Question

At t1 = 1.00 s, the acceleration of a particle moving at constant speed in counterclockwise circular motion is At t2 = 2.00 s (less than one period later), the acceleration is The period is more than 1.00 s. What is the radius of the circle?

Answer 1

Answer:

$2.925$ meters

Explanation:

See the remaining part of required information in the attached image

Solution

The formula for calculating velocity is equal to

$v = \frac{r\theta}{t_2 - t_1}$

Where v is the velocity, $t_1, t_2,\theta$ are time range for two acceleration vectors and angle between two acceleration vector respectively and r is the radius of the circle

Putting the give values in above equation we get

$v = \frac{r * \frac{\pi }{2} }{2 -1} \\v = \frac{r\pi }{2}$

Acceleration is equal to

$\sqrt{6^2 + 4^2} \\\sqrt{36+16} \\\sqrt{52} \\= 7.21$

Radius is equal to

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

Putting the given values we get -

$r = \frac{(\frac{r\pi }{2})^2}{7.211} \\r = \frac{4 * 7.211 }{\pi^2 } \\r = 2.925$