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

Aug 8, 2018

Frequency of rotation $n = 2 r p s$

Angular velocity of the rotating particle

$\omega = 2 \pi n = 4 \pi$ rad/s

$r = 25$ cm.

So radial acceleration of the particle

${a}_{\text{radial"=omega^2r=(4pi)^2*25=400pi^2" }} c m {s}^{-} 2 = 4 {\pi}^{2} \approx 39.48 m {s}^{-} 2$

Aug 8, 2018

$a = {\text{39.5 m/s}}^{2}$

Explanation:

Formulas needed:

$C = 2 \pi r$

$a = {v}^{2} / r$

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

$C = 2 \pi \cdot \text{0.25 m}$

$C = \pi \cdot \text{0.5 m}$

$C = \text{1.571 m}$

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

$a \approx {\text{39.5 m/s}}^{2}$

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