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Chapter 3: Problem 320

A particle of mass \(m\) is attached to the end of the light rigid rod of length \(L,\) and the assembly rotates freely about a horizontal axis through the pivot \(O .\) The particle is given an initial speed \(v_{0}\) when the assembly is in the horizontal position \(\theta=0 .\) Determine the speed \(v\) of the particle as a function of \(\theta\).

Short Answer

Expert verified
The speed \(v\) as a function of \(\theta\) is \(v = \sqrt{v_0^2 - 2gL(1 - \cos\theta)}\).

Step by step solution

01

Define the problem

We are given a particle of mass \(m\) attached to a rod of length \(L\). The particle is given an initial speed \(v_0\) at \(\theta = 0\), and we want to determine its speed \(v\) as a function of \(\theta\).
02

Apply conservation of mechanical energy

Since the system is rotating about a pivot and no external work is done, mechanical energy is conserved. Initially, at \(\theta = 0\), the total mechanical energy consists of kinetic energy (due to \(v_0\)) and gravitational potential energy (taken as zero at this point). As the particle moves, its speed changes and it gains potential energy.
03

Write the conservation of energy equation

The initial kinetic energy at \(\theta = 0\) is \(\frac{1}{2} m v_0^2\), and the potential energy is zero. At an angle \(\theta\), the kinetic energy is \(\frac{1}{2} m v^2\) and the potential energy with respect to the horizontal reference is \(mgL(1 - \cos\theta)\). The conservation equation can be written as:\[\frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + mgL(1 - \cos\theta)\]
04

Solve for the speed \(v\)

Rearrange the energy conservation equation to solve for \(v\):\[\frac{1}{2} m v^2 = \frac{1}{2} m v_0^2 - mgL(1 - \cos\theta)\]\[v^2 = v_0^2 - 2gL(1 - \cos\theta)\]\[v = \sqrt{v_0^2 - 2gL(1 - \cos\theta)}\]This is the expression for the speed \(v\) as a function of \(\theta\).

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is the energy that an object has due to its motion. When a particle is moving, it possesses kinetic energy simply because of its velocity. This form of energy is given by the formula: \[KE = \frac{1}{2} m v^2\] where:
  • \(m\) is the mass of the object,
  • \(v\) is its speed.
For a particle on a rotating rod, its speed may vary depending on its position during rotation. Initially, at \(\theta = 0\), the particle has a velocity \(v_0\) and kinetic energy \(\frac{1}{2} m v_0^2\).
As the particle rotates, its speed and consequently its kinetic energy will change. Using the concept of the conservation of energy, we can establish a relation between the kinetic energy at different points of rotation.
Exploring Potential Energy
Potential energy is energy stored in an object due to its position relative to some reference point. For a particle attached to a rod, this potential energy arises because of gravity. The formula to find the gravitational potential energy is: \[PE = mgh\] where:
  • \(m\) is the mass of the particle,
  • \(g\) is the acceleration due to gravity,
  • \(h\) is the height of the particle above the reference plane.

In the original setup, when the rod is horizontal, \(\theta = 0\) and the potential energy is marked as zero. As the rod rotates, the height \(h\) changes
depending on the angle \(\theta\). The potential energy expression when the rod is at an angle \(\theta\) is \(mgL(1 - \cos\theta)\). Here, the additional factor reflects how the height changes with \(\theta\). This allows us to see how energy is transferred between kinetic and potential forms as the rod swings.
Understanding Rotational Dynamics
Rotational dynamics involves the motion of objects rotating about an axis. For the problem at hand, the assembly rotates around a horizontal axis at the pivot point. In such scenarios, it鈥檚 critical to understand how forces and energy behave in rotational systems. In this context, the conservation of mechanical energy principle applies:
  • The total mechanical energy (sum of kinetic and potential energy) in an isolated system remains constant when only conservative forces like gravity act on it.
As the particle鈥檚 position described by \(\theta\) changes, its kinetic and potential energy change, reflecting the shift between the two forms of energy.
Rotational dynamics facilitate the understanding of these transformations and explains how the initial kinetic energy given to the particle
converts partly into potential energy and back into kinetic energy as it moves. By using the energy conservation equation: \[\frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + mgL(1 - \cos\theta)\] we relate the speed of the rotating particle at any angle \(\theta\) back to how energy flows in a rotating motion.

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Most popular questions from this chapter

The third and fourth stages of a rocket are coasting in space with a velocity of \(18000 \mathrm{km} / \mathrm{h}\) when a small explosive charge between the stages separates them. Immediately after separation the fourth stage has increased its velocity to \(v_{4}=18060 \mathrm{km} / \mathrm{h} .\) What is the corresponding velocity \(v_{3}\) of the third stage? At separation the third and fourth stages have masses of 400 and \(200 \mathrm{kg},\) respectively.

The retarding forces which act on the race car are the drag force \(F_{D}\) and a nonaerodynamic force \(F_{R}\) The drag force is \(F_{D}=C_{D}\left(\frac{1}{2} \rho v^{2}\right) S,\) where \(C_{D}\) is the drag coefficient, \(\rho\) is the air density, \(v\) is the car speed, and \(S=30 \mathrm{ft}^{2}\) is the projected frontal area of the car. The nonaerodynamic force \(F_{R}\) is constant at 200 lb. With its sheet metal in good condition, the race car has a drag coefficient \(C_{D}=0.3\) and it has a corresponding top speed \(v=200 \mathrm{mi} / \mathrm{hr}\) After a minor collision, the damaged front-end sheet metal causes the drag coefficient to be \(C_{D}^{\prime}=\) \(0.4 .\) What is the corresponding top speed \(v^{\prime}\) of the race car?

An escalator handles a steady load of 30 people per minute in elevating them from the first to the second floor through a vertical rise of 24 ft. The average person weighs 140 lb. If the motor which drives the unit delivers 4 hp, calculate the mechanical efficiency \(e\) of the system.

The 0.5 -kg cylinder \(A\) is released from rest from the position shown and drops the distance \(h_{1}=\) \(0.6 \mathrm{m} .\) It then collides with the 0.4 -kg block \(B ;\) the coefficient of restitution is \(e=0.8 .\) Determine the maximum downward displacement \(h_{2}\) of block \(B\) Neglect all friction and assume that block \(B\) is initially held in place by a hidden mechanism until the collision begins. The two springs of modulus \(k=500 \mathrm{N} / \mathrm{m}\) are initially unstretched, and the distance \(d=0.8 \mathrm{m}\).

A particle with a mass of \(4 \mathrm{kg}\) has a position vector in meters given by \(\mathbf{r}=3 t^{2} \mathbf{i}-2 t \mathbf{j}-3 \mathbf{k},\) where \(t\) is the time in seconds. For \(t=5\) s determine the angular momentum of the particle and the moment of all forces on the particle, both about the origin \(O\) of coordinates.
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