Chapter 6 Lectures: Applying Newston's Laws: Uniform Circular Motion: Centripetal Acceleration Derived

Kinematic Foundations of Uniform Circular Motion

• Kinematics establishes fundamental calculus relationships between an object's position, velocity, and acceleration over time:

  • The instantaneous velocity vector $\mathbf{v}(t)$ is the first time derivative of the position vector $\mathbf{r}(t)$:     $\mathbf{v}(t) = \frac{d\mathbf{r}}{dt}$

  • The instantaneous acceleration vector $\mathbf{a}(t)$ is the second time derivative of the position vector $\mathbf{r}(t)$, or equivalently, the first time derivative of the velocity vector $\mathbf{v}(t)$:     $\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{r}}{dt^2}$

• In uniform circular motion, an object travels along a circular trajectory of fixed radius $r$ at a constant speed $v$:

  • The radius $r$ is constant because the path is strictly circular.

  • The speed $v$ is constant by definition of uniform motion; speed is a scalar representing the magnitude of velocity, whereas the velocity vector $\mathbf{v}(t)$ continuously changes direction.

  • The ratio $\frac{v}{r}$ represents a constant angular parameter relating linear speed and path radius.

• The position vector of an object in uniform circular motion in Cartesian coordinates is given by:   $\mathbf{r}(t) = r\cos\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + r\sin\left(\frac{v}{r}t\right)\mathbf{\hat{j}}$

  • The unit vectors $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$ define fixed, invariant orthogonal directions along the $x$-axis and $y$-axis, respectively.

  • Because $\mathbf{\hat{i}}$ and $\mathbf{\hat{j}}$ do not rotate or change magnitude over time, their time derivatives are zero:     $\frac{d\mathbf{\hat{i}}}{dt} = 0$     $\frac{d\mathbf{\hat{j}}}{dt} = 0$

  • The radius $r$ is invariant with respect to time and can be factored out of the differential operator.

Derivation of the Velocity Vector

• Computing the velocity vector requires differentiating the position function with respect to time:   $\mathbf{v}(t) = \frac{d}{dt}\left[r\cos\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + r\sin\left(\frac{v}{r}t\right)\mathbf{\hat{j}}\right]$

• Factoring out the constant scalar $r$ gives:   $\mathbf{v}(t) = r\left[\frac{d}{dt}\left(\cos\left(\frac{v}{r}t\right)\right)\mathbf{\hat{i}} + \frac{d}{dt}\left(\sin\left(\frac{v}{r}t\right)\right)\mathbf{\hat{j}}\right]$

• Applying standard calculus chain-rule derivative identities where $a$ is a constant coefficient:   $\frac{d}{dx}\left[\cos(ax)\right] = -a\sin(ax)$   $\frac{d}{dx}\left[\sin(ax)\right] = a\cos(ax)$

• Setting the constant coefficient to $a = \frac{v}{r}$:

  • Differentiating the cosine term yields:     $\frac{d}{dt}\left[\cos\left(\frac{v}{r}t\right)\right] = -\frac{v}{r}\sin\left(\frac{v}{r}t\right)$

  • Differentiating the sine term yields:     $\frac{d}{dt}\left[\sin\left(\frac{v}{r}t\right)\right] = \frac{v}{r}\cos\left(\frac{v}{r}t\right)$

• Substituting these derivatives back into the factored velocity equation:   $\mathbf{v}(t) = r\left[-\frac{v}{r}\sin\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + \frac{v}{r}\cos\left(\frac{v}{r}t\right)\mathbf{\hat{j}}\right]$

• Distributing the factor of $r$ cancels the denominator $r$ in each component:   $\mathbf{v}(t) = -v\sin\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + v\cos\left(\frac{v}{r}t\right)\mathbf{\hat{j}}$

Verification Using Initial Conditions and Angular Substitution

• Initial condition evaluation at $t = 0$ confirms physical consistency:

  • At time $t = 0$, the object is located on the positive $x$-axis and travels in a counterclockwise trajectory at speed $v$.

  • Evaluating trigonometric values at zero argument:     $\sin(0) = 0$     $\cos(0) = 1$

  • Graphically, the sine function starts at the origin ($0$) and moves upward, whereas the cosine function starts at its peak value ($1$) and decreases.

  • Evaluating the velocity vector at $t = 0$:     $\mathbf{v}(0) = -v\sin\left(\frac{v}{r}\cdot 0\right)\mathbf{\hat{i}} + v\cos\left(\frac{v}{r}\cdot 0\right)\mathbf{\hat{j}}$     $\mathbf{v}(0) = -v(0)\mathbf{\hat{i}} + v(1)\mathbf{\hat{j}} = v\mathbf{\hat{j}}$

  • This matches physical expectations: on the positive $x$-axis moving counterclockwise, the velocity is purely tangential, oriented strictly vertically along the positive $y$-direction with magnitude $v$.

• Parameterization by angle $\theta$:

  • The angle as a function of time is given by:     $\theta(t) = \frac{v}{r}t$

  • The velocity vector expressed as a direct function of $\theta$ is:     $\mathbf{v}(\theta) = -v\sin(\theta)\mathbf{\hat{i}} + v\cos(\theta)\mathbf{\hat{j}}$

  • Evaluating this expression at cardinal trajectory angles demonstrates tangential motion at each point:

  • At $\theta = 90^\circ$: $\mathbf{v} = -v\mathbf{\hat{i}}$

  • At $\theta = 180^\circ$: $\mathbf{v} = -v\mathbf{\hat{j}}$

  • At $\theta = 270^\circ$: $\mathbf{v} = v\mathbf{\hat{i}}$

  • At $\theta = 360^\circ$: $\mathbf{v} = v\mathbf{\hat{j}}$

Derivation of the Acceleration Vector

• Acceleration is derived by differentiating the velocity vector function with respect to time:   $\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}\left[-v\sin\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + v\cos\left(\frac{v}{r}t\right)\mathbf{\hat{j}}\right]$

• Factoring out the constant scalar speeds and differentiating each component:

  • For the $\mathbf{\hat{i}}$ component:     $\frac{d}{dt}\left[-v\sin\left(\frac{v}{r}t\right)\right] = -v\left(\frac{v}{r}\cos\left(\frac{v}{r}t\right)\right) = -\frac{v^2}{r}\cos\left(\frac{v}{r}t\right)$

  • For the $\mathbf{\hat{j}}$ component:     $\frac{d}{dt}\left[v\cos\left(\frac{v}{r}t\right)\right] = v\left(-\frac{v}{r}\sin\left(\frac{v}{r}t\right)\right) = -\frac{v^2}{r}\sin\left(\frac{v}{r}t\right)$

• Combining the differentiated components yields:   $\mathbf{a}(t) = -\frac{v^2}{r}\cos\left(\frac{v}{r}t\right)\mathbf{\hat{i}} - \frac{v^2}{r}\sin\left(\frac{v}{r}t\right)\mathbf{\hat{j}}$

• Factoring out the common factor $-\frac{v^2}{r}$ produces:   $\mathbf{a}(t) = -\frac{v^2}{r}\left[\cos\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + \sin\left(\frac{v}{r}t\right)\mathbf{\hat{j}}\right]$

Emergence of the Radial Coordinate and Unit Vectors

• Inspecting mathematical derivations often reveals embedded geometric structures:

  • In advanced problem solving, such as in graduate electromagnetic theory, an effective strategy after reaching an intermediate derivation is to pause and inspect the mathematical expression for recognizable patterns and established identities.

  • The bracketed term in the acceleration equation matches the form of the original position vector:     $\mathbf{r}(t) = r\left[\cos\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + \sin\left(\frac{v}{r}t\right)\mathbf{\hat{j}}\right]$

• Dividing the position vector by its scalar magnitude $r$ ($|\mathbf{r}| = r$):   $\frac{\mathbf{r}(t)}{r} = \cos\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + \sin\left(\frac{v}{r}t\right)\mathbf{\hat{j}}$

• Definition of unit vectors:

  • A unit vector is defined as any vector divided by its own scalar magnitude.

  • In three-dimensional Cartesian coordinates, unit vectors along the coordinate axes are defined by:     $\mathbf{\hat{i}} = \frac{\mathbf{x}}{|\mathbf{x}|}$     $\mathbf{\hat{j}} = \frac{\mathbf{y}}{|\mathbf{y}|}$     $\mathbf{\hat{k}} = \frac{\mathbf{z}}{|\mathbf{z}|}$

  • Applying this definition to the position vector defines the radial unit vector $\mathbf{\hat{r}}$, which points outward from the origin along the radius:     $\mathbf{\hat{r}} = \frac{\mathbf{r}}{|\mathbf{r}|} = \cos\left(\frac{v}{r}t\right)\mathbf{\hat{i}} + \sin\left(\frac{v}{r}t\right)\mathbf{\hat{j}}$

• Expressing position and acceleration using radial vector notation:

  • The position vector is expressed compactly as:     $\mathbf{r}(t) = r\mathbf{\hat{r}}$

  • Substituting $\mathbf{\hat{r}}$ into the derived acceleration equation gives:     $\mathbf{a}(t) = -\frac{v^2}{r}\mathbf{\hat{r}}$

  • This demonstrates that the acceleration in uniform circular motion lies entirely along the radial line.

  • Because the positive radial unit vector $\mathbf{\hat{r}}$ points radially outward, the leading negative sign indicates that the acceleration vector points in the opposite direction: radially inward toward the center of the circle.

Centripetal Acceleration and Vector Relationships

• Definition and properties of centripetal acceleration:

  • Centripetal translates literally to "center-seeking."

  • The acceleration vector points toward the center of the circular trajectory at all points in time.

  • The magnitude of centripetal acceleration, denoted $a_c$, is given by:     $a_c = |\mathbf{a}| = \frac{v^2}{r}$

  • The magnitude depends solely on the constant tangential speed $v$ and the constant path radius $r$, making $a_c$ constant in magnitude while continuously changing in spatial direction.

• Summary of velocity and acceleration vector orientations in kinematics:

  • Parallel vectors ($\mathbf{a} \parallel \mathbf{v}$):

  • When acceleration is oriented parallel to velocity (in the same direction), the object speeds up.

  • Antiparallel vectors ($\mathbf{a}$ antiparallel to $\mathbf{v}$):

  • When acceleration is oriented antiparallel to velocity (in the opposite direction), the object slows down.

  • Perpendicular vectors ($\mathbf{a} \perp \mathbf{v}$):

  • When acceleration is strictly perpendicular to velocity, the speed remains constant while only the direction of the velocity vector changes.

  • Application to uniform circular motion:

  • The velocity vector $\mathbf{v}$ is always tangent to the circular trajectory.

  • The centripetal acceleration vector $\mathbf{a}$ is always directed inward toward the center, perpendicular to the tangent line.

  • Because $\mathbf{a} \cdot \mathbf{v} = 0$ at every instant, the tangential speed remains constant while the direction of travel continuously rotates.