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Chapter 4: Problem 64

A particle moves horizontally in uniform circular motion, over a horizontal \(x y\) plane. At one instant, it moves through the point at coordinates \((4.00 \mathrm{~m}, 4.00 \mathrm{~m})\) with a velocity of \(-5.00 \mathrm{i} \mathrm{m} / \mathrm{s}\) and an acceleration of \(+12.5 \mathrm{j} \mathrm{m} / \mathrm{s}^{2} .\) What are the (a) \(x\) and (b) \(y\) coordinates of the center of the circular path?

Short Answer

Expert verified
The coordinates of the center are (2.00 m, 6.00 m).

Step by step solution

01

Understand the Circular Motion Concepts

In uniform circular motion, the velocity vector is always tangent to the path, and the acceleration vector points towards the center of the circle. This means that if the velocity is in the negative x-direction (-i), the center of the circle is to the left of the point under consideration. The acceleration is orthogonal to the velocity and provides the centripetal acceleration needed to maintain circular motion.
02

Use Acceleration to Locate the Center

The acceleration vector of the particle provides information about the direction towards the center of the circle. Given acceleration is in the positive j-direction, which means the center is positioned vertically above the particle's current position.
03

Calculate the Radius of the Circular Path

In uniform circular motion, the centripetal acceleration is given by the formula: \( a_c = \frac{v^2}{r} \). With \( a_c = 12.5 \, \text{m/s}^2 \) and \( v = 5.00 \, \text{m/s} \), we solve for \( r \):\[ r = \frac{v^2}{a_c} = \frac{(5.00)^2}{12.5} = 2.00 \, \text{m} \].
04

Find the x-coordinate of the Center

Since the velocity is solely in the negative i-direction, the x-coordinate of the center is simply 2.00 m to the left of the x-coordinate of the particle: \( x_{center} = 4.00 \, \text{m} - 2.00 \, \text{m} = 2.00 \, \text{m} \).
05

Find the y-coordinate of the Center

Given acceleration is in the positive j-direction, the center is directly above the current y-position by the radius. Hence, \( y_{center} = 4.00 \, \text{m} + 2.00 \, \text{m} = 6.00 \, \text{m} \).
06

Conclusion

The center of the circular path is at the coordinates \((2.00 \, \text{m}, 6.00 \, \text{m})\).

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Key Concepts

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

Centripetal Acceleration
In uniform circular motion, centripetal acceleration is a key concept that describes the acceleration of an object moving along a circular path. It's important because it keeps the object moving in a circle rather than in a straight line.
Centripetal acceleration always points towards the center of the circle. This direction is perpendicular to the object's velocity but collinear with the radius of the path. The formula for calculating it is:
  • \(a_c = \frac{v^2}{r}\)
      • V is the speed and r is the radius.
In our exercise, the acceleration \(a_c\) is given as 12.5 m/s虏. Using this value along with the speed of the particle, we calculate the radius of the circular path, which we found to be 2.00 m. This calculation tells us how sharply the particle is turning at any given point.
Velocity Vector
The velocity vector in circular motion has an interesting behavior. It is always tangent to the circle, meaning it points in the direction the object is moving at any moment. It depicts both speed and direction. When a particle moves in a uniform circular path, its speed remains constant, but its direction changes continuously.
In our scenario, the velocity was given as \(-5.00 \text{i} \text{m/s}\). This indicates the object is moving to the left along the x-axis because the direction is negative.
Understanding the velocity vector helps determine the circle's center. Since the velocity points in the negative i-direction, we know that the center of the circle must be on the left side of the particle's position on the x-axis. This leads us to find the x-coordinate of the center of the circle using the radius we calculated earlier.
Radius of Circular Path
The radius of a circular path is the distance from the center of the circle to any point on its circumference. It's essential in understanding the circular motion, as it determines the size and shape of the path the object follows.
From the formula for centripetal acceleration \(a_c = \frac{v^2}{r}\), we can solve for the radius to get:
  • \(r = \frac{v^2}{a_c} \)
Using the values given in our problem鈥攕peed \(v = 5.00 \text{m/s}\) and acceleration \(a_c = 12.5 \text{m/s}^2\)鈥攚e calculated the radius to be 2.00 m. This computation is crucial because it allows us to find the position of the center of the circle relative to the particle. It tells us how far horizontally and vertically the center is from the particle's current position, leading us to understand its motion fully.

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Most popular questions from this chapter

An iceboat sails across the surface of a frozen lake with constant acceleration produced by the wind. At a certain instant the boat's velocity is \((6.30 \hat{\mathrm{i}}-8.42 \mathrm{j}) \mathrm{m} / \mathrm{s}\). Three seconds later, because of a wind shift, the boat is instantaneously at rest. What is its average acceleration for this \(3,00 \mathrm{~s}\) interval?

An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are the (a) speed and (b) magnitude of the centripetal acceleration of the satellite?

What is the magnitude of the acceleration of a sprinter running at \(10 \mathrm{~m} / \mathrm{s}\) when rounding a turn of radius \(25 \mathrm{~m}\) ?

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)?

An ion's position vector is initially \(\vec{r}=5.0 \hat{\mathrm{i}}-6.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{k}}\), and \(10 \mathrm{~s}\) later it is \(\vec{r}=-2.0 \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}}-2.0 \hat{\mathrm{k}}\), all in meters. In unit- vector notation, what is its \(\vec{v}_{\text {ave }}\) during the \(10 \mathrm{~s}\) ?
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