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Chapter 8: Problem 49

A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as the drawing shows. The centripetal acceleration measured at corner \(A\) is \(n\) times as great as that measured at corner \(B\). What is the ratio \(L_{1} / L_{2}\) of the lengths of the sides of the rectangle when \(n=2.00 ?\)

Short Answer

Expert verified
The ratio \( \frac{L_1}{L_2} \) is \( \sqrt{3} \).

Step by step solution

01

Understanding the Problem

A rectangular plate rotates around one of its corners at a constant angular speed \( \omega \). The centripetal acceleration at corner \( A \) is \( n \) times that at corner \( B \), and \( n = 2.00 \). We need to find the ratio \( \frac{L_1}{L_2} \) of the sides of the rectangle.
02

Expressing Centripetal Acceleration

The centripetal acceleration \( a_c \) at any point on a rotating object is given by \( a_c = \omega^2 r \), where \( r \) is the distance from the rotation axis. For corner \( A \), \( r = \sqrt{L_1^2 + L_2^2} \), and for corner \( B \), \( r = L_2 \).
03

Formulating the Ratio Equation

Given that \( a_A = n \times a_B \), we express this as: \( \omega^2 \sqrt{L_1^2 + L_2^2} = n \times \omega^2 L_2 \). This simplifies to \( \sqrt{L_1^2 + L_2^2} = nL_2 \).
04

Isolating Terms

By squaring both sides, we have: \( L_1^2 + L_2^2 = n^2 L_2^2 \). Simplifying, it becomes \( L_1^2 = (n^2 - 1) L_2^2 \).
05

Finding the Ratio

Taking the square root of both sides: \( L_1 = \sqrt{n^2 - 1}L_2 \). Therefore, the ratio \( \frac{L_1}{L_2} = \sqrt{n^2 - 1} \).
06

Substituting the Value of n

With \( n = 2.00 \), we substitute into the equation: \( \frac{L_1}{L_2} = \sqrt{2^2 - 1} = \sqrt{4 - 1} = \sqrt{3} \).

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Key Concepts

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

Rotational motion
Rotational motion refers to the movement of an object around a fixed point or axis. When an object spins around this axis, every point on the object moves in a circular path.
Rotational motion is different from linear motion where objects move along a straight line. In the context of our rectangular plate, rotational motion occurs because it spins around one of its corners. Here are some key points to understand about rotational motion:
  • Axis of rotation: This is the line around which the object rotates. It can pass through any part of the object, in this problem, it passes perpendicularly through one corner of the rectangular plate.
  • Path of rotation: Each point on the plate moves in a circle around this axis.
  • Angular speed: This is a measure of how fast the object is rotating, expressed in radians per second.
Understanding the concepts of rotational motion helps determine other aspects like centripetal acceleration and angular momentum.
Centripetal acceleration
Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It is required to keep the object moving in its curved trajectory.The formula for centripetal acceleration is given by:\[ a_c = \omega^2 r \]where:
  • \( a_c \) is the centripetal acceleration.
  • \( \omega \) is the angular speed.
  • \( r \) is the distance from the axis of rotation.
In our problem, the plate's corner A has a centripetal acceleration that is twice that of corner B. This difference is due to their distances from the axis of rotation. Understanding this concept is key to solving the problem and identifying how acceleration impacts both corners differently.
Rectangular plate
A rectangular plate, in this scenario, is a flat, two-dimensional shape with opposite sides equal in length. The salient features that are relevant to rotational motion are:
  • Dimensions: It has two sides of lengths \( L_1 \) and \( L_2 \), creating the rectangle's width and height.
  • Mass distribution: The entire mass of the rectangle contributes to its rotational inertia, which affects how it rotates around the corner.
  • Rotation: When the rectangle rotates about a point - such as a corner - each point on the rectangle traces out its own circular path.
The problem focuses on how the lengths of the sides affect the rotation and thus influence the ratio \( L_1 / L_2 \). By understanding these characteristics, we can see how changes in dimensions affect their rotational characteristics when fixed at a corner.
Angular speed
Angular speed is a measure of how fast an object rotates or revolves around an axis. It describes the rate of rotation and is commonly measured in radians per second. For the rectangular plate in this exercise:
  • Constant angular speed: The plate rotates at a constant rate; meaning the speed at which it turns is unchanged over time.
  • Influence on acceleration: As the angular speed is used to calculate centripetal acceleration, an understanding of its constancy helps predict how forces act on different parts of the plate.
The concept of angular speed links to how all points on the rotating body maintain their path speed relative to each other, which is crucial in evaluating the overall motion dynamics and solving rotational problems.

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Most popular questions from this chapter

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.00 \(\mathrm{rad} / \mathrm{s}^{2} .\) Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of \(-25.0 \mathrm{rad} / \mathrm{s} .\) While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

A compact disc (CD) contains music on a spiral track. Music is put onto a CD with the assumption that, during playback, the music will be detected at a constant tangential speed at any point. since \(v_{\mathrm{T}}=r \omega, \mathrm{a} \mathrm{CD}\) rotates at a smaller angular speed for music near the outer edge and a larger angular speed for music near the inner part of the disc. For music at the outer edge \((r=0.0568 \mathrm{m}),\) the angular speed is 3.50 rev/s. Find \((\mathrm{a})\) the constant tangential speed at which music is detected and (b) the angular speed (in rev/s) for music at a distance of \(0.0249 \mathrm{m}\) from the center of a CD.

The table that follows lists four pairs of initial and fi nal angular velocities for a rotating fan blade. The elapsed time for each of the four pairs of angular velocities is 4.0 s. For each of the four pairs, fi nd the average angular acceleration (magnitude and direction as given by the algebraic sign of your answer). $$ \begin{array}{lcc} & \text { Initial angular } & \text { Final angular } \\ & \text { velocity } \omega_{0} & \text { velocity } \omega \\ \text { (a) } & +2.0 \mathrm{rad} / \mathrm{s} & +5.0 \mathrm{rad} / \mathrm{s} \\ \hline \text { (b) } & +5.0 \mathrm{rad} / \mathrm{s} & +2.0 \mathrm{rad} / \mathrm{s} \\ \hline \text { (c) } & -7.0 \mathrm{rad} / \mathrm{s} & -3.0 \mathrm{rad} / \mathrm{s} \\ \hline \text { (d) } & +4.0 \mathrm{rad} / \mathrm{s} & -4.0 \mathrm{rad} / \mathrm{s} \\ \hline \end{array} $$

A racing car travels with a constant tangential speed of \(75.0 \mathrm{m} / \mathrm{s}\) around a circular track of radius \(625 \mathrm{m}\). Find (a) the magnitude of the car's total acceleration and (b) the direction of its total acceleration relative to the radial direction.

A motorcycle, which has an initial linear speed of \(6.6 \mathrm{m} / \mathrm{s}\), decelerates to a speed of \(2.1 \mathrm{m} / \mathrm{s}\) in \(5.0 \mathrm{s}\). Each wheel has a radius of \(0.65 \mathrm{m}\) and is rotating in a counterclockwise (positive) direction. What are (a) the constant angular acceleration (in \(\mathrm{rad} / \mathrm{s}^{2}\) ) and (b) the angular displacement (in rad) of each wheel?
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