Question

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

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

Convert angular speed from rpm to rad/s

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

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.

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.

Key concepts

Angular Speed

Angular speed is a measure of how quickly an object rotates or revolves relative to another point, typically the center of a circle or axis of rotation. In the context of a rotating lawn mower blade, angular speed is quantified in terms of revolutions per minute (rpm) or radians per second (rad/s). It answers "how fast" something is spinning.

• To calculate angular speed in rad/s, you must convert from rpm to rad/s using the relationship: \[ \omega = ext{rpm} \times \frac{2 \pi \, \text{rad/rev}}{60 \, \text{s/min}} \]
• This conversion is key since most formulas in rotational dynamics use rad/s instead of rpm.
• In our example, the lawn mower blade's angular speed was given as 3400 rpm, which converted to 356.5 rad/s.

Understanding angular speed is fundamental when dealing with problems involving rotational motion, as it forms the basis for calculating other dynamic aspects such as linear speed and angular acceleration.

Linear Speed

Linear speed defines how fast an object is moving along a path, in a straight line, and it connects intimately with rotational motion through a seemingly simple concept. For objects rotating around an axis, like a lawn mower blade, the linear speed at a point on the edge of the rotating object can be calculated using the relationship:

• Linear speed \( v \) is found through the formula: \( v = r\omega \), where \( r \) is the radius from the center of rotation, and \( \omega \) is angular speed.
• This formula shows how rotational and translational motion relate, indicating that a point further away from the axis moves faster than one closer to it.
• In the solved example, the radius of the lawn mower blade is half its length, 0.265 meters, and when multiplied by the angular speed (356.5 rad/s), the linear speed of the blade's tip is 94.47 m/s.

Linear speed is crucial in understanding the efficiency and safety of rotating machinery, providing insights into how fast a point on the edge moves through space.

Angular Acceleration

Angular acceleration explains how the angular speed of a rotating object changes over time. It tells us how quickly an object speeds up or slows down its rotational motion. For cases like the lawn mower blade needing to stop, angular acceleration is critical to determine how fast it can come to a complete stop.

• Angular acceleration \( \alpha \) can be deduced using the angular kinematic equation: \( \omega_f = \omega_i + \alpha \Delta t \).
  • \( \omega_f \) is the final angular speed, and \( \omega_i \) is the initial angular speed.
  • \( \Delta t \) is the time interval over which the change occurs, and \( \alpha \) is what we solve for.
• If a blade needs to stop, set \( \omega_f = 0 \), rearrange the equation \( \alpha = \frac{-\omega_i}{\Delta t} \).
• In the lawn mower problem, given a time frame of 3 seconds and an initial angular speed of 356.5 rad/s, the calculated angular acceleration is -118.83 rad/s², indicating a deceleration.

Understanding angular acceleration helps predict how quickly machines or systems can alter their motion, which is vital for safety and efficiency.