diff --git a/content/physics/Forces.md b/content/physics/Forces.md index 49174f4..fb9ba45 100644 --- a/content/physics/Forces.md +++ b/content/physics/Forces.md @@ -203,7 +203,38 @@ $$ # Circular motion -**TODO!** +An object that moves in a circle at a constant speed $||\vec{v}||$ experiences **uniform circular motion**. The most important part of that sentence is the constant speed part; the measurement $||\vec{v}||$ is reference to the center of the circle. + +![](circular-motion.png) + +As seen above, a velocity for every point on the path of the circle's motion exits, and the change in the two velocity vectors from any two points separated by $\Delta t$ with respect to time computes the acceleration. As these two vectors become closer in size and more parallel, the difference becomes the vertical difference between the two showed in Figure 5-2 B. The acceleration vector is always directed to the center of the circle. +$$ +a_R=\lim_{\Delta t\to0}{\frac{\Delta \vec{v}}{\Delta t}}=\frac{v^2}{r} +$$ +$a_R$ is the radial acceleration, otherwise known as the centripetal acceleration. It's easier not to consider centripetal acceleration as a result of the imaginary centripetal force if instead thought of as acceleration experienced by the object radially, or on the radius of motion. +$$ +T=\frac{1}{f} +$$ + +$f$ is the revolutions per second of the object, usually expressed in its SI units: hertz. In radians: + +$$ +v=\frac{\text{distance}}{\text{time}}=\frac{2\pi r}{T} +$$ + +It's important to note that the acceleration and velocity vectors at an instant are always perpendicular. Additionally, in uniform circular motion, the magnitude of the acceleration vector is always constant. + +![](perp-vectors.png) + +> [!FAQ] Question +> A $150$ gram ball at the end of a string is revolving uniformly in a horizontal circle of radius $0.6$ m. The ball makes 2 revolutions in a second. What is its centripetal acceleration? + +> [!NOTE]- Answer +> First recognize the centripetal acceleration formula is $a_R=v^2/r$. $r$ is give. Additionally, we must use the velocity formula for fixed periods: \ +> $v=\frac{2\pi r}{T}$. \ +> Since the ball makes 2 revolutions in a second, the period $T$ must be $\tfrac{1}{2}$. $v=\frac{2\pi(0.6\mathrm{m})}{0.5}=7.54$, the unit of that output being meters/second. Substituting that value for the speed $v$ in the acceleration formula, acceleration is $94.7 \mathrm{m}/\mathrm{s}^2$ + + diff --git a/content/physics/circular-motion.png b/content/physics/circular-motion.png new file mode 100644 index 0000000..b8a8fce Binary files /dev/null and b/content/physics/circular-motion.png differ diff --git a/content/physics/perp-vectors.png b/content/physics/perp-vectors.png new file mode 100644 index 0000000..e5b3a68 Binary files /dev/null and b/content/physics/perp-vectors.png differ