Question

Consider a \(53.0-\mathrm{cm}\) -long lawn mower blade rotating about its center at \(3400 . \mathrm{rpm}\) a) Calculate the linear speed of the tip of the blade. b) If safety regulations require that the blade be stoppable within \(3.00 \mathrm{~s}\), what minimum angular acceleration will accomplish this? Assume that the angular acceleration is constant.

Short answer

The linear speed of the tip of the blade is calculated as follows: \(v = \omega_\text{radian} \times \frac{L}{2} = \frac{3400}{60} \times 2\pi \times \frac{0.53}{2} \mathrm{m/s}\) = \(35.49\mathrm{m/s}\) b) What is the minimum angular acceleration required to stop the blade within 3.00 seconds? The minimum angular acceleration required to stop the blade is calculated as follows: \(\alpha = \frac{\omega_\text{final} - \omega_\text{initial}}{t} = \frac{0 - \frac{3400}{60} \times 2\pi}{3.00} \mathrm{rad/s^2}\) = \(-35.66\mathrm{rad/s^2}\) Note that the negative sign indicates that the angular acceleration is in the opposite direction of the initial positive rotation, acting to slow the blade down.

Step-by-step solution

Identify the knowns and the unknowns

For part a), we know the following: - Length of the blade: \(L = 53.0 \mathrm{cm}\) (I will convert it to meters later) - Rotations per minute: \(3400 \mathrm{rpm}\) The unknown we want to find is the linear speed of the tip of the blade: \(v\) For part b), we already know the rotational speed and now we want to find the minimum angular acceleration required to stop the blade within 3.00 seconds: \(\alpha\)

Convert the length of the blade to meters

We will need to work with the length in meters. To convert the length from centimeters to meters, we will divide by 100: \(L = 53.0 \mathrm{cm} \times \frac{1 \mathrm{m}}{100 \mathrm{cm}} = 0.53 \mathrm{m}\)

Convert rotations per minute to radians per second

We are given the rotational speed in rpm, but we need it to be in radians per second. First, we will convert it to rotations per second, and then to radians per second. rotations per second: \(\omega_\text{second} = 3400 \mathrm{rpm} \times \frac{1 \mathrm{rotation}}{60 \mathrm{s}} = \frac{3400}{60} \mathrm{rotation/s}\) Now we'll convert it to radians per second by multiplying by \(2\pi\) (since there are \(2\pi\) radians in one rotation): \(\omega_\text{radian} = \omega_\text{second} \times 2\pi = \frac{3400}{60} \times 2\pi \mathrm{rad/s}\)

Calculate the linear speed of the tip of the blade

We can calculate the linear speed of the tip of the blade using the following formula: \(v = \omega \times r\) where \(r\) is the radius of the circle created by the tip of the blade (which is half of the length of the blade). \(v = \omega_\text{radian} \times \frac{L}{2} = \frac{3400}{60} \times 2\pi \times \frac{0.53}{2} \mathrm{m/s}\) Now, calculate the result.

Calculate the angular acceleration required to stop the blade

We are given the required time to stop the blade: \(t = 3.00\mathrm{s}\) When the blade is stopped, the final angular velocity will be zero: \(\omega_\text{final} = 0\) To find the angular acceleration required to reach a final angular velocity within the given time, we can use the following equation: \(\omega_\text{final} = \omega_\text{initial} + \alpha \cdot t\) Rearrange to solve for the unknown angular acceleration: \(\alpha = \frac{\omega_\text{final} - \omega_\text{initial}}{t} = \frac{0 - \frac{3400}{60} \times 2\pi}{3.00} \mathrm{rad/s^2}\) Calculate the result. Now we have completed the exercise and solved for the linear speed of the tip of the blade in part a) and the minimum angular acceleration required to stop the blade in part b).

Key concepts

Linear Speed

Linear speed refers to the distance traveled by a point moving along a path in a given time interval. It is particularly relevant when analyzing the movement of objects in rotational motion, such as the tip of a lawnmower blade. Calculating the linear speed involves determining how fast the point is moving along the circumference of the circle it creates as it rotates.

Using the formula \(v = \text{omega} \times r\), where \(v\) is linear speed, \(\text{omega}\) is the angular velocity in radians per second, and \(r\) is the radius, students can understand the direct relationship between linear speed and radial distance from the center of rotation. As the radius increases, so does the linear speed for a constant angular velocity.

For a blade of length 53.0 cm, which is the diameter of the rotational path, calculating the radius as half the length gives us 0.53/2 meters. By finding the angular velocity in radians per second, which is another crucial concept, and multiplying by the radius, one can calculate the linear speed of the tip of the blade, thus applying the relationship between rotational movement and linear speed.

Rotational Motion

Rotational motion is the movement of an object around a center or pivot point. In this context, it is the circular motion enacted by a lawn mower blade spinning about its axis. Each point on the blade follows a circular track, with points further from the pivot covering more distance compared to those closer to the center, despite sharing the same angular velocity.

In rotational motion, the object's speed can be described either in terms of linear speed, as addressed previously, or angular velocity, which emphasizes the rate of rotation. Understanding the linkage between these two descriptions is essential for grasping concepts in physics concerns with rotating objects. Such comprehension is pivotal when interpreting scenarios like the stopping time of a mower's blade due to safety regulations.

Considering the angular acceleration and its effect on the blade's rotational motion, the frame of reference is generally the axis of rotation. The blade's velocity is tangential to its circular path, while the acceleration directs towards the center (centripetal) or away from the center when slowing down (centrifugal).

Radians per Second

Radians per second is a measure of angular velocity, representing the rate of change of the angular displacement over time. A full rotation corresponds to an angle of \(2\pi\) radians. When working with rotational systems like the lawn mower blade, conversion from practical units like revolutions per minute (rpm) to radians per second is a common requirement.

Understanding the conversion process is essential. For instance, the given angular velocity of 3400 rpm is first converted to rotations per second by dividing by 60, since there are 60 seconds in a minute. This intermediate step is followed by multiplying by \(2\pi\) to transition from rotations to radians, exploiting the fact that one complete rotation encompasses \(2\pi\) radians.

Having a firm grasp of this measure allows one to work seamlessly within the framework of rotational dynamics, notably when it comes to quantifying angular acceleration, which similarly uses the unit radians per second squared (\(\mathrm{rad/s^2}\)). Comprehension of these concepts facilitates solving problems involving changing rotational speeds, such as determining the necessary angular acceleration to halt a rotating blade within a specified time frame.