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Chapter 3: Problem 34

A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its \(\mathrm{CM}\) is traveling vertically up at speed \(v_{\mathrm{o}}\) and it is rotating with angular velocity \(\omega_{\mathrm{o}} .\) To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should \(v_{\mathrm{o}}\) be, if the rod is to have made exactly \(n\) rotations when it returns to his hand?

Short Answer

Expert verified
\(v_0 = \frac{\pi n g}{\omega_0}\)

Step by step solution

01

Analyzing the Problem

We are given a rod with its center of mass moving vertically upward at speed \(v_0\) and rotating with angular velocity \(\omega_0\). The goal is for the rod to make \(n\) complete rotations and return to the juggler's hand.
02

Calculating Time for Rod to Rise and Fall

The rod is thrown up with an initial vertical speed \(v_0\). To find how long it takes until it returns to the original height, use the kinematic equation for vertical motion: \(t = \frac{2v_0}{g}\), where \(g\) is the acceleration due to gravity.
03

Relating Rotations to Angular Speed

The number of rotations the rod makes is given by the angular displacement \(\theta = \omega_0 \cdot t\). For \(n\) complete rotations, we need \(\theta = 2\pi n\).
04

Solving for v_0

Substitute \(t = \frac{2v_0}{g}\) into the equation \(\theta = \omega_0 \cdot t = 2\pi n\): \[ \omega_0 \cdot \frac{2v_0}{g} = 2\pi n \] leading to \[ v_0 = \frac{\pi n g}{\omega_0} \].

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Key Concepts

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

Angular Velocity
Angular velocity is an important concept when understanding rotational motion. It describes how fast an object rotates or spins around an axis. In our exercise, the juggler is dealing with a uniform rod rotating as it's tossed. The angular velocity, denoted by \(\omega_0\), represents the rate at which the rod is spinning at the start of its journey.
In practical terms, angular velocity is measured in radians per second. When something completes one full rotation, it has moved through an angular path of \(2\pi\) radians. So, if the rod completes 1 rotation every second, its angular velocity is \(2\pi\, \text{rad/s}\). Understanding this helps us predict how many times an object like the rod will rotate in a given time.
In the context of the exercise, the juggler wants the rod to complete exactly \(n\) full rotations. By setting up the equation \(\theta = \omega_0 \cdot t\), we are using angular velocity to determine how long the rod will spin in the air, ensuring that \(\theta\) is equal to \(2\pi \times n\).
Kinematic Equations
Kinematic equations are crucial tools in solving problems involving motion, whether linear or rotational. They help us predict future motion based on initial conditions. In this exercise, we used one such equation to calculate the time it takes for the rod to rise and fall back to the juggler's hand.
The particular kinematic equation employed here is \(t = \frac{2v_0}{g}\), where \(v_0\) is the initial velocity upwards and \(g\) is the gravitational acceleration. This equation provides the total time of flight for a projectile, which in this case is the rod thrown by the juggler.
Given that gravity acts consistently downward with acceleration \(g = 9.81 \text{m/s}^2\), this formula helps us calculate that complete trip time. The juggler uses this time period to synchronize the rotational aspect of the rod, ensuring that the angular displacement corresponds with the desired number of rotations \(n\) before catching it.
Uniform Rod
Understanding the characteristics of a uniform rod is vital in analyzing its motion. A uniform rod means that the mass is evenly distributed along its length. This property significantly influences the dynamics, as it affects how the rod will rotate and balance in the air.
Because the mass distribution is even, the center of mass (CM) of the rod is exactly in the middle. This means the motion of the rod, when thrown, is predictable and can be analyzed using simple rotational dynamics. The balance of forces and torque is symmetric with respect to the center, simplified by its uniform nature.
For the juggler, knowing these details allows them to anticipate how the rod will behave in flight and upon catching. Any changes in the mass distribution, such as a non-uniform rod, could cause unexpected behavior, making the task more challenging. Hence, considering the uniformity of the rod is a crucial aspect of solving the problem seamlessly.

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Most popular questions from this chapter

The first couple of minutes of the launch of a space shuttle can be described very roughly as follows: The initial mass is \(2 \times 10^{6} \mathrm{kg}\), the final mass (after 2 minutes) is about \(1 \times 10^{6} \mathrm{kg}\), the average exhaust speed \(v_{\mathrm{ex}}\) is about \(3000 \mathrm{m} / \mathrm{s},\) and the initial velocity is, of course, zero. If all this were taking place in outer space, with negligible gravity, what would be the shuttle's speed at the end of this stage? What is the thrust during the same period and how does it compare with the initial total weight of the shuttle (on earth)?

Consider a uniform solid disk of mass \(M\) and radius \(R\), rolling without slipping down an incline which is at angle \(\gamma\) to the horizontal. The instantaneous point of contact between the disk and the incline is called \(P\). (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration \(\dot{v}\) of the disk by applying the result \(\dot{\mathbf{L}}=\Gamma^{\text {ext }}\) for rotation about \(P .\) (Remember that \(L=I \omega\) and the moment of inertia for rotation about a point on the circumference is \(\frac{3}{2} M R^{2} .\) The condition that the disk not slip is that \(v=R \omega \text { and hence } \dot{v}=R \dot{\omega} .)\) (c) Derive the same result by applying \(\dot{\mathbf{L}}=\mathbf{\Gamma}^{\mathrm{ext}}\) to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton's second law to the motion of the CM. The moment of inertia for rotation about the \(\mathrm{CM}\) is \(\frac{1}{2} M R^{2}\).)

Consider a gun of mass \(M\) (when unloaded) that fires a shell of mass \(m\) with muzzle speed \(v\). (That is, the shell's speed relative to the gun is \(v\).) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is \(v /(1+m / M).\)

Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the \(x y\) plane, with the sun at the origin, and label the planet's position by polar coordinates \((r, \phi)\). (a) Show that the planet's angular momentum has magnitude \(\ell=m r^{2} \omega,\) where \(\omega=\dot{\phi}\) is the planet's angular velocity about the sun. (b) Show that the rate at which the planet "sweeps out area" (as in Kepler's second law) is \(d A / d t=\frac{1}{2} r^{2} \omega,\) and hence that \(d A / d t=\ell / 2 m .\) Deduce Kepler's second law. SECTION 3.5 Angular Momentum for Several Particles

A uniform spherical asteroid of radius \(R_{\mathrm{o}}\) is spinning with angular velocity \(\omega_{\mathrm{o}}\). As the aeons go by, it picks up more matter until its radius is \(R\). Assuming that its density remains the same and that the additional matter was originally at rest relative to the asteroid (anyway on average), find the asteroid's new angular velocity. (You know from elementary physics that the moment of inertia is \(\frac{2}{5} M R^{2}\).) What is the final angular velocity if the radius doubles?
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