent operation.

Solution 3 (b)

(c) Now, that we know the components of velocity, we can obtain components of acceleration by taking derivatives. Thus,

\begin{align*} \amp a_x = \frac{dv_x}{dt} = (-v_0 \sin\theta)\frac{d\theta}{dt} \\ \amp a_y = \frac{dv_y}{dt} = (-v_0\cos\theta)\frac{d\theta}{dt} \end{align*}

Replacing \(d\theta/dt\) by \(v_0/R\) as found in (b), we get

\begin{equation*} a_x = -\frac{v_0^2}{R}\;\sin\theta;\ \ a_y = -\frac{v_0^2}{R}\;\cos\theta. \end{equation*}

Therefore, magnitude of acceleration is

\begin{equation*} a = \sqrt{a_x^2 + a_y^2} = \frac{v_0^2}{R}. \end{equation*}

The direction with respect to \(x\) axis will again be obtained by using arctangent.

\begin{equation*} \theta_x = \tan^{-1}(a_y/a_x) = \tan^{-1}(\cot(\theta)). \end{equation*}

Of course, we can write \(cot(\theta)\) as \(\tan(\pi/2 - \theta)\text{.}\) Hence

\begin{equation*} \theta_x = \frac{\pi}{2} -\theta. \end{equation*}

This is actually direction towards the moving center of the tire. Take values of \(\theta=0, \pi/2, \pi, 3\pi/2, 2\pi\) and draw them on the corresponding \(Oxy\) plane. You will see that the direction is always towards the center of the tire as shown in Figure 5.6.32.

Figure 5.6.32.