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

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

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Answer 1 · 2018-08-08T07:39:08

Frequency of rotation $n = 2 r p s$ $n=2rps$

Angular velocity of the rotating particle

$ω = 2 π n = 4 π$ $omega=2pin=4pi$ rad/s

Radius of the circular path

$r = 25$ $r=25$ cm.

So radial acceleration of the particle

$a_{radial} = ω^{2} r = {(4 π)}^{2} \cdot 25 = 400 π^{2} c m s^{- 2} = 4 π^{2} \approx 39.48 m s^{- 2}$ $a_"radial"=omega^2r=(4pi)^2*25=400pi^2" "cms^-2=4pi^2~~39.48ms^-2$

Answer 2 · 2018-08-08T07:54:05

$a = {39.5 m/s}^{2}$ $a="39.5 m/s"^2$

Explanation:

Formulas needed:

$C = 2 π r$ $C=2pir$

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

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

$C = 2 π \cdot 0.25 m$ $C=2pi*"0.25 m"$

$C = π \cdot 0.5 m$ $C=pi*"0.5 m"$

$C = 1.571 m$ $C="1.571 m"$

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

If we use the equation for centripetal acceleration

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

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

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

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

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

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