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Chapter 9: Problem 1

Be sure you understand why a pendulum in equilibrium hanging in a car that is accelerating forward tilts backward, and then consider the following: A helium balloon is anchored by a massless string to the floor of a car that is accelerating forward with acceleration \(A\). Explain clearly why the balloon tends to tilt forward and find its angle of tilt in equilibrium. [Hint: Helium balloons float because of the buoyant Archimedean force, which results from a pressure gradient in the air. What is the relation between the directions of the gravitational field and the buoyant force?]

Short Answer

Expert verified
The balloon tilts forward at an angle \(\theta = \tan^{-1}\left( \frac{A}{g} \right)\).

Step by step solution

01

Understanding Forces on the Balloon

A helium balloon experiences two main forces in still air: the buoyant force and gravity. The buoyant force acts upward, while gravity pulls the balloon downward.
02

Analyze Forces in an Accelerating Car

When a car accelerates forward with acceleration \(A\), an additional pseudo-force acts on the balloon in the opposite direction of the acceleration due to the frame of reference. This pseudo-force is \(F_p = m \, A\), where \(m\) is the mass of the displaced air.
03

Determine the Net Force

The pseudo-force and gravity alter the effective 'gravity' the balloon experiences. The effective gravitational field tilts from the vertical axis due to acceleration, pointing toward the back of the car.
04

Calculate the Tilt Angle

At equilibrium, the buoyant force that counteracts gravity and pseudo-force aligns along the effective gravitational field. The angle \(\theta\) of tilt can be found using \( \tan(\theta) = \frac{A}{g} \), where \(g\) is the acceleration due to gravity.
05

Find the Angle of Tilt

Using \( \tan(\theta) = \frac{A}{g} \), calculate the angle of tilt \(\theta\) by taking the arctangent: \[ \theta = \tan^{-1}\left( \frac{A}{g} \right). \]

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Key Concepts

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

Buoyant Force
The concept of buoyant force is essential to understanding why objects, such as helium balloons, float. When a body is submerged in a fluid, whether completely or partially, it experiences an upward force. This force opposes the weight of the body in the fluid. This upward push is called the buoyant force.

Originating from a pressure difference within the fluid, the buoyant force follows Archimedes' principle. According to this principle, the buoyant force is equal to the weight of the fluid displaced by the submerged part of the object.
  • For a balloon, this means that it will rise if the buoyant force exceeds its weight.
  • The greater density of surrounding air compared to the inside of a helium balloon causes this upward force.
In essence, buoyant force keeps the balloon rising and floating unless counteracted by another force, such as when it is tied down.
Pseudo-Force
Pseudo-forces, also known as fictitious forces, arise when observing dynamics from an accelerating frame of reference. They aren't real forces in the Newtonian sense but rather corrections made in non-inertial reference frames.
  • In an accelerating car, as our reference frame accelerates, a pseudo-force seems to act on objects inside.
  • Its magnitude is proportional to both the mass of the affected body and the acceleration of the reference frame itself.
As for our helium balloon in the car scenario, a pseudo-force acts backward on the balloon due to the forward acceleration of the car. This backward force and gravity peel away the anchoring balance of forces, making the balloon tilt forward as if pushed from behind.

Understanding pseudo-forces is crucial because it helps describe movements and tilts in non-inertial, accelerating frames, just like in our accelerating car example.
Equilibrium
Equilibrium in physics refers to a state where all forces acting on a system are balanced, with no net force causing a change in motion. For an object, achieving equilibrium means holding a consistent velocity or remaining static.
  • In our accelerating car scenario, equilibrium occurs when the balloon stops tilting further and stabilizes in its position, despite ongoing forces.
  • At equilibrium, all acting forces like buoyant, gravitational, and pseudo-forces find a balanced net effect.
In analyzing equilibrium for the balloon, we consider the effective gravitational field. This field results from a combination of true gravitational forces and pseudo-forces creating an inclined effect. When these forces are balanced, the balloon does not accelerate relative to the car, achieving equilibrium.

To find the angle of tilt, we examine geometry and trigonometric relationships among forces, leading to the conclusion that the tilt is at an angle \( \theta = \tan^{-1}\left( \frac{A}{g} \right) \), where the effective balance is visualized.

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Most popular questions from this chapter

The center of a long frictionless rod is pivoted at the origin and the rod is forced to rotate at a constant angular velocity \(\Omega\) in a horizontal plane. Write down the equation of motion for a bead that is threaded on the rod, using the coordinates \(x\) and \(y\) of a frame that rotates with the rod (with \(x\) along the rod and \(y\) perpendicular to it). Solve for \(x(t)\). What is the role of the centrifugal force? What of the Coriolis force?

I am spinning a bucket of water about its vertical axis with angular velocity \(\Omega\). Show that, once the water has settled in equilibrium (relative to the bucket), its surface will be a parabola. (Use cylindrical polar coordinates and remember that the surface is an equipotential under the combined effects of the gravitational and centrifugal forces.)

The Coriolis force can produce a torque on a spinning object. To illustrate this, consider a horizontal hoop of mass \(m\) and radius \(r\) spinning with angular velocity \(\omega\) about its vertical axis at colatitude \(\theta\). Show that the Coriolis force due to the earth's rotation produces a torque of magnitude \(m \omega \Omega r^{2} \sin \theta\) directed to the west, where \(\Omega\) is the earth's angular velocity. This torque is the basis of the gyrocompass.

v9.34 \star\star\star At a point \(P\) on the earth's surface, an enormous perfectly flat and frictionless platform is built. The platform is exactly horizontal \(-\) that is, perpendicular to the local free-fall acceleration \(\mathbf{g}_{P}\). Find the equation of motion for a puck sliding on the platform and show that it has the same form as (9.61) for the Foucault pendulum except that the pendulum's length \(L\) is replaced by the earth's radius R. What is the frequency of the puck's oscillations and what is that of its Foucault precession? [Hints: Write the puck's position vector, relative to the earth's center \(O\) as \(\mathbf{R}+\mathbf{r},\) where \(\mathbf{R}\) is the position of the point \(P\) and \(\mathbf{r}=(x, y, 0)\) is the puck's position relative to \(P .\) The contribution to the centrifugal force involving \(\mathbf{R}\) can be absorbed into \(\mathbf{g}_{P}\) and the contribution involving \(\mathbf{r}\) is negligible. The restoring force comes from the variation of \(g\) as the puck moves. \(J\) To check the validity of your approximations, compare the approximate size of the gravitational restoring force, the Coriolis force, and the neglected term \(m(\boldsymbol{\Omega} \times \mathbf{r}) \times \boldsymbol{\Omega}\) in the centrifugal force.

The Compton generator is a beautiful demonstration of the Coriolis force due to the earth's rotation, invented by the American physicist A. H. Compton (1892-1962, best known as author of the Compton effect) while he was still an undergraduate. A narrow glass tube in the shape of a torus or ring (radius \(R\) of the ring \(\gg\) radius of the tube) is filled with water, plus some dust particles to let one see any motion of the water. The ring and water are initially stationary and horizontal, but the ring is then spun through \(180^{\circ}\) about its east-west diameter. Explain why this should cause the water to move around the tube. Show that the speed of the water just after the \(180^{\circ}\) turn should be \(2 \Omega R \cos \theta,\) where \(\Omega\) is the earth's angular velocity, and \(\theta\) is the colatitude of the experiment. What would this speed be if \(R \approx 1 \mathrm{m}\) and \(\theta=40^{\circ} ?\) Compton measured this speed with a microscope and got agreement within 3\%.
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