Question

A circular saw blade with radius 0.120 m starts from rest and turns in a vertical plane with a constant angular acceleration of 2.00 rev/s\(^2\). After the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

Short answer

The piece travels approximately 0.409 times the tangential velocity horizontally.

Step-by-step solution

Convert Revolutions to Radians

To analyze rotational motion, we first need to convert revolutions to radians because angular displacement is typically measured in radians. Since there are \( 2\pi \) radians in one revolution, we multiply the given number of revolutions by \( 2\pi \) to find the angular displacement in radians:\[ \theta = 155 \times 2\pi = 310\pi \text{ rad} \]

Calculate Final Tangential Velocity

Using the formula for rotational kinematics, we can find the final angular velocity \( \omega_f \) after 155 revolutions. The kinematic equation is:\[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]where \( \omega_i = 0 \) (starting from rest), \( \alpha = 2 \times 2\pi \text{ rad/s}^2\), and \( \theta = 310\pi \text{ rad} \). Thus,\[ \omega_f^2 = 0 + 2 \times 2 \times 2\pi \times 310\pi = 8\pi^2 \times 310 \]\[ \omega_f = \sqrt{8 \pi^2 \times 310} \]\[ \omega_f = \sqrt{2480\pi^2} \]The tangential velocity \( v_t \) at the rim of the circular saw is given by \( v_t = r \times \omega_f \), where \( r = 0.120 \text{ m} \):\[ v_t = 0.120 \times \omega_f \]

Use Tangential Velocity in Projectile Motion

The tangential velocity \( v_t \) is the initial horizontal velocity of the piece as it breaks off the blade. We need to find out how long it takes for the piece to hit the floor using the vertical motion. The piece falls a vertical distance of 0.820 m, and using the equation of motion under gravity:\[ h = \frac{1}{2} g t^2 \]where \( h = 0.820 \) m and \( g = 9.81 \text{ m/s}^2 \):\[ 0.820 = \frac{1}{2} \times 9.81 \times t^2 \]Solve for \( t \):\[ t^2 = \frac{0.820}{4.905} \]\[ t = \sqrt{0.1671} \]\[ t \approx 0.409 \text{ s} \]

Calculate Horizontal Distance Traveled

Now that we have the time \( t \) the piece is in the air, we use the earlier found tangential velocity to find the horizontal distance \( d \):The formula for horizontal distance is:\[ d = v_t \times t \]Substitute the values for \( v_t \) and \( t \) to find \( d \).

Key concepts

Angular Displacement

Angular displacement is crucial to understanding how much an object has rotated around a circle. It's typically expressed in radians, a unit based on the radius of a circle. In this exercise, the rotational movement of the circular saw blade is analyzed through 155 revolutions. Since one full revolution is equal to \(2\pi\) radians, we need to convert revolutions into radians to make calculations possible.
For instance, converting 155 revolutions to radians involves multiplying the number of revolutions by \(2\pi\): \(\theta = 155 \times 2\pi = 310\pi\) radians. This conversion gives a precise measure of the angular displacement and sets the stage for further calculations involving rotational kinematics.
Grasping angular displacement is key for further understanding how rotational motion translates into tangential or linear movements, especially when dealing with rotating objects transitioning into different forms of motion, like our saw blade piece.

Angular Velocity

Angular velocity describes how quickly an object rotates. In the context of the saw blade, it started from rest and accelerated due to a constant angular acceleration. Calculating the angular velocity at any given point requires understanding initial conditions and the effect of angular acceleration over time.
Using the equation \(\omega_f^2 = \omega_i^2 + 2\alpha\theta\), where \(\omega_i\) is the initial angular velocity (here, zero), \(\alpha\) is the angular acceleration, and \(\theta\) is the angular displacement, we solve for the final angular velocity \(\omega_f\).
This final angular velocity can be converted to tangential velocity using the relationship \(v_t = r \times \omega_f\), where \(r\) is the radius. Making these conversions connects rotational motion with linear, setting up for analyzing when the saw blade piece breaks off and becomes a projectile, with its initial horizontal velocity being this tangential velocity.

Projectile Motion

Projectile motion is all about understanding how objects move when they are launched into the air and acted upon by gravity alone. For the saw blade piece, once it breaks off, it transitions from rotational to linear motion, traveling horizontally while falling vertically towards the ground.
Key to solving projectile motion problems is separating the horizontal and vertical components of motion. The horizontal distance traveled depends on the initial horizontal speed and the time in the air. Using the equation \(h = \frac{1}{2} g t^2\), the time required for the piece to hit the floor can be calculated. Here, \(h\) is the vertical distance (0.820 m), and \(g\) is the acceleration due to gravity (9.81 m/s²).
After finding the time the piece stays in the air, we calculate its horizontal travel using \(d = v_t \times t\). This formula uses the initial horizontal velocity (equal to the tangential velocity from the saw blade) and the time to determine how far it moves horizontally before hitting the ground. Combining these calculations reveals the path traveled by the piece once it shifts from circular to straight-line motion.