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To make a pendulum, a \(300 \mathrm{~g}\) ball is attached to one end of a string that has a length of \(1.4 \mathrm{~m}\) and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of \(30.0^{\circ}\) with the vertical; then (with the string taut) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of \(20.0^{\circ}\) with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

Short Answer

Expert verified
Found speed at 20°: calculate; maximum speed with full energy conversion; angle using energy at one-third speed.

Step by step solution

01

Analyze the Problem

To solve this, we use energy conservation principles. The potential energy (PE) at the release point is converted into kinetic energy (KE) when the pendulum swings down. The total mechanical energy at both positions must be equal, excluding any losses to friction.
02

Calculate Initial Potential Energy (PE_initial)

The ball is initially raised to a height difference, determined by the length of the string and the angles involved. We use: \[ h_i = L(1 - \cos \theta_0) \]\[ \theta_0 = 30.0^{\circ}, L = 1.4 \, \mathrm{m} \]\[ h_i = 1.4 \times (1 - \cos 30^{\circ}) \]Calculate \( h_i \).
03

Calculate Potential at Angle of 20° (PE_20°)

Use the formula again to calculate height at 20°:\[ h_{20} = L(1 - \cos \theta) \]\[ \theta = 20.0^{\circ} \]Calculate \( h_{20} \).
04

Apply Energy Conservation to Find Speed at 20°

Total energy is conserved, so initial potential energy equals the sum of potential and kinetic energy at 20°:\[ \mathrm{PE}_\mathrm{initial} = \mathrm{PE}_{20^\circ} + \mathrm{KE}_{20^\circ} \]\[ mgh_{i} = mgh_{20} + \frac{1}{2}mv^2 \]\[ v = \sqrt{2g(h_i - h_{20})} \]Calculate \( v \).
05

Determine Maximum Speed of the Ball

The maximum speed occurs at the lowest point where all initial potential energy is converted into kinetic energy:\[ \mathrm{PE}_\mathrm{initial} = \mathrm{KE}_{\mathrm{max}} \]\[ mgh_i = \frac{1}{2}mv_{\mathrm{max}}^2 \]\[ v_{\mathrm{max}} = \sqrt{2gh_i} \]Calculate \( v_{\mathrm{max}} \).
06

Calculate Speed at One-third Maximum Speed

Speed \( v = \frac{1}{3} v_{\mathrm{max}} \). Use energy conservation:\[ \frac{1}{2}mv_{\mathrm{full}}^2 = \frac{1}{2}m\left(\frac{1}{3}v_{\mathrm{max}}\right)^2 + mgh \]Solve for \( h \) to find the angle \( \theta \) where this condition is met.
07

Determine Angle for One-third Maximum Speed

Calculate angle using:\[ L \cdot \cos \theta = L - h \]Find \( \theta \) using inverse cosine:\[ \theta = \cos^{-1}\left(1 - \frac{(v_{\mathrm{full}}^2/9)}{2g}\right) \]

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Key Concepts

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

Potential Energy
In the context of a pendulum, potential energy refers to the stored energy due to its position in a gravitational field. When the pendulum is at its highest point, it possesses maximum potential energy. This occurs because the ball is elevated above the lowest point in its swing.

Potential energy ( PE ) can be calculated using the formula:
  • PE = mgh
where m is the mass of the object, g is the acceleration due to gravity, and h is the height relative to some reference point.

In our exercise, the ball starts at an angle of 30 degrees from the vertical, meaning it is raised to a certain height above the lowest point. The potential energy at this point can be calculated as the ball is initially still, so all its energy is stored as potential energy. This height, relative to the lowest point in the swing, provides us with the maximum potential energy.
Kinetic Energy
Kinetic energy is the energy an object has because of its motion. As the pendulum swings downwards from its initial position, the potential energy is converted into kinetic energy.

The formula for kinetic energy (KE) is:
  • KE = \( \frac{1}{2}mv^2 \)
where m is the mass of the object and v is its velocity.

In our exercise, as the pendulum swings down and reaches the angle of 20 degrees, some of the initial potential energy has been converted into kinetic energy. The formula above helps us calculate the resultant speed of the ball. Kinetic energy is at its maximum when the pendulum is at its lowest point in the arc since all the potential energy has been converted into kinetic energy, resulting in the fastest speed.
Mechanical Energy Conservation
Mechanical energy conservation is a principle stating that if no external forces are acting on a mechanical system (like a pendulum), its total mechanical energy remains constant. This total mechanical energy is the sum of potential and kinetic energy.

In terms of the pendulum:
  • Total Mechanical Energy = Potential Energy + Kinetic Energy
  • At any point in its swing: PE_initial = PE_current + KE_current
As the pendulum moves, energy is transferred back and forth between potential and kinetic forms. Initially, when the ball is released from a height, potential energy is high, and kinetic energy is zero. As it swings down, potential energy decreases while kinetic energy increases until the lowest point, where potential energy is minimized, and kinetic energy is maximized.

This exercise demonstrates mechanical energy conservation through the equations of energy balance. When solving for speeds at different angles, we use the conservation principle to equate the initial potential energy to the kinetic plus residual potential energy at any other point. This helps us find the velocities and respective angles during the pendulum's movement.

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Most popular questions from this chapter

A \(3.2 \mathrm{~kg}\) sloth hangs \(3.0 \mathrm{~m}\) above the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point \(y=0\) to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

We move a particle along an \(x\) axis, first outward from \(x=1.0 \mathrm{~m}\) to \(x=4.0 \mathrm{~m}\) and then back to \(x=1.0 \mathrm{~m}\), while an external force acts on it. That force is directed along the \(x\) axis, and its \(x\) component can have different values for the outward trip and for the return trip. Here are the values (in newtons) for four situations, where \(x\) is in meters: $$ \begin{array}{ll} \hline \text { Outward } & \text { Inward } \\ \hline \text { (a) }+3.0 & -3.0 \\ \text { (b) }+5.0 & +5.0 \\ \text { (c) }+2.0 x & -2.0 x \\ \text { (d) }+3.0 x^{2} & +3.0 x^{2} \\ \hline \end{array} $$ Find the net work done on the particle by the external force for the round trip for each of the four situations. (e) For which, if any, is the external force conservative?

A \(5.0 \mathrm{~kg}\) block is projected at \(5.0 \mathrm{~m} / \mathrm{s}\) up a plane that is inclined at \(30^{\circ}\) with the horizontal. How far up along the plane does the block go (a) if the plane is frictionless and (b) if the coefficient of kinetic friction between the block and the plane is \(0.40 ?(\mathrm{c})\) In the latter case, what is the increase in thermal energy of block and plane during the block's ascent? (d) If the block then slides back down against the frictional force, what is the block's speed when it reaches the original projection point?

A \(0.63\) kg ball thrown directly upward with an initial speed of \(14 \mathrm{~m} / \mathrm{s}\) reaches a maximum height of \(8.1 \mathrm{~m}\). What is the change in the mechanical energy of the ball-Earth system during the ascent of the ball to that maximum height?

A machine pulls a \(40 \mathrm{~kg}\) trunk \(2.0 \mathrm{~m}\) up a \(40^{\circ}\) ramp at constant velocity, with the machine's force on the trunk directed parallel to the ramp. The coefficient of kinetic friction between the trunk and the ramp is \(0.40\). What are (a) the work done on the trunk by the machine's force and (b) the increase in thermal energy of the trunk and the ramp?

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