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

2 Answers
Aug 8, 2018

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

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 .