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Chapter 5: Problem 4

An object thrown straight up in the air is momentarily at rest when it reaches the top of its motion. Does that mean that it is in equilibrium at that point? Explain.

Short Answer

Expert verified
The object is not in equilibrium at the top because gravity is unbalanced; it just temporarily stops.

Step by step solution

01

Understanding the Problem

We need to determine if an object at the top of its motion, momentarily at rest, is in equilibrium. Equilibrium is defined as when all the forces acting on an object are balanced, resulting in no net force.
02

Identify Forces Acting on Object

At the top of its motion, the object is momentarily at rest, meaning its velocity is zero. However, at this point, the only force acting on the object is gravity, pulling it back down to the ground.
03

Analyze the Condition of Equilibrium

For an object to be in equilibrium, the net force acting on it must be zero. Since gravity is acting downwards on the object and there is no upward force to counterbalance it, the net force is not zero.
04

Conclusion

Even though the object is momentarily at rest, it is not in equilibrium because the force of gravity is unbalanced, causing it to start moving downward as soon as that moment passes.

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Key Concepts

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

Net Force
When it comes to understanding whether an object is in equilibrium, the concept of net force is crucial. Net force refers to the sum of all the forces acting on an object.
If the net force is zero, then the forces are balanced and the object is in equilibrium, meaning it remains at rest or moves with constant velocity. However, if the net force is not zero, it means there's an imbalance in the forces, causing the object to accelerate.

For instance, consider an object thrown straight up. At its highest point, it momentarily stops before gravity pulls it back down. Although it's not moving for a brief instant, gravity is still acting on it. So, the net force isn't zero at that moment. Since net force causes acceleration, the object soon begins to fall back to the ground.
  • Net force = Sum of all individual forces
  • Zero net force = Equilibrium (no acceleration)
  • Non-zero net force = No equilibrium (acceleration occurs)
Gravity
Gravity is a compelling force that affects every object with mass. It always pulls objects toward the center of the Earth. The force of gravity is what makes a dropped apple fall to the ground and keeps the planets in their orbits around the Sun.
When an object is thrown into the air, gravity continues to act on it throughout its flight. Even when the object reaches the peak of its trajectory and momentarily pauses, gravity hasn't ceased its pull.

This constant action of gravity means there is always a force acting on the object. Since there鈥檚 no opposing force to balance it at the object's highest point, gravity pulls it down, thus creating a net force. This action ensures the object isn鈥檛 in equilibrium.
  • Gravity acts downward, towards the Earth
  • Unopposed, it results in a net force
  • Keeps objects like satellites and apples moving toward Earth
Motion
Motion is the change in an object's position over time due to forces acting on it. Net force is integral to any motion because it determines whether an object starts, stops, or changes direction.
In the case of an object thrown upward, motion is initiated by the initial force exerted. Once in air, gravity takes over, slowing it down until it stops momentarily at the peak of its motion. This stoppage is temporary as gravity, an unbalanced force, causes it to resume motion downwards.

Being at rest momentarily doesn鈥檛 imply equilibrium. Instead, it's a transient phase between the upward and downward journey:
  • Motion ceases when forces are perfectly balanced
  • Unbalanced forces like gravity change motion
  • A temporary stop doesn鈥檛 equal equilibrium

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Most popular questions from this chapter

The sun turns on its axis once every \(26.0\) days. Its mass is \(2.0 \times 10^{30} \mathrm{~kg}\) and its radius is \(7.0 \times 10^{8} \mathrm{~m}\). Assume it is a rigid sphere of uniform density. (a) What is the sun's angular momentum? (answer check available at lightandmatter.com) In a few billion years, astrophysicists predict that the sun will use up all its sources of nuclear energy, and will collapse into a ball of exotic, dense matter known as a white dwarf. Assume that its radius becomes \(5.8 \times 10^{6} \mathrm{~m}\) (similar to the size of the Earth.) Assume it does not lose any mass between now and then. (Don't be fooled by the photo, which makes it look like nearly all of the star was thrown off by the explosion. The visually prominent gas cloud is actually thinner than the best laboratory vacuum ever produced on earth. Certainly a little bit of mass is actually lost, but it is not at all unreasonable to make an approximation of zero loss of mass as we are doing.) (b) What will its angular momentum be? (c) How long will it take to turn once on its axis? (answer check available at lightandmatter.com)

Show that a sphere of radius \(R\) that is rolling without slipping has angular momentum and momentum in the ratio \(L / p=(2 / 5) R .\)

(a) Prove the identity \(\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\mathbf{b}(\mathbf{a} \cdot \mathbf{c})-\mathbf{c}(\mathbf{a} \cdot \mathbf{b}) \quad\) by expanding the product in terms of its components. Note that because the \(x, y\), and \(z\) components are treated symmetrically in the definitions of the vector cross product, it is only necessary to carry out the proof for the \(x\) component of the result. (b) Applying this to the angular momentum of a rigidly rotating body, \(L=\int \mathbf{r} \times(\boldsymbol{\omega} \times \mathbf{r}) d m\), show that the diagonal elements of the moment of inertia tensor can be expressed as, e.g., \(I_{x x}=\int\left(y^{2}+z^{2}\right) d m\). (c) Find the diagonal elements of the moment of inertia matrix of an ellipsoid with axes of lengths \(a, b\), and \(c\), in the principalaxis frame, and with the axis at the center.(answer check available at lightandmatter.com)

The nucleus \({ }^{168} \mathrm{Er}\) (erbium-168) contains 68 protons (which is what makes it a nucleus of the element erbium) and 100 neutrons. It has an ellipsoidal shape like an American football, with one long axis and two short axes that are of equal diameter. Because this is a subatomic system, consisting of only 168 particles, its behavior shows some clear quantummechanical properties. It can only have certain energy levels, and it makes quantum leaps between these levels. Also, its angular momentum can only have certain values, which are all multiples of \(2.109 \times 10^{-34} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). The table shows some of the observed angular momenta and energies of \({ }^{168} \mathrm{Er}\), in SI units \(\left(\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}\right.\) and joules). $$ \begin{array}{ll} \hline L \times 10^{34} & E \times 10^{14} \\ 0 & 0 \\ 2.109 & 1.2786 \\ 4.218 & 4.2311 \\ 6.327 & 8.7919 \\ 8.437 & 14.8731 \\ 10.546 & 22.3798 \\ 12.655 & 31.135 \\ 14.764 & 41.206 \\ 16.873 & 52.223 \end{array} $$ (a) These data can be described to a good approximation as a rigid end-over- end rotation. Estimate a single best-fit value for the moment of inertia from the data, and check how well the data agree with the assumption of rigid-body rotation. hwhint \(\\{\) hwhint:erbium \(\\}\) (answer check available at lightandmatter.com) (b) Check whether this moment of inertia is on the right order of magnitude. The moment of inertia depends on both the size and the shape of the nucleus. For the sake of this rough check, ignore the fact that the nucleus is not quite spherical. To estimate its size, use the fact that a neutron or proton has a volume of about \(1 \mathrm{fm}^{3}\) (one cubic femtometer, where \(1 \mathrm{fm}=10^{-15} \mathrm{~m}\) ), and assume they are closely packed in the nucleus.

An object is observed to have constant angular momentum. Can you conclude that no torques are acting on it? Explain. [Based on a problem by Serway and Faughn.]
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