Question

a tire of 0.61 m diameter is placed on a balancing machine where it is spun so that its tread is moving at a constant speed of 26.8 m/s. a small stone is stuck in the tread of the tire. what is the acceleration of the stone as the tire is being balanced?

Answer 1

The acceleration of the stone stuck in the tire tread is 720.33 m/s^2.

Step-by-step explanation:

The problem gives us the diameter of the tire (0.61 m) and the speed of the tire tread (26.8 m/s), and asks us to find the acceleration of the stone stuck in the tread.

Since the stone is stuck in the tread, we can assume that the stone moves with the same speed as the tread.

We can use the formula for acceleration to solve this problem. The formula for acceleration is a = (v^2)/r, where v is the velocity and r is the radius of the circular path.

In this case, the velocity is 26.8 m/s and the radius is half of the diameter, which is 0.305 m.

Plugging in these values, we get:

a = (26.8 m/s)^2 / 0.305 m

= 720.33 m/s^2

The acceleration of the stone is 720.33 m/s^2.