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Everyday application: Suppose a yo-yo has a center shaft that has a \(0.250 \mathrm{cm}\) radius and that its string is being pulled. (a) If the string is stationary and the yo-yo accelerates away from it at a rate of \(1.50 \mathrm{m} / \mathrm{s}^{2},\) what is the angular acceleration of the yo-yo? (b) What is the angular velocity after 0.750 s if it starts from rest? (c) The outside radius of the yo-yo is \(3.50 \mathrm{cm}\). What is the tangential acceleration of a point on its edge?

Short Answer

Expert verified
The angular acceleration of the yo-yo is \(600 \mathrm{rad/s^2}\). The angular velocity after 0.750 s, starting from rest, is \(450 \mathrm{rad/s}\). The tangential acceleration of a point on the yo-yo's edge is \(21 \mathrm{m/s^2}\).

Step by step solution

01

Convert radius to meters

To work with standard SI units, convert the radius of the center shaft from centimeters to meters. Since there are 100 cm in a meter, the radius in meters is: \(0.250 \mathrm{cm} \times (1 \mathrm{m}/100 \mathrm{cm}) = 0.00250 \mathrm{m}\).
02

Calculate angular acceleration

Using the relationship between tangential acceleration \(a_t\) and angular acceleration \(\alpha\), where \(a_t = \alpha \times r\), solve for \(\alpha\): \(\alpha = a_t / r = 1.50 \mathrm{m/s^2} / 0.00250 \mathrm{m} = 600 \mathrm{rad/s^2}\).
03

Determine angular velocity after given time

Use the kinematic equation for angular motion, \(\omega = \omega_0 + \alpha t\), where \(\omega_0\) is the initial angular velocity which is 0 since it starts from rest: \(\omega = 0 + (600 \mathrm{rad/s^2}) \times (0.750 \mathrm{s}) = 450 \mathrm{rad/s}\).
04

Convert outside radius to meters

Convert the outside radius of the yo-yo from centimeters to meters similarly as in Step 1: \(3.50 \mathrm{cm} \times (1 \mathrm{m}/100 \mathrm{cm}) = 0.0350 \mathrm{m}\).
05

Calculate tangential acceleration at the outside edge

Apply the same tangential-to-angular relationship as in Step 2 at the outside edge: \(a_t = \alpha \times r_{outside} = 600 \mathrm{rad/s^2} \times 0.0350 \mathrm{m} = 21 \mathrm{m/s^2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations
Understanding kinematic equations is essential for solving problems involving motion. These equations relate the variables of motion: displacement, velocity, acceleration, and time. When dealing with rotational motion, we have analogous kinematic equations, where angular displacement, angular velocity, and angular acceleration are used.
In the yo-yo problem, we used the rotational kinematic equation ewline ewline \[ \omega = \omega_0 + \alpha t \]ewline ewline to find the angular velocity after 0.750 seconds. Here, \( \omega \) stands for final angular velocity, \( \omega_0 \) is the initial angular velocity, \( \alpha \) represents angular acceleration, and \( t \) is the time elapsed. Since the yo-yo starts from rest, \( \omega_0 \) is zero. By substituting the given values, we can solve for the final angular velocity.
Tangential Acceleration
Tangential acceleration is the rate at which the tangential velocity of a point in a rotational system changes. In simpler terms, it's how quickly a point on a rotating object is speeding up or slowing down along its path of motion. It's called 'tangential' because the direction of acceleration is tangent to the circle of rotation at any given point.
To find the tangential acceleration, we relate it to angular acceleration using the formula:ewline ewline \[ a_t = \alpha \times r \]ewline ewline where \( a_t \) is tangential acceleration, \( \alpha \) is angular acceleration, and \( r \) is the radius of rotation. In the exercise, by calculating the angular acceleration and knowing the radius of the yo-yo's outside edge, we found the tangential acceleration, demonstrating the direct proportionality between angular acceleration and tangential acceleration.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves relative to another point, namely the center of rotation. It's usually measured in radians per second (rad/s). Angular velocity is a vector, meaning it has both a magnitude (how fast) and a direction (clockwise or counter-clockwise).
In the context of our yo-yo example, we calculated the angular velocity after a specific time from rest. The initial angular velocity (\( \omega_0 \)) was zero, as the yo-yo was not moving initially. By using angular acceleration (\( \alpha \)) which we previously found and the time (\( t \)), we applied the angular version of the kinematic equation to find the final angular velocity. The time factor in the equation illustrates that angular velocity increases as time passes with a constant angular acceleration.

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Most popular questions from this chapter

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of \(2.00 \times 10^{3} \mathrm{N}\) with an effective perpendicular lever arm of \(3.00 \mathrm{cm},\) producing an angular acceleration of the forearm of \(120 \mathrm{rad} / \mathrm{s}^{2}\). What is the moment of inertia of the boxer's forearm?

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