Question

A particle moves in a circle of radius 25cm covering 2 revolutions per second what will be the radial acceleration of that particle?

Answer 1

Frequency of rotation `n=2rps`

Angular velocity of the rotating particle

`omega=2pin=4pi` rad/s

Radius of the circular path

`r=25` cm.

So radial acceleration of the particle

`a_"radial"=omega^2r=(4pi)^2*25=400pi^2" "cms^-2=4pi^2~~39.48ms^-2`

Answer 2

`a="39.5 m/s"^2`

Explanation:

Formulas needed:

`C=2pir`

`a=v^2/r`

Here the circumference is calculated after everything has been converted into usable units (meters, seconds, Kelvin, etc)

`C=2pi*"0.25 m"`

`C=pi*"0.5 m"`

`C="1.571 m"`

So If the particle goes around the circle twice per second, then `(2*"1.571 m")/"1 s"` can be used to find `v`, the velocity.

If we use the equation for centripetal acceleration

`a=v^2/r`

`a = ("3.142 m/s")^2/"0.25 m" = 4xx 3.142^2 \ "m/s"^2`

`a~~"39.5 m/s"^2`

we get that the angular acceleration is equal to `39.48` meters per second per second .