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Chapter 11: Problem 10

Vertical pendulum. A pendulum consists of a sphere of mass \(m\) attached to a massless rope of length \(L\). You start with the rope in your hand and the sphere hanging down. You hit the sphere, giving it a horizontal velocity \(v_{0}\) - (a) What is the speed of the sphere at the top of the path? (b) How large must \(v_{0}\) be for the sphere to be able to make a complete revolution with the rope tight at all times?

Short Answer

Expert verified
The speed of the sphere at the top is \sqrt{v_0^2 - 4gL}. The initial speed \(v_0\) must be at least \sqrt{5gL\) for the sphere to complete a full revolution.

Step by step solution

01

Analyze the motion at the top of the path

The sphere's speed at the top of the pendulum's path can be found using energy conservation. The total mechanical energy at the starting point equals the total mechanical energy at the top of the path.
02

Write the energy conservation equation

The kinetic energy at the start is \[ E_{k0} = \frac{1}{2}mv_0^2 \] and the potential energy at the start is zero because it is our reference point. At the top, the sphere has both potential and kinetic energy: \[ E_{ptop} = mg(2L) \] and \[ E_{ktop} = \frac{1}{2}mv^2 \] where v is the speed at the top.
03

Set the energies equal

Set the total energy at the start equal to the energy at the top: \[ \frac{1}{2}mv_0^2 = mg(2L) + \frac{1}{2}mv^2 \] Divide through by m to simplify: \[ \frac{1}{2}v_0^2 = 2gL + \frac{1}{2}v^2 \]
04

Solve for the speed at the top

Isolate v by subtracting 2gL from both sides: \[ \frac{1}{2}v_0^2 - 2gL = \frac{1}{2}v^2 \] Multiply through by 2: \[ v_0^2 - 4gL = v^2 \] Take the square root: \[ v = \sqrt{v_0^2 - 4gL} \]
05

Determine the condition for a complete revolution

For the sphere to make a complete revolution, the centripetal force at the top must be at least the gravitational force. Thus, \[ m\frac{v^2}{L} \geq mg \] Simplify to \[ \frac{v^2}{L} \geq g \]
06

Substitute v with the previously found expression

\[ \frac{(\sqrt{v_0^2 - 4gL})^2}{L} \geq g \] This simplifies to: \[ v_0^2 - 4gL \geq gL \] Adding 4gL to both sides: \[ v_0^2 \geq 5gL \]
07

Solve for initial speed

Take the square root of both sides: \[ v_0 \geq \sqrt{5gL} \]}],

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Key Concepts

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

energy conservation
Energy conservation is a powerful principle in physics. It tells us that energy cannot be created or destroyed, only transformed from one type to another. In the pendulum problem, we deal with two main types of energy: kinetic energy and potential energy. At the start, the sphere has kinetic energy due to its velocity, and no potential energy because it's our reference point. As the sphere swings upward, some of its kinetic energy transforms into potential energy. At the top, it has a mix of both kinetic and potential energy.
mechanical energy
Mechanical energy is the sum of kinetic energy and potential energy in a system. For the pendulum, the total mechanical energy remains constant throughout the motion, provided there's no energy loss. Initially, the sphere's mechanical energy is entirely kinetic. As it moves upwards, its mechanical energy is a combination of kinetic and potential energy. The sphere's speed at any point depends on this total mechanical energy.
centripetal force
Centripetal force is the force that keeps an object moving in a circular path. For the pendulum to complete a full revolution, the centripetal force at the top must be at least equal to the gravitational force pulling it down. This means that the sphere must be moving fast enough to provide this force. The condition given by the formula \( \frac{v^2}{L} \geq g \) ensures that the required centripetal force is maintained.
kinetic energy
Kinetic energy is the energy an object possesses due to its motion. It's given by the formula \( E_k = \frac{1}{2}mv^2 \). When the pendulum starts, all its energy is kinetic because it's moving and at its lowest point. As it rises, some of this energy converts into potential energy. At the top, its speed drops, reducing its kinetic energy but increasing its potential energy correspondingly.
potential energy
Potential energy is the energy stored in an object due to its position. For the pendulum, potential energy is highest at the top of its swing because it's at the maximum height. It's given by \( E_p = mgh \), where 'h' is the height above the reference point. In our problem, at the top of the swing, the height is 2L, so the potential energy is \( 2mgL \), showing a clear transformation from kinetic energy at the start to potential energy at the top.

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Most popular questions from this chapter

A bouncing ball-part 2. You lift a ball of radius \(R\) and mass \(m\) vertically up, until its center is a height \(h\) above the ground. Here, you hit the ball, giving it an initial horizontal velocity \(v_{0}\). You may model the collision between the ball and the ground using a spring force acting in the direction normal to the ground. You can assume that all the deformation occurs in the ball and that the spring constant is \(k\). There is no friction between the ball and the ground. (a) What is the (vector) velocity of the ball when it comes in contact with the ground? (b) What is the (vector) velocity of the ball when it reaches its maximum compression? (c) What is the maximum deformation, \(\delta y\), of the ball during the collision with the ground? (d) What is the (vector) velocity of the ball as it loses contact with the ground? (e) Based on your results, can you propose a law for how the velocity of a ball changes during a collision with a (frictionless) wall?

Lennard-Jones Potential. The Lennard-Jones potential is often used to describe the interaction between two atoms in a diatomic molecule. The potential energy of the molecule is: $$ U(r)=U_{0}\left(\left(\frac{a}{r}\right)^{12}-\left(\frac{b}{r}\right)^{6}\right) $$ where \(r\) is the distance between the atoms. (a) What is the force acting on one of the atoms from the other atom? (b) Sketch the potential \(U(r)\). (c) Find and classify the equilibrium points.

A bouncing ball-part 1. You lift a ball of radius \(R\) and mass \(m\) vertically up, until its center is a height \(h\) above the ground, and let it go. The ball starts from rest. You may model the collision between the ball and the ground using a spring force, where all the deformation occurs in the ball. The spring constant is \(k\). (a) What is the height above ground of the center of the ball as the ball comes in contact with the ground? (b) What is the speed of the ball when it comes in contact with the ground? (c) What is the maximum deformation, \(\delta y\), of the ball during the collision with the ground?
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