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Chapter 9: Problem 58

Consider a \(53-\mathrm{cm}\) -long lawn mower blade rotating about its center at 3400 rpm. a) Calculate the linear speed of the tip of the blade. b) If safety regulations require that the blade be stoppable within \(3.0 \mathrm{~s}\), what minimum angular acceleration will accomplish this? Assume that the angular acceleration is constant.

Short Answer

Expert verified
Answer: The linear speed of the tip of the lawn mower blade is 94.47 m/s. The minimum angular acceleration needed to stop the blade within 3 seconds is -118.83 rad/s².

Step by step solution

01

Convert angular speed from rpm to rad/s

To convert the given angular speed of 3400 rpm, we use the conversion factor \(\frac{2 \pi \text{ rad/rev}}{60 \, \text{s/min}}\): \(\omega = 3400 \, \text{rpm} \times \frac{2 \pi \text{ rad/rev}}{60 \, \text{s/min}} = 356.5\, \text{rad/s}\)
02

Calculate the linear speed of the tip of the blade

Using the formula for linear speed: \(v = r\omega\), where \(r\) is the radius of the blade (half its length) and \(\omega\) is the angular speed, we have: \(r = \frac{53 \, \text{cm}}{2} = 26.5 \, \text{cm} = 0.265 \, \text{m}\) (converting cm to m) \(v = (0.265 \, \text{m})(356.5 \, \text{rad/s}) = 94.47 \, \text{m/s}\), which is the linear speed of the tip of the blade.
03

Use the angular kinematic equation to find the minimum angular acceleration

The angular kinematic equation is: \(\omega_f = \omega_i + \alpha \Delta t\). Since the blade needs to be stoppable, the final angular speed (\(\omega_f\)) should be 0 rad/s. We are given the initial angular speed (\(\omega_i = 356.5\, \text{rad/s}\)) and the time interval (\(\Delta t = 3.0 \, \text{s}\)). Solving for \(\alpha\): \(0 = 356.5\, \text{rad/s} + \alpha (3.0 \, \text{s})\) \(\alpha = - \frac{356.5\, \text{rad/s}}{3.0 \, \text{s}} = -118.83 \, \text{rad/s}^2\) The minimum angular acceleration needed to stop the blade within 3 seconds is -118.83 rad/s².

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