Question

A pendulum of mass 5.0 kg hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal force of 30.0 ????. The force is applied over 0.3 seconds. How long does the pendulum have to be to have a period of 5.0 seconds? What is the maximum angle of displacement of the swinging pendulum?

Answer 1

Answer:

The length is $L = 6.206m$ and the angle is $\theta = 37.752^o.$

Explanation:

The period $T$ of the pendulum is related to its length $L$ by

$T = 2\pi \sqrt{\dfrac{L}{g} },$

where $g =9.8m/s^2$ is the acceleration due to gravity.

Solving for $L$ we get

$L = \dfrac{T^2g}{4\pi^2}$

putting in $T =5.0s$ and $g =9.8m/s^2$ we get:

$L = \dfrac{(5.0s)^2*9.8m/s^2}{4\pi^2}$

$\boxed{L = 6.206m.}$

There are two forces acting on the pendulum: The gravitational force $mg$ and the $F = 30N$ student's force. Therefore, the angular displacement $\theta$ that these forces give is

$sin(\theta ) = \dfrac{F}{mg}$

$\theta = sin^{-1}( \dfrac{F}{mg})$

putting in $F =30N$, $m =5.0kg$, and $g =9.8m/s^2$ we get

$\theta = sin^{-1}( \dfrac{30N}{5.0kg*9.8m/s^2})$

$\boxed{\theta = 37.752^o.}$