Question

a wheel with a tire mounted on it rotates at the constant rate of 2.63 revolutions per second. a tack is stuck in the tire at a distance of 31.9 cm from the rotation axis. what is the tack’s tangential speed v. v = \boxed{ } m/s what is the tack’s acceleration magnitude a? a = \boxed{ } m/s²

Explanation:

Tangential Speed Calculation

Step1: Find angular velocity (ω)

Angular velocity ω is related to frequency f by \( \omega = 2\pi f \). Given \( f = 2.63 \, \text{rev/s} \), so \( \omega = 2\pi \times 2.63 \, \text{rad/s} \).
\( \omega = 2 \times 3.1416 \times 2.63 \approx 16.52 \, \text{rad/s} \)

Step2: Convert radius to meters

Radius \( r = 31.9 \, \text{cm} = 0.319 \, \text{m} \)

Step3: Calculate tangential speed (v)

Tangential speed \( v = r\omega \). Substitute \( r = 0.319 \, \text{m} \) and \( \omega \approx 16.52 \, \text{rad/s} \).
\( v = 0.319 \times 16.52 \approx 5.27 \, \text{m/s} \)

Centripetal Acceleration Calculation

Step1: Use centripetal acceleration formula

Centripetal acceleration \( a = \frac{v^2}{r} \) (or \( a = r\omega^2 \)). We can use \( a = r\omega^2 \) with \( r = 0.319 \, \text{m} \) and \( \omega \approx 16.52 \, \text{rad/s} \).

Step2: Substitute values

\( a = 0.319 \times (16.52)^2 \)
First, calculate \( (16.52)^2 \approx 272.91 \)
Then, \( a = 0.319 \times 272.91 \approx 87.1 \, \text{m/s}^2 \)

Answer:

(Tangential Speed):
\( \boxed{5.27} \) (approximate, depending on π precision)