A particle moves at a constant speed in a circular path with a radius of 2.06 cm. If the particle makes four revolutions each second, what is the magnitude of its acceleration? A) 20 m/s² B) 18 m/s² C) 13 m/s² D) 15 m/s² E) 24 m/s²

Question

Quizplus · Accepted Answer

The acceleration of an object in uniform circular motion is given by the equation $a=\frac{v^2}{r}$, where $v$ is the speed of the object and $r$ is the radius of the circular path.

In this problem, the particle is moving at a constant speed and making four revolutions per second, which means it completes one revolution in $\frac{1}{4}$ seconds. The distance traveled in one revolution is $2\pi r = 2\pi (2.06	ext{ cm}) \approx 12.95	ext{ cm}$. Therefore, the speed of the particle is $v=\frac{12.95	ext{ cm}}{\frac{1}{4}	ext{ s}} \approx 51.8	ext{ cm/s}$.

Plugging in $v=51.8	ext{ cm/s}$ and $r=2.06	ext{ cm}$ into the equation for acceleration gives:

$a=\frac{(51.8	ext{ cm/s})^2}{2.06	ext{ cm}} \approx 1300	ext{ cm/s}^2 = 13	ext{ m/s}^2$

Therefore, the magnitude of the acceleration is $\boxed{	extbf{(C)}\ 13	ext{ m/s}^2}$.

A Particle Moves at a Constant Speed in a Circular