Question

Explain formula for conical pendulum ​

Answer 1

Explanation:

A conical pendulum is a string with a mass attached at the end. The mass moves in a horizontal circle. In this lesson, we'll analyze a conical pendulum and derive equations for its angle and height.

Answer 2

Answer:

The formula of the conical pendulum is

                                        ${\displaystyle t\approx 2\pi {\sqrt {\frac {L}{g}}}}{\displaystyle$

Explanation:

• When the conical pendulum is an extension of a simple pendulum in which the bob, instead of moving back and forth, moves at a constant speed in a circle in a horizontal plane.
• The force exerted by the string can be resolved into a horizontal component, $T sin$(θ) toward the center of the circle, and a vertical component, $T cos$(θ), in the upward direction.  

                         $T sin$(θ)$=\frac{mv^{2} }{r}$

• Since there is no acceleration in the vertical direction, the vertical component of the tension in the string is equal and opposite to the weight of the bob:

                          $T cos$ (θ)$=mg$

•   These two equations can be solved for T/m and equated, thereby eliminating T and m:

                  ${\displaystyle {\frac {g}{\cos \theta }}={\frac {v^{2}}{r\sin \theta }}}{\frac {g}{\cos \theta }}={\frac {v^{2}}{r\sin \theta }}$

•     Since the speed of the pendulum bob is constant, it can be expressed as the circumference $2\pi r$ divided by the time t required for one revolution of the bob:

                                  ${\displaystyle v={\frac {2\pi r}{t}}}$

• Substituting the right side of this equation for v in the previous equation, we find:

                           ${\displaystyle {\frac {g}{\cos \theta }}={\frac {({\frac {2\pi r}{t}})^{2}}{r\sin \theta }}={\frac {(2\pi )^{2}r}{t^{2}\sin \theta }}}{\frac {g}{\cos \theta }}={\frac {({\frac {2\pi r}{t}})^{2}}{r\sin \theta }}={\frac {(2\pi )^{2}r}{t^{2}\sin \theta }}$

• Using the trigonometric identity tan(θ) = sin(θ) / cos(θ) and solving for t, the time required for the bob to travel one revolution is
•                         ${\displaystyle t=2\pi {\sqrt {\frac {r}{g\tan \theta }}}}t\\=2\pi {\sqrt {{\frac {r}{g\tan \theta }}}}$

• In a practical experiment, r varies and is not as easy to measure as the constant string length L. r can be eliminated from the equation by noting that r, h, and L form a right triangle, with θ being the angle between the leg h and the hypotenuse L (see diagram). Therefore,

                                ${\displaystyle r=L\sin \theta \,}$

• Substituting this value for r yields a formula whose only varying parameter is the suspension angle:

                        ${\displaystyle \,t=2\pi {\sqrt {\frac {L\cos \theta }{g}}}\,}{\displa$

• For small angles θ, cos(θ) ≈ 1; in which case

                           ${\displaystyle t\approx 2\pi {\sqrt {\frac {L}{g}}}}{\displaystyle$