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question 7
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a wheel with a 0.10 - m radius is rotating at 35 rev/s. it then slows uniformly to 15 rev/s over a 3.0 - s interval. what is the angular acceleration of a point on the wheel?
0.67 rev/s²
42 rev/s²

• 17 rev/s²
• 6.7 rev/s²
• 2.0 rev/s²

Explanation:

Step1: Recall angular acceleration formula

The formula for angular acceleration \(\alpha\) in rotational motion (when angular velocity changes uniformly) is \(\alpha=\frac{\omega_f - \omega_i}{t}\), where \(\omega_i\) is the initial angular velocity, \(\omega_f\) is the final angular velocity, and \(t\) is the time interval.

Step2: Identify given values

We are given \(\omega_i = 35\space rev/s\), \(\omega_f=15\space rev/s\), and \(t = 3.0\space s\).

Step3: Substitute values into the formula

Substitute the values into the formula: \(\alpha=\frac{15 - 35}{3.0}\)
First, calculate the numerator: \(15 - 35=- 20\)
Then, divide by the time: \(\alpha=\frac{-20}{3.0}\approx - 6.7\space rev/s^2\)

Answer:

\(-6.7\space rev/s^2\) (corresponding to the option with \(-6.7\space rev/s^2\))