91Ó°ÊÓ

Angular acceleration, denoted as \( \alpha \), is the rate at which the angular velocity of an object changes with time. It's a crucial concept in understanding circular motion physics.
In this exercise, we relate angular acceleration to tangential acceleration \( a_t \) through the formula:

• \( a_t = \alpha \times r \)

where \( r \) is the radius of the circular path. This equation indicates how a change in speed along a circle's edge affects rotational movement.
Given that for our car \( a_t = 2.00 \, \text{m/s}^2 \) and \( r = 60.0 \, \text{m} \), we find \( \alpha \) by rearranging the equation to:

• \( \alpha = \frac{a_t}{r} \)
• \( \alpha = 0.0333 \, \text{rad/s}^2 \)

This result tells us that for every second, the angular velocity of the car increases by \(0.0333\) radians per second.
Understanding how forces and motions work on a curved path helps demystify many real-world applications, from car dynamics to amusement park rides.

Radial acceleration, also known as centripetal acceleration, is the acceleration that is directed towards the center of the circle along which an object is moving. It does not change the speed but rather the direction of the object in motion.
The formula to calculate radial acceleration \( a_r \) is:

• \( a_r = \omega^2 \times r \)

where \( \omega \) is the angular speed, and \( r \) is the radius. After 6 seconds, our car reaches an angular speed of \( 0.20 \, \text{rad/s} \), leading to:

• \( a_r = (0.20)^2 \times 60.0 = 2.40 \, \text{m/s}^2 \)

This acceleration keeps the car moving in a circular path instead of flying off tangent. It's fascinating to think about how such forces constantly act on objects, ensuring they maintain their circular motion.

In physics, net force refers to the overall force acting on an object. For an object moving in a circle, both tangential and radial accelerations combine to produce a total accelerative effect.
To find the total acceleration \( a_{\text{total}} \), we combine the tangential and radial components:

• \( a_{\text{total}} = \sqrt{a_t^2 + a_r^2} \)
• \( a_{\text{total}} = \sqrt{(2.00)^2 + (2.40)^2} = 3.13 \, \text{m/s}^2 \)

The net force \( F \), acting on the car is then given by:

• \( F = m \times a_{\text{total}} \)
• \( F = 1240 \times 3.13 = 3881.2 \, \text{N} \)

This force ensures the car remains in circular motion while accelerating. Understanding net force is essential as it translates theoretical physics concepts into real-world implications, like ensuring the stability and safety of vehicles on circular tracks.