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

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

Step-by-step explanation:

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

`T = 2\pi \sqrt{(L)/(g) },`

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

Solving for
`L` we get

`L = (T^2g)/(4\pi^2)`

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

`L = ((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 ) = (F)/(mg)`

`\theta = sin^(-1)( (F)/(mg))`

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

`\theta = sin^(-1)( (30N)/(5.0kg*9.8m/s^2))`

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