Find the linear speed \( \nu \) for the following. a point on the edge of a flywheel of radius 2 m , rotating 42 times per min The linear speed \( \nu \) for a point on the edge of a flywheel of radius 2 m , rotating 42 times per min is (Simplify your answer. Type an exact answer, using \( \pi \) as needed. Use integers or fractions for any numbers in the expression.

To find the linear speed \( \nu \) of a point on the edge of a flywheel, we can use the formula: \[ \nu = r \cdot \omega \] where: - \( r \) is the radius of the flywheel, - \( \omega \) is the angular speed in radians per second. First, we need to convert the rotational speed from revolutions per minute (rpm) to radians per second. 1. **Convert revolutions per minute to radians per second:** - There are \( 2\pi \) radians in one revolution. - There are 60 seconds in one minute. Thus, the angular speed \( \omega \) in radians per second is given by: \[ \omega = \text{(revolutions per minute)} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} \] Substituting the given value of 42 revolutions per minute: \[ \omega = 42 \times \frac{2\pi}{60} \] Calculating this: \[ \omega = 42 \times \frac{2\pi}{60} = \frac{84\pi}{60} = \frac{14\pi}{10} = \frac{7\pi}{5} \text{ radians per second} \] 2. **Now, substitute \( r \) and \( \omega \) into the linear speed formula:** - The radius \( r = 2 \) m. Thus, the linear speed \( \nu \) is: \[ \nu = r \cdot \omega = 2 \cdot \frac{7\pi}{5} \] Calculating this gives: \[ \nu = \frac{14\pi}{5} \text{ m/s} \] Therefore, the linear speed \( \nu \) for a point on the edge of the flywheel is: \[ \boxed{\frac{14\pi}{5}} \text{ m/s} \]