/*! 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 46 A small sphere with mass \(m\) i... [FREE SOLUTION] | 91Ó°ÊÓ

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A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\Theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\Theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

Short Answer

Expert verified
Acceleration starts as \( g \sin \Theta \). At \( \Theta/2 \), it is \( g \sin(\Theta/2) \). Speed is \( \sqrt{2gL(1 - \cos \Theta)} \) when vertical.

Step by step solution

01

Understand the Forces Acting on the Sphere

The forces acting on the sphere are gravity, represented by vector \( mg \) pointing directly downward, and the tension in the rod pointing towards the pivot. As the sphere is released from an angle \( \Theta \), its linear acceleration is directed tangentially to the arc of the swing.
02

Determine the Components of Motion

The gravitational force can be broken into two components relative to the pendulum's motion: a component along the rod which is balanced by the tension and does not affect acceleration, and a component perpendicular to the rod causing the sphere's acceleration. This perpendicular component is \( mg \sin \Theta \).
03

Calculate the Linear Acceleration at \(\Theta\)

Using the perpendicular component, the linear (tangential) acceleration \( a \) of the sphere can be calculated using \( a = g \sin \Theta \). This represents the acceleration at the beginning when the angle is \( \Theta \).
04

Diagram of Forces and Acceleration at \( \Theta/2 \)

At angle \( \Theta/2 \), the forces remain the same as in Step 1, but the angle with the vertical has changed. The perpendicular force component is now \( mg \sin( \Theta/2 ) \), leading to a new acceleration \( a = g \sin( \Theta/2 ) \).
05

Calculate Linear Speed When the Rod is Vertical

When the pendulum is vertical, all potential energy is converted to kinetic energy. Using conservation of energy, \( \frac{1}{2} m v^2 = mgL(1 - \cos \Theta) \). Solve for \( v \), the speed of the sphere, using \( v = \sqrt{2gL(1 - \cos \Theta)} \).

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Key Concepts

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

Linear Acceleration
Linear acceleration refers to the rate of change of velocity in a particular direction. In the context of a simple pendulum, the linear acceleration is tangential to the arc of the pendulum's swing. When a pendulum is released from an angle \( \Theta \), the pendulum begins to accelerate due to gravitational forces pulling it downward. This linear acceleration is caused by the component of gravity that acts perpendicular to the rod.
To calculate this acceleration, we use the formula \( a = g \sin \Theta \), where \( a \) is the acceleration, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth), and \( \Theta \) is the angle from the vertical. This expression shows that the acceleration is directly proportional to the sine of the angle.
It's important to note that as the pendulum moves through its arc, \( \Theta \) changes, thus changing the amount of tangential acceleration. This is why the pendulum starts to accelerate at different rates at different points in its swing.
Gravitational Force Components
The gravitational force acting on the pendulum's sphere can be separated into two components relative to the pendulum's position.
  • The component along the rod: This acts towards the pivot point and is countered by the tension in the rod. This component does not contribute to the motion of the pendulum because it does not cause any tangential acceleration.
  • The perpendicular component: This component acts perpendicular to the rod and is responsible for the pendulum's motion. It is given by \( mg \sin \Theta \), where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity.
This perpendicular component is crucial because it provides the force that propels the sphere in its curved path along the arc. It illustrates why the pendulum accelerates as it swings downward and decelerates as it climbs back up.
Conservation of Energy
The principle of conservation of energy is a key element in understanding pendulum motion. When dealing with a simple pendulum, energy transformations occur continuously during its swing.
At the highest point of the swing, the pendulum has maximum potential energy because of its height. This potential energy is calculated using \( mgh \), where \( h \) is the height corresponding to the angle \( \Theta \). As the pendulum swings down toward its lowest point, this potential energy is gradually converted into kinetic energy, described by \( \frac{1}{2} mv^2 \).
When the pendulum reaches the vertical position, theoretically all the potential energy is converted into kinetic energy, making the speed of the sphere the greatest at this point. The conservation of energy can be mathematically expressed as:\[ mgh = \frac{1}{2} mv^2 \] Solving for the velocity when the rod is vertical involves using the converted kinetic energy: \[ v = \sqrt{2gL(1 - \cos \Theta)} \].
This showcases the beautifully balanced exchange between potential and kinetic energy in pendulum motions.
Pendulum Motion
Pendulum motion refers to the repetitive oscillatory movement of the pendulum, caused by gravitational forces acting on it. This motion is a classic example of harmonic motion and is characterized by specific properties.
Simple pendulum motion depends on factors such as length \( L \) and the gravitational constant \( g \). Contrary to common belief, the mass of the pendulum and the amplitude (within reasonable small angles) do not significantly affect the period of the swing. This unique feature makes pendulums extremely useful in timekeeping devices like clocks.
  • The period \( T \) of a simple pendulum is given by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), which shows that the period is proportional to the square root of its length.
  • The pendulum reaches a maximum angle \( \Theta \) where its speed is zero before reversing direction.
This rhythmic and predictable motion is not just fascinating, but it's a powerful educational model for illustrating principles of physics such as inertia, energy conservation, and angular motion.

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Most popular questions from this chapter

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.

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; \((6)\) the acceleration and speed when the poin-poin's coordinate is \(x=+9.0 \mathrm{cm}\); (c) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asproach used parts \((\mathrm{a}),(\mathrm{b}),\) and \((\mathrm{c})\) can be found using the energy approach used in Section \(13.3,\) and which cannot? Explain.

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.

The balance wheel of a watch vibrates with an angular amplitude \(\Theta,\) angular frequency \(\omega\) , and phase angle \(\phi=0\) . (a) Find expressions for the angular velocity \(d \theta / d t\) and angular acceleration \(d^{2} \theta / d t^{2}\) as functions of time. (b) Find the balance wheel's angular velocity and angular acceleration when its angular displacement is \(\Theta,\) and when its angular displacement is \(\Theta / 2\) and \(\theta\) is decreasing. (Hint: Sketch a graph of \(\theta\) versus \(t\) )

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