Question

1. a tire 0.500 m in radius rotates at a constant rate of 200 rev/min. find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge).

Explanation:

Step1: Convert angular velocity

The angular velocity \(\omega\) is given in revolutions per minute. First, convert it to radians per second.
We know that \(1\) revolution \( = 2\pi\) radians and \(1\) minute \(=60\) seconds.
\(\omega=200\frac{\text{rev}}{\text{min}}\times\frac{2\pi\text{ rad}}{1\text{ rev}}\times\frac{1\text{ min}}{60\text{ s}}=\frac{20\pi}{3}\text{ rad/s}\)

Step2: Calculate the linear speed \(v\)

The formula for linear speed \(v = r\omega\), where \(r = 0.500\) m and \(\omega=\frac{20\pi}{3}\text{ rad/s}\)
\(v=(0.500)\times\frac{20\pi}{3}=\frac{10\pi}{3}\approx10.5\text{ m/s}\)

Step3: Calculate the centripetal acceleration \(a_c\)

The formula for centripetal acceleration \(a_c=r\omega^{2}\)
Substitute \(r = 0.500\) m and \(\omega=\frac{20\pi}{3}\text{ rad/s}\)
\(a_c=(0.500)\times(\frac{20\pi}{3})^{2}=\frac{0.500\times400\pi^{2}}{9}=\frac{200\pi^{2}}{9}\approx219\text{ m/s}^{2}\)

Answer:

The speed of the stone is approximately \(10.5\) m/s and the acceleration is approximately \(219\) m/s².