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#

Answer:

#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 .