/*! 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 65 A purse at radius \(2.00 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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A purse at radius \(2.00 \mathrm{~m}\) and a wallet at radius \(3.00 \mathrm{~m}\) travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is \(\left(2.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} .\) At that instant and in unit-vector notation, what is the acceleration of the wallet?

Short Answer

Expert verified
The acceleration of the wallet is \(3\sqrt{5} \hat{i} + 3\sqrt{5} \hat{j}.\)

Step by step solution

01

Understanding Variables and Given Information

First, identify the given values and variables for the problem. The radius for the purse is \(R_p = 2.00 \ ext{m}\) and for the wallet, \(R_w = 3.00 \ ext{m}\). We are given the acceleration of the purse as \(\vec{a}_p = (2.00 \ ext{m/s}^2) \hat{i} + (4.00 \ ext{m/s}^2) \hat{j}\). Both objects are on the same radial line, and we are asked to find the acceleration of the wallet \(\vec{a}_w\).
02

Relating Acceleration of Both Objects

Since both objects are in uniform circular motion and located on the same radial line, they have the same angular velocity \(\omega\). This implies that their accelerations can be related through the centripetal acceleration formula \(a = \omega^2 r\).
03

Expressing Acceleration in Radial and Tangential Components

The acceleration vector can be broken into radial and tangential components. For uniform circular motion, the acceleration is purely radial with \(\vec{a}_i = -\omega^2 \vec{r}_i\), where \(\vec{r}_i\) is the position vector for either object. The angle \(\theta\) for the purse is the same as for the wallet since they share a radial line, only the radius changes.
04

Calculating Angular Velocity from Purse's Acceleration

Using the magnitude of the given acceleration of the purse, \(a_p = \sqrt{(2.00)^2 + (4.00)^2} = \sqrt{20} \ ext{m/s}^2\). Since \(a_p = \omega^2 R_p\), solve for \(\omega\): \[ \omega^2 = \frac{a_p}{R_p} = \frac{\sqrt{20}}{2} = \sqrt{5}.\]Thus, \(\omega = \sqrt[4]{5}.\)
05

Calculating Acceleration of the Wallet

Using the same angular velocity for the wallet: \[ a_w = \omega^2 R_w = \sqrt{5} \times 3 = 3\sqrt{5}\]Since direction is the same as the purse's acceleration, the unit vector direction doesn't change, thus:\[\vec{a}_w = 3\sqrt{5} \hat{i} + 3\sqrt{5} \hat{j}.\]

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Key Concepts

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

Angular Velocity
Angular velocity is a fundamental concept when talking about objects in uniform circular motion. It describes how fast an object is rotating around a fixed point, which in this case, is the center of the merry-go-round. Angular velocity is denoted by the symbol \( \omega \) and has units of radians per second. A unique feature of angular velocity in uniform circular motion is that it remains constant.

Since the purse and wallet are on the same radial line and moving in uniform circular motion, they share the same angular velocity. This means that regardless of how far they are from the center (2.00 meters for the purse and 3.00 meters for the wallet), their rotation speed around the center is identical. In the problem, you find \( \omega \) by realizing that the purse’s measured acceleration is linked to its known radius and using the relation \( a = \omega^2 r \). This relation helps establish that the magnitude of acceleration depends on both the radius and the square of the angular velocity.

Remember, same \( \omega \) for different radii leads to different linear speeds, but in uniform circular motion, the angular speed stays consistent across all radii.
Centripetal Acceleration
Centripetal acceleration is crucial for understanding circular motion. It is the acceleration directed towards the center of the circular path that ensures the object stays in motion along the circular path. For any object traveling in a circle, whether it’s the purse or the wallet on the merry-go-round, centripetal acceleration ensures that the object doesn't fly off in a tangent direction.

In this exercise, you can see that the centripetal acceleration formula \( a = \omega^2 r \) is central to finding the acceleration of the wallet. The acceleration of the purse was given, allowing us to find \( \omega \) using that formula. Then we applied the same angular velocity \( \omega \) to the wallet which is located at a different radius. This shows how centripetal acceleration is directly proportional to both \( r \) and \( \omega^2 \), thereby changing with radius but maintaining uniformity in path and direction.

Thus, centripetal acceleration keeps both the purse and wallet moving along their circular tracks, ensuring they remain equidistant from their radial point at every instance.
Radial Line Position
The radial line position is important in problems dealing with uniform circular motion because it helps identify objects' relative positions and rotations. A radial line originates from the center of rotation and extends outward, passing through both the purse and the wallet in this scenario.

Having both objects on the same radial line implies they experience their accelerations along parallel lines but at different magnitudes due to varying radii. Understanding their radial line position helps us apply the correct formulas for their motions, as it indicates they share the same angular velocity and similar directional properties for their accelerations.

In this exercise, knowing their radial positioning allows correct application of the relation \( \vec{a}_i = -\omega^2 \vec{r}_i \) to find each object's radial and tangential components. This helps us solve for the wallet's acceleration using established constraints on the radial line, knowing that for uniform circular motion, tangential acceleration is zero, and only radial (centripetal) acceleration is considered.

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Most popular questions from this chapter

A transcontinental flight of \(4350 \mathrm{~km}\) is scheduled to take 50 min longer westward than eastward. The airspeed of the airplane is \(966 \mathrm{~km} / \mathrm{h},\) and the jet stream it will fly through is presumed to move due east. What is the assumed speed of the jet stream?

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During a tennis match, a player serves the ball at \(23.6 \mathrm{~m} / \mathrm{s},\) with the center of the ball leaving the racquet horizontally \(2.37 \mathrm{~m}\) above the court surface. The net is \(12 \mathrm{~m}\) away and \(0.90 \mathrm{~m}\) high. When the ball reaches the net, (a) does the ball clear it and (b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at \(5.00^{\circ}\) below the horizontal. When the ball reaches the net, (c) does the ball clear it and (d) what now is the distance between the center of the ball and the top of the net?

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