Question

A racing car travels on a circular track of radius 250 m. assuming the car moves with a constant speed of 45.0 m/s, find (a) its angular speed and (b) the magnitude and direction of its acceleration

Answer 1

(a) The relationship between angular speed and tangential speed in a uniform circular motion is given by
$\omega= \frac{v}{r}$
where
$\omega$ is the angular speed
v is the tangential speed
r is the radius of the orbit

Using v=45.0 m/s and r=250 m, we find
$\omega= \frac{45.0 m/s}{250 m}=0.18 rad/s$


b) The car is moving in circular motion with constant tangential speed; it means there is no acceleration in the tangential direction. Instead, because it is a uniform circular motion, there is a centripetal acceleration toward the center of the circle, and this acceleration is given by:
$a_c = \frac{v^2}{r}= \frac{(45.0m/s)^2}{250 m}=8.1 m/s^2$
This is the magnitude of the acceleration, and its direction is toward the center of the orbit.

Answer 2

a. The angular speed of the car is equal to 0.18 rad/s.

b. The magnitude and direction of its acceleration is 8.1 $m/s^2$ towards the center of its orbit.

Given the following data:

• Radius of circular track = 250 meters.

• Speed = 45.0 m/s.

a. To calculate the angular speed of the car:

Mathematically, angular speed is given by this formula:

$\omega = \frac{V}{r}$

Where:

• $\omega$ is the angular speed.

• V is the speed of an object.

• r is the radius.

Substituting the given parameters into the formula, we have;

$\omega =\frac{45}{250} \\\\\omega =0.18\;rad/s$

b. To calculate the magnitude and direction of its acceleration:

Mathematically, centripetal acceleration is given by this formula:

$A = \frac{V^2 }{r}$

Where:

• r is the radius.

• V is the speed.

Substituting the given parameters into the formula, we have;

$A_c =\frac{45^2}{250} \\\\A_c =\frac{2025}{250}\\\\A_c =8.1\;m/s^2$

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