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Chapter 4: Problem 66

A particle moves along a circular path over a horizontal \(x y\) coordinate system, at constant speed. At time \(t_{1}=4.00 \mathrm{~s}\), it is at point \((5.00 \mathrm{~m}, 6.00 \mathrm{~m})\) with velocity \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and acceleration in the positive \(x\) direction. At time \(t_{2}=10.0 \mathrm{~s}\), it has velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and acceleration in the positive \(y\) direction. What are the (a) \(x\) and (b) \(y\) coordinates of the center of the circular path if \(t_{2}-t_{1}\) is less than one period?

Short Answer

Expert verified
The center of the circular path is at (5.00 m, 6.00 m).

Step by step solution

01

Understand the Problem

The particle moves along a circular path with a constant speed. At two different times, we know its velocity and acceleration. We need to find the center of the circular path given the velocities and accelerations in different directions.
02

Identify Known Values

At time \( t_1 = 4.00 \mathrm{~s} \), the position is \((5.00 \mathrm{~m}, 6.00 \mathrm{~m})\), the velocity is \((3.00 \mathrm{~m}/\mathrm{s}) \hat{\mathrm{j}}\), and acceleration is in the positive \(x\) direction.At time \( t_2 = 10.0 \mathrm{~s} \), the velocity is \((-3.00 \mathrm{~m}/\mathrm{s}) \hat{\mathrm{i}}\), and acceleration is in the positive \(y\) direction.
03

Relate Acceleration and Radius

From uniform circular motion, we know acceleration is perpendicular to velocity and points to the center of the circle. At \(t_1\), acceleration is along \(x\), so the center lies on the \(x\)-axis.Similarly, at \(t_2\), acceleration is along \(y\), so the center lies on the \(y\)-axis.
04

Calculate Center Coordinates

Using the condition that acceleration points to the center:- From \(t_1\), velocity \( (3.00 \mathrm{~m/s}) \hat{\mathrm{j}} \) implies radial velocity component relative to center is zero along \( x \), center's \(x\)-coordinate is \( 5.00 \mathrm{~m} \).- From \(t_2\), velocity \((-3.00 \mathrm{~m/s}) \hat{\mathrm{i}}\) implies radial velocity component relative to center is zero along \( y \), center's \(y\)-coordinate is \( 6.00 \mathrm{~m} \).

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Key Concepts

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

Uniform Circular Motion
Uniform circular motion refers to the movement of a particle along a circular path at a constant speed. This movement means that while the speed remains the same, the direction of travel changes continuously.
This change of direction results in an acceleration directed towards the center of the circle, known as centripetal acceleration.
In the context of our example, the particle's consistent speed along the circular path ensures that even as the direction changes, the magnitude of speed (velocity without its directional component) stays fixed.
Velocity and Acceleration
In circular motion, velocity is a vector quantity with both magnitude and direction. The acceleration, known as centripetal acceleration, always points towards the center of the circle.
  • At time \( t_1 = 4.00 \mathrm{~s} \), the velocity is \((3.00 \mathrm{~m/s}) \hat{\mathrm{j}}\), indicating motion upwards in the y-direction.
  • At time \( t_2 = 10.0 \mathrm{~s} \), the velocity changes to \((-3.00 \mathrm{~m/s}) \hat{\mathrm{i}}\), indicating motion in the negative x-direction.
The precise nature of these vectors and their continuous influence leads us to consider accelerations occurring in different directions at different times. The acceleration at each time is moving toward the center of the circular path from the particle's position.
Coordinate System
The coordinate system is essential for describing the position and movement of objects in physics. In this scenario, the particle is moving within an \( xy \) coordinate plane.
  • At \( t_1 = 4.00 \mathrm{~s} \), the particle is located at \((5.00 \mathrm{~m}, 6.00 \mathrm{~m})\).
  • Position gives us a reference to understand the motion's path.
  • The velocity vector direction aligns with the axes, simplifying calculations for determining motion mechanics.
Understanding how position vectors interact with velocity and acceleration helps us derive other crucial pieces of circular motion. The coordinate system transforms complex motions into accessible mathematical analysis.
Centripetal Force
Centripetal force is the force required to keep a particle moving in a circular path. This force always acts inward, toward the center of the circle, ensuring the particle stays in motion along the curved path.
  • In the context of our example, at different times, the direction of centripetal force and acceleration points respectively in line with the changing positions of the particle.
  • It's the reason behind the so-called centripetal acceleration essential for circular motion.
Effectively, without centripetal force, a moving particle would continue in a straight line due to inertia, exiting the circular path. The centripetal force in this exercise can be linked to a conceptual tether, keeping the particle on its circular path.

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Most popular questions from this chapter

An airport terminal has a moving sidewalk to speed passengers through a long corridor. Larry does not use the moving sidewalk; he takes \(150 \mathrm{~s}\) to walk through the corridor. Curly, who simply stands on the moving sidewalk, covers the same distance in \(70 \mathrm{~s}\). Moe boards the sidewalk and walks along it. How long does Moe take to move through the corridor? Assume that Larry and Moe walk at the same speed.

A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. Figure \(4-51\) shows a cross section of Mt. Fuji, in Japan. (a) At what initial speed would a bomb have to be ejected, at angle \(\theta_{0}=35^{\circ}\) to the horizontal, from the vent at \(A\) in order to fall at the foot of the volcano at \(B\), at vertical distance \(h=\) \(3.30 \mathrm{~km}\) and horizontal distance \(d=9.40 \mathrm{~km}\) ? Ignore, for the moment, the effects of air on the bomb's travel. (b) What would be the time of flight? (c) Would the effect of the air increase or decrease your answer in (a)?

At \(t_{1}=2.00 \mathrm{~s}\), the acceleration of a particle in counterclockwise circular motion is \(\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right)\) ) . It moves at constant speed. At time \(t_{2}=5.00 \mathrm{~s}\), the particle's acceleration is \(\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{1}+\left(-6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{j} .\) What is the radius of the path taken by the particle if \(t_{2}-t_{1}\) is less than one period?

The current world-record motorcycle jump is \(77,0 \mathrm{~m}\), set by Jason Renie. Assume that he left the take-off ramp at \(12.0^{\circ}\) to the horizontal and that the take-off and landing heights are the same. Neglecting air drag, determine his take-off speed.
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