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Chapter 3: Problem 114

The 4 -kg ball and the attached light rod rotate in the vertical plane about the fixed axis at \(O .\) If the assembly is released from rest at \(\theta=0\) and moves under the action of the 60 -N force, which is maintained normal to the rod, determine the velocity \(v\) of the ball as \(\theta\) approaches \(90^{\circ} .\) Treat the ball as a particle.

Short Answer

Expert verified
Find the velocity using \( v = \sqrt{60 \cdot r \cdot \pi} \).

Step by step solution

01

Identify the Forces and Parameters

The ball has a mass of 4 kg and is subjected to a continuous force of 60 N, normal to the rod. It starts from rest when \( \theta = 0 \) and we need to find its velocity at \( \theta = 90^{\circ} \). The force causes rotational motion about point \( O \).
02

Apply Work-Energy Principle

The work-energy principle states that the work done by all forces acting on a particle equals the change in kinetic energy of the particle. The force is always perpendicular to the rod, so work is done as the ball rotates from \( \theta = 0 \) to \( \theta = 90^{\circ} \).
03

Calculate the Work Done by the Force

The work done by the force as the rod rotates through an angle \( \theta \) is given by:\[ W = \int_{0}^{\frac{\pi}{2}} F \cdot ds = F \cdot r \cdot \theta \]Here, \( F = 60 \) N, and \( r \) is the length of the rod. But since \( \theta \to \frac{\pi}{2} \), we find:\[ W = 60 \cdot r \cdot \frac{\pi}{2} \].
04

Analyze Kinetic Energy

The initial kinetic energy at \( \theta = 0 \) is zero because the ball starts from rest. The final kinetic energy when \( \theta = 90^{\circ} \) is \( \frac{1}{2}mv^2 \), where \( v \) is the velocity we need to find. So, the equation becomes:\[ W = \frac{1}{2}mv^2 \]
05

Solve for Velocity

Substituting \( W = 60 \cdot r \cdot \frac{\pi}{2} \) and the mass \( m = 4 \text{ kg} \) into the kinetic energy equation:\[ 60 \cdot r \cdot \frac{\pi}{2} = \frac{1}{2} \times 4 \times v^2 \]Solve for \( v \):\[ v^2 = \frac{240 \cdot r \cdot \pi}{4} \]\[ v = \sqrt{60 \cdot r \cdot \pi} \]This gives \( v \) as a function of the rod length \( r \).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that connects the work done on an object to its change in kinetic energy. When a force is applied to an object, work is done if there's movement in the direction of the force. This principle helps us understand how energy transfers between an object's kinetic and potential states.
In the given problem, the assembly of the ball and rod experiences a force of 60 N, which does work on the system as it moves from one position to another. This force acts perpendicular to the rod, causing it to rotate. The total work done translates into a change in the system's kinetic energy.
To put it simply, as the ball moves from rest (where its kinetic energy is zero) to the position where it is at 90 degrees to the starting position, it gains kinetic energy due to the work done by the force. By equating the work done to the change in kinetic energy, we are able to solve for the velocity of the ball at a specific point in its motion.
Kinetic Energy
Kinetic energy is the energy of motion. Any object that moves possesses kinetic energy, which depends on its mass and velocity. For a particle-like object, the kinetic energy (KE) is given by the formula:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity.
In this exercise, the ball starts at rest, meaning its initial kinetic energy is zero. As it rotates, work is done on the ball, specifically increasing its velocity and thereby its kinetic energy. By the time it reaches the angle of 90 degrees, its velocity has increased due to this force-induced motion.
The final kinetic energy helps us determine the speed at which the ball is moving just by understanding the relationship between work done by forces and the resulting kinetic energy. This shift in energy from a state of rest to motion is a crucial aspect to grasp when analyzing many physical systems in rotational dynamics.
Rotational Motion
Rotational motion involves objects that turn around an axis. In this context, it's particularly fascinating because it combines linear motion principles with rotation about a point or an axis. For any object undergoing rotational motion, there are additional considerations such as torque and angular displacement.
The problem involves a ball attached to a rod, rotating about a fixed point \( O \). The force applied normal to the rod introduces torque, causing the assembly to rotate. Even though the rod moves through angles, the principles of linear work and energy still apply. Here, the distance along the arc of the circle (traversed by the force's point of application) is essential when calculating work done. This distance, combined with the force, provides the rotational work.
Understanding rotational motion connects concepts of linear motion with rotational analogs such as angular velocity and torque. For this problem, thinking of the ball's path as an arc simplifies the math and helps in determining how fast the ball is moving at various points in its path.

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Most popular questions from this chapter

The bungee jumper, an 80 -kg man, falls from the bridge at \(A\) with the bungee cord secured to his ankles. He falls \(20 \mathrm{m}\) before the 17 -m length of elastic bungee cord begins to stretch. The \(3 \mathrm{m}\) of rope above the elastic cord has no appreciable stretch. The man is observed to drop a total of \(44 \mathrm{m}\) before being projected upward. Neglect any energy loss and calculate \((a)\) the stiffness \(k\) of the bungee cord (increase in tension per meter of elongation) (b) the maximum velocity \(v_{\max }\) of the man during his fall, and \((c)\) his maximum acceleration \(a_{\max } .\) Treat the man as a particle located at the end of the bungee cord.

Car \(B\) is initially stationary and is struck by car \(A\) which is moving with speed \(v\). The mass of car \(B\) is \(p m,\) where \(m\) is the mass of car \(A\) and \(p\) is a positive constant. If the coefficient of restitution is \(e=\) \(0.1,\) express the speeds \(v_{A}^{\prime}\) and \(v_{B}^{\prime}\) of the two cars at the end of the impact in terms of \(p\) and \(v .\) Evaluate your expressions for \(p=0.5\).

The third stage of a rocket fired vertically up over the north pole coasts to a maximum altitude of \(500 \mathrm{km}\) following burnout of its rocket motor. Calculate the downward velocity \(v\) of the rocket when it has fallen \(100 \mathrm{km}\) from its position of maximum altitude. (Use the mean value of \(9.825 \mathrm{m} / \mathrm{s}^{2}\) for \(g\) and \(6371 \mathrm{km}\) for the mean radius of the earth.)

Car \(A\) weighing 3200 lb and traveling north at \(20 \mathrm{mi} / \mathrm{hr}\) collides with car \(B\) weighing \(3600 \mathrm{lb}\) and traveling at \(30 \mathrm{mi} / \mathrm{hr}\) as shown. If the two cars become entangled and move together as a unit after the crash, compute the magnitude \(v\) of their common velocity immediately after the impact and the angle \(\theta\) made by the velocity vector with the north direction.

The hollow tube assembly rotates about a vertical axis with angular velocity \(\omega=\dot{\theta}=4 \quad \mathrm{rad} / \mathrm{s}\) and \(\dot{\omega}=\ddot{\theta}=-2 \operatorname{rad} / \mathrm{s}^{2} .\) A small \(0.2-\mathrm{kg}\) slider \(P\) moves inside the horizontal tube portion under the control of the string which passes out the bottom of the assembly. If \(r=0.8 \mathrm{m}, \dot{r}=-2 \mathrm{m} / \mathrm{s},\) and \(\ddot{r}=4 \mathrm{m} / \mathrm{s}^{2}\) determine the tension \(T\) in the string and the horizontal force \(F_{\theta}\) exerted on the slider by the tube.
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