Uniform Circular Motion (UCM) - Conical Pendulum

Imagine a ball tied to a string, but instead of swinging back and forth (like a simple pendulum), you push it so it moves in a perfect horizontal circle.

• As the mass (bob) moves, the string makes a constant angle (θ) with the vertical.
• The string traces out the shape of a cone in the air, which is why it's called a Conical Pendulum!

Only two forces act on the mass (m):

• Gravity (mg): Acts vertically downward.
• Tension (T): Acts upward along the string, towards the fixed support.

For the bob to move in a stable horizontal circle, two conditions must be met:

• Vertical Balance: The net force in the vertical direction must be zero.
• Horizontal Net Force: The net force in the horizontal direction must provide the Centripetal Force (F_c) needed for circular motion.

To find the speed (v) and the time period (T), we combine the two force equations:

Step 1: Divide the two key equations.
\[\frac{T\sin\theta}{T\cos\theta}=\frac{(mv^2/r)}{mg}\]

Step 2: Simplify and find tan θ.
tan θ = \[\frac{v^{2}}{rg}\Longrightarrow v^{2}=rg\tan\theta\]

Step 3: Relate v to the Time Period (T). The speed is v = \[\frac {Distance}{Time}\] = \[\frac {2πr}{t}\].

Step 4: Substitute v into the tanθ equation.
tan θ = \[\frac{(2\pi r/T)^2}{rg}=\frac{4\pi^2r^2}{T^2rg}\]
tan θ = \[\frac{4\pi^2r}{T^2g}\]

Step 5: Isolate the Time Period (T).
\[T^2=\frac{4\pi^2r}{g\tan\theta}\]
T = \[2\pi\sqrt{\frac{r}{g\tan\theta}}\]

Step 6: Final Simplification (using r = l sin θ and tan θ = \[\frac {sinθ}{cosθ}\]).
T = \[2\pi\sqrt{\frac{l\sin\theta}{g(\sin\theta/\cos\theta)}}=2\pi\sqrt{\frac{l\cos\theta}{g}}\]

Final Formula: Since h = l cos θ (the vertical height from the bob's plane to the support):
T = \[2\pi\sqrt{\frac{\mathbf{h}}{\mathbf{g}}}\]

This shows the time period depends only on the vertical height (h) and is independent of the mass (m)!