Question

A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each t = 1 second, what is the magnitude of its acceleration?

Answer 1

The centripetal acceleration is $13.0 m/s^2$

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

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

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference ($2\pi r$) and the period of revolution (T), so it can be rewritten as

$v=\frac{2\pi r}{T}$

Therefore we can rewrite the acceleration as

$a=\frac{4\pi^2 r}{T^2}$

For the particle in this problem,

r = 2.06 cm = 0.0206 m

While it makes 4 revolutions each second, so the period is

$T=\frac{1}{4}s = 0.25 s$

Substituting into the equation, we find the acceleration:

$a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2$

Learn more about centripetal acceleration:

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