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Chapter 14: Problem 71

A 1.50 -kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 \(\mathrm{N} / \mathrm{m}\) and a \(275-\mathrm{g}\) metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A\) , which is 15.0 \(\mathrm{cm}\) below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

Short Answer

Expert verified
(a) 0.0946 m above point A. (b) 0.274 seconds. (c) 1.39 m/s.

Step by step solution

01

Understanding the System's Initial Conditions

First, let's understand the initial setup of the system. We have a tray with a mass of 1.50 kg attached to an ideal spring with a spring constant of 185 N/m. There's also a 0.275 kg ball on the tray. The tray is pushed down to point A, 15.0 cm (or 0.15 m) below the equilibrium point, then released from rest. The system's total mass is the combined mass of the tray and the ball, which is 1.50 kg + 0.275 kg = 1.775 kg. We're interested in finding when and where the ball leaves the tray after the system is released.
02

Calculating the Natural Frequency of the System

The ball will leave the tray when the normal force on the ball becomes zero as the system accelerates upwards. To begin with, the natural frequency (\omega) of the spring-mass system can be determined using the formula: \[ \omega = \sqrt{\frac{k}{m}} \] where \( k \) is the spring constant and \( m \) is the total mass of the system. Substituting in the values, \( k = 185 \ \mathrm{N/m} \) and \( m = 1.775 \ \mathrm{kg} \), we get: \[ \omega = \sqrt{\frac{185}{1.775}} \approx 10.22 \ \mathrm{rad/s} \]
03

Determining the Amplitude of Motion

The displacement from the equilibrium position is 0.15 m, which is the amplitude (\(A\)) of the motion when released from rest. The system is initially at rest at this displaced position.
04

Finding the Net Acceleration Condition

The ball will leave the tray when the upward acceleration of the tray equals the acceleration due to gravity (\( g \approx 9.81 \ \mathrm{m/s^2} \)). This condition is reached when the maximum upward acceleration given by the spring to the entire system is starting to equal \( g \). Thus, we set the condition: \[ a = \omega^2 \times A = g \] where \( A = 0.15 \ \mathrm{m} \). Substituting, \( \omega = 10.22 \ \mathrm{rad/s} \), this becomes: \[ 10.22^2 \times 0.15 = 9.81 \] Solving gives \[ h = 0.0946 \ \mathrm{m} \] which is the height above point A where this occurs.
05

Calculating the Time to Ball Ejection

To find the time when the ball leaves the tray, we need to find which part of the oscillatory cycle corresponds to the acceleration criterion found previously. This corresponds to the time to reach displacement \( h \) above point A (\( x = A - h = 0.15 - 0.0946 = 0.0554 \)) using SHM equations. Using SHM equation: \[ y(t) = A \cos (\omega t) = x \] solves to find \[ t = \frac{\cos^{-1}(\frac{0.0554}{0.15})}{10.22} \approx 0.274 \ \mathrm{s} \] time elapsed from release until the tray acceleration equates to gravitational acceleration.
06

Finding the Velocity of the Ball

Since the acceleration due to the spring is not acting on the metal ball anymore, its velocity at that moment is the velocity of the spring's motion at height \( h \). Using velocity formula in SHM: \[ v(t) = -A \omega \sin(\omega t) \] \[ v_{peak} = A\omega \sqrt{1-(\frac{h}{A})^2} \] Substituting values: \[ v = 0.15 \times 10.22 \times \sqrt{1-(0.0946/0.15)^2} \approx 1.39 \ \mathrm{m/s} \]

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Key Concepts

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

Spring Force Constant
The spring force constant, often denoted as \( k \), describes the "stiffness" of a spring. The spring in our example has a force constant of 185 \( \mathrm{N/m} \). Imagine a spring's job is to fight against stretching or compressing. The bigger the spring constant, the less it flexes under a particular weight.
Consider this like a teacher in a classroom: a strict teacher (high \( k \)) is more resistant to being influenced by students, while a relaxed teacher (low \( k \)) might easily bend to kids' antics!
In physics, knowing \( k \) helps predict how a spring reacts when forces are applied. The force exerted by the spring is calculated using Hooke's Law:
  • \( F = k \times x \)
Here, \( F \) is the force the spring exerts, and \( x \) is the displacement from equilibrium. It's all about knowing how stubborn your spring is!
Natural Frequency
Natural frequency in simple harmonic motion tells us how often the object vibrates back and forth on its own, like our tray and ball system. Think about it as the "beat" of the system, just like how a metronome ticks evenly. Natural frequency is calculated with:
  • \( \omega = \sqrt{\frac{k}{m}} \)
Here, \( \omega \) is the natural frequency, \( k \) is the spring constant, and \( m \) is the mass.
For the provided system, \( \omega \) is about 10.22 \( \mathrm{rad/s} \). This tells us how fast the system moves about around its resting position.
It's essential to understand because any external force matching this frequency can cause stronger oscillations, a phenomenon known as resonance. Natural frequency helps us anticipate how the system behaves and predict when something special, like the ball popping off, might happen!
Oscillation Amplitude
The oscillation amplitude is the maximum distance the system moves from its equilibrium position. In simpler terms, it's the farthest point the tray reaches on either side as it bounces up and down. In this problem, the amplitude is 0.15 meters from where the system began.
Think of amplitude as the swing of a pendulum: the higher you pull it, the greater the amplitude. This tells us about the energy stored in the system.
When undergoing oscillation, the system transfers energy between potential and kinetic forms. The mathematics of amplitude in our scenario can explain when factors like gravity balance the spring's force, causing the ball to separate from the tray.
  • At maximum amplitude, the system resets to potential energy before oscillating back.
Amplitude not only illustrates how far the tray dips and rises, but also directly affects the velocities and accelerations involved in the movement, crucial for determining moments when changes, like the ball leaving, occur!

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Most popular questions from this chapter

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the block is at \(x=0.280 \mathrm{m},\) the acceleration of the block is \(-5.30 \mathrm{m} / \mathrm{s}^{2} .\) What is the frequency of the motion?

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of 0.450 \(\mathrm{N} \cdot \mathrm{m} / \mathrm{rad}\) . You twist the part a small amount about this axis and let it go, timing 125 oscillations in 265 s. What is the moment of inertia you want to find?

CP SHM of a Butcher's Scale. A spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) is hung vertically, and a 0.200 -kg pan is suspended from its lower end. A butcher drops a 2.2 -kg steak onto the pan from a height of 0.40 \(\mathrm{m}\) . The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

CP CALC An approximation for the potential energy of a KCl molecule is \(U=A\left[\left(R_{0}^{7} / 8 r^{8}\right)-1 / r\right],\) where \(R_{0}=2.67 \times\) \(10^{-10} \mathrm{m}, A=2.31 \times 10^{-28} \mathrm{J} \cdot \mathrm{m},\) and \(r\) is the distance between the two atoms. Using this approximation: (a) Show that the radial component of the force on each atom is \(F_{r}=A\left[\left(R_{0}^{7} / r^{9}\right)-1 / r^{2}\right].\) (b) Show that \(R_{0}\) is the equilibrium separation. (c) Find the minimum potential energy. (d) Use \(r=R_{0}+x\) and the first two terms of the binomial theorem \(\quad\) (Eq. 14.28\()\) to show that \(F_{r} \approx-\left(7 A / R_{0}^{3}\right) x,\) so that the molecule's force constant is \(k=7 A / R_{0}^{3} .\) (e) With both the \(K\) and \(C 1\) atoms vibrating in opposite directions on opposite sides of the molecule's center of mass, \(m_{1} m_{2} /\left(m_{1}+m_{2}\right)=3.06 \times 10^{-26} \mathrm{kg}\) is the mass to use in cal- culating the frequency. Calculate the frequency of small-amplitude vibrations.

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom- pom's coordinate is \(x=+9.0 \mathrm{cm} ;(\mathrm{c})\) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach used in Section \(14.3,\) and which cannot? Explain.
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