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Chapter 13: Problem 67

A \(1.50-\mathrm{kg},\) 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 15.0 \(\mathrm{cm}\) below its equilibrium point (call this point \(A\) ) 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 oceur 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) Tray is 0.0927 m above point A. (b) Time is 0.133 s. (c) Velocity is 0.76 m/s.

Step by step solution

01

Determine the system's equilibrium

The equilibrium position is where the force of the spring balances the gravitational force. The force due to gravity on the tray and ball is \( (1.50 \, \text{kg} + 0.275 \, \text{kg}) \times 9.81 \, \text{m/s}^2 = 17.31 \, \text{N} \). Thus, the spring compression at equilibrium, \( x_{eq} \), satisfies \( k \cdot x_{eq} = 17.31 \, \text{N} \). Solving gives \( x_{eq} = \frac{17.31}{185} \approx 0.0935 \, \text{m} \).
02

Define the amplitude of oscillation

When the system is pushed down by 15.0 cm and released, this distance becomes the amplitude (A) of oscillation relative to equilibrium. Hence, \( A = 0.15 \, \text{m} \).
03

Apply free-fall condition for ball leaving the tray

The ball leaves the tray when the normal force is zero, which occurs when the acceleration of the system equals \( g \). From the equation \( a = -蠅^2 y \), setting \( g = 蠅^2 y \), we solve for \( y \). With \( 蠅 = \sqrt{\frac{k}{m}} = \sqrt{\frac{185}{1.775}} \), find \( y = A \sin(蠅t) = \frac{g}{蠅^2} \). Then calculate \( y = \frac{9.81}{\left(\sqrt{\frac{185}{1.775}}\right)^2} \approx 0.0927 \, \text{m} \) above point A.
04

Calculate time when ball leaves tray

Using \( y = A \sin(蠅t) \), if \( y = 0.0927 \, \text{m} \), solve: \( 0.0927 = 0.15 \sin(蠅t) \). Therefore, \( \sin(蠅t) = \frac{0.0927}{0.15} \). \( t = \frac{\arcsin\left(\frac{0.0927}{0.15}\right)}{蠅} \approx \frac{\arcsin(0.618)}{\sqrt{\frac{185}{1.775}}} \approx 0.133 \text{ s} \).
05

Calculate velocity of the ball just as it leaves

The velocity can be found using the derivative of position: \( v = 蠅A \cos(蠅t) \). From Step 4, use \( t \approx 0.133 \, \text{s} \) to calculate \( v \). Thus, \( v = \omega \cdot 0.15 \cdot \cos(\omega \cdot 0.133) \) where \( \omega = \sqrt{\frac{185}{1.775}} \). Numerically, \( v \approx 0.76 \, \text{m/s} \).

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Key Concepts

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

Simple Harmonic Motion
In physics, many systems oscillate back and forth around an equilibrium position. When the restoring force is directly proportional to the displacement and acts in the direction opposite to the displacement, the motion is specifically termed Simple Harmonic Motion (SHM). In our system, the tray attached to the spring moves in SHM after it is displaced from its equilibrium position and then released.
- The spring force is a conservative force described by Hooke's Law: \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.
- The system proceeds to oscillate about the equilibrium due to the presence of this restoring force.
- The amplitude of this motion is determined by how far the tray was initially displaced, in this example, 0.15 m below its equilibrium.
SHM can be characterized by its properties of amplitude, period, and phase. The angular frequency \( \omega \) (in radians per second) is defined as \( \omega = \sqrt{\frac{k}{m}} \), where \( m \) is the mass of the system. This frequency determines how quickly the system oscillates through its cycle.
Equilibrium Position
The equilibrium position is the point at which the net force on the tray is zero, allowing the system to be at rest if undisturbed. In our system, it is achieved when the force exerted by the spring is equal and opposite to the gravitational force pulling the tray and ball downward.
- The gravitational force on the system here equals \( (m_{tray} + m_{ball}) \times g = 17.31 \, \text{N} \).
- At equilibrium, the spring force equals this gravitational force: \( k \cdot x_{eq} = 17.31 \, \text{N} \). Solving gives \( x_{eq} = \frac{17.31}{185} \approx 0.0935 \, \text{m} \).
This equilibrium position is significant because all the oscillatory motion described rests about this point. When displaced from here, the system strives to return, setting off harmonic motion. Thus, understanding the equilibrium position is crucial as it is the starting point of the movement.
Free-Fall Condition
In our problem scenario, the metal ball sits on the oscillating tray. The ball will lose contact with the tray and experience a free-fall motion when the normal force supporting it becomes zero. This situation happens when the downward acceleration of the system equals the acceleration due to gravity.
- The condition is expressed as \( g = \omega^2 y \), where \( y \) is the vertical distance above point A.
- Solving \( g = \omega^2 y \) provides \( y = \frac{g}{\omega^2} \). The ball will leave the tray when it reaches this height, computed to be \( 0.0927 \, \text{m} \) above the baseline point A.
This concept is essential for calculating the precise point at which free-fall begins, influencing the system's dynamics. The intricacies lie in calculating this exact height through precise measurement of forces and system properties.

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Most popular questions from this chapter

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 \(R_{0}=2.67 \times 10^{-10} \mathrm{m}\) and \(A=2.31 \times 10^{-28} \mathrm{J} \cdot \mathrm{m}\) . 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 (Eq. 13.28 ) 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 \(C l\) 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 calculating the frequency (see Problem 13.86\() .\) Calculate the frequency of small-amplitude vibrations.

A thin metal disk with mass \(2.00 \times 10^{-3} \mathrm{kg}\) and radius 2.20 \(\mathrm{cm}\) is attached at its center to a long fiber (Fig. 13.32\() .\) The disk, when twisted and released, oscillates with a period of 1.00 s. Find the torsion constant of the fiber.

A \(0.0200-\mathrm{kg}\) bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{c})\) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;(\mathrm{d})\) the speed of the bolt when \(x=-0.180 \mathrm{m} .\)

A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 \(\mathrm{mm}\) . (a) What is the maximum speed of the tip of a prong? (b) A housefly (Musca domestica) with mass 0.0270 \(\mathrm{g}\) is holding on to the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

Weighing Astronauts. This procedure has actually been used to "weigh" astronauts in space. A \(42.5-\mathrm{kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?
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