/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 (II) A small rubber wheel is use... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The small wheel has a radius of 2.0 \(\mathrm{cm}\) and accelerates at the rate of 7.2 \(\mathrm{rad} / \mathrm{s}^{2}\) and it is in contact with the pottery wheel (radius 21.0 \(\mathrm{cm} )\) without slipping. Calculate \((a)\) the angular acceleration of the pottery wheel, and \((b)\) the time it takes the pottery wheel to reach its required speed of 65 rpm.

Short Answer

Expert verified
(a) The angular acceleration of the pottery wheel is 0.6857 rad/s². (b) It takes approximately 9.93 seconds to reach 65 rpm.

Step by step solution

01

Understand the Relationship Between Wheels

Both wheels are in contact without slipping. This means that their linear velocities at the edge are the same. Given that the small wheel's angular velocity is denoted by \( \omega_s \) and that of the pottery wheel by \( \omega_p \), and using the radius \( r_s = 2.0\, \text{cm} \) and \( r_p = 21.0\, \text{cm} \), we have the linear velocity equation:\[\omega_s \cdot r_s = \omega_p \cdot r_p\]This implies a relationship between the angular accelerations, \( \alpha_s = 7.2\, \text{rad/s}^2 \) and \( \alpha_p \).
02

Calculate the Angular Acceleration of the Pottery Wheel

Using the relationship derived in Step 1: \[\alpha_s \cdot r_s = \alpha_p \cdot r_p\]we solve for \( \alpha_p \):\[\alpha_p = \frac{\alpha_s \cdot r_s}{r_p} = \frac{7.2\, \text{rad/s}^2 \times 2.0\, \text{cm}}{21.0\, \text{cm}} = \frac{14.4}{21} = 0.6857\, \text{rad/s}^2\]Thus, the angular acceleration of the pottery wheel is \( 0.6857\, \text{rad/s}^2 \).
03

Convert Desired Speed to Radians per Second

The required speed of the pottery wheel is given as \( 65 \text{ rpm} \). We convert this to \( \text{rad/s} \) using the conversion factor \( 2\pi/60 \):\[\omega_p = 65 \times \frac{2\pi}{60} = \frac{65 \times 2\pi}{60} = 6.8068 \text{ rad/s}\]This is the final angular velocity of the pottery wheel.
04

Calculate the Time to Reach Final Speed

Using the equation of motion for angular acceleration:\[\omega_p = \alpha_p \cdot t + \omega_{p0}\]where \( \omega_{p0} = 0 \) (initial angular velocity), we find:\[6.8068 = 0.6857 \times t\]Solving for \( t \):\[t = \frac{6.8068}{0.6857} \approx 9.93\, \text{seconds}\]Thus, it takes approximately 9.93 seconds to reach the required speed.

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Key Concepts

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

Linear Velocity
Linear velocity is a fundamental concept when analyzing systems involving motion, especially in cases like interconnected wheels in machinery. It describes the speed and direction at which a point on the perimeter of a rotating object moves. In cases like the rubber wheel and pottery wheel situation described in the exercise, linear velocity at the point of contact is identical for both wheels because they are slipping-free.
When a circular object rotates, every point on its circumference experiences linear velocity. This velocity, denoted by \( v \), is related to the angular velocity \( \omega \) and the radius of the circle \( r \) through the equation:\[v = \omega \cdot r\]This equation signifies that the linear velocity is directly proportional to both the angular velocity and the radius. Consequently, when two wheels touch, their linear velocities at the edge are equal, leading to the relationship \( \omega_s \cdot r_s = \omega_p \cdot r_p \) observed in the exercise.
Understanding this relationship allows one to derive other important rotational parameters of the system, such as angular acceleration, as seen in the solution steps.
Rotational Motion
Rotational motion refers to the movement of an object in a circular path around a central point. This type of motion is highly prevalent in mechanical systems and understanding it is crucial for solving problems like the one involving both the small rubber wheel and the large pottery wheel.
When an object experiences rotational motion, there are a few key parameters to consider:
  • Angular Velocity (\( \omega \)): The rate at which an object rotates, typically measured in radians per second. It describes how fast the object spins around the axis.
  • Angular Acceleration (\( \alpha \)): The rate of change of angular velocity. It tells us how quickly an object's rotation speed increases or decreases.
  • Moment of Inertia: A property that quantifies how difficult it is to change an object's rotation. However, this isn't explicitly needed for this particular problem.
In the exercise, we dealt with angular acceleration, calculated by considering the relation between the small wheel and the pottery wheel due to their contact. By understanding rotational motion, we can predict system behaviors such as reaching a specific angular velocity over time, as solved in the solution.
Radians per Second
Radians per second (\( \text{rad/s} \)) is a unit of measure for angular velocity, which indicates how much angle an object traverses per unit time during rotational motion. It is the standard unit for these measurements because it directly relates the angular displacement and the time interval.
Radians themselves are a natural unit for measuring angular distances, as they relate the arc length of a circle to the radius. One complete revolution around a circle equals \( 2\pi \) radians, thus making it a convenient unit for motion calculations.
In rotational motion, converting units like rotations per minute (rpm) into radians per second is common. For example, in our exercise, the pottery wheel’s speed is given as 65 rpm, which must be converted to radians per second for accurate calculations. The conversion formula:\[\omega = \text{rpm} \times \frac{2\pi}{60}\]ensures the calculations stay consistent with standard scientific measurements, allowing us to work seamlessly across different fields of physics and engineering. Understanding this conversion and unit helps solve rotational problems accurately.

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