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45\. The Bob The bob on a simple pendulum of length \(R\) moves in an arc of a circle. (a) By considering that the radial acceleration of the bob as it moves through its equilibrium position is that for uniform circular motion \(\left(v^{2} / R\right)\), show that the tension in the string at that position is \(m g\left(1+\Theta^{2}\right)\) if the angular amplitude \(\Theta\) is small. (See "Trigonometric Expansions" in Appendix E.)

Short Answer

Expert verified
The tension in the string at the equilibrium position of the pendulum is defined as \( T_r = mg (1 + 2\theta^2 ) \).

Step by step solution

01

- Understand the radial acceleration

For a pendulum bob moving through its equilibrium position, the radial (or centripetal) acceleration is given by the formula for uniform circular motion: \[ a_r = \frac{v^2}{R} \]where \(v\) is the instantaneous speed of the bob at the equilibrium position and \(R\) is the length of the pendulum.
02

- Calculate the velocity at equilibrium position

The energy of the pendulum is conserved. At the maximum displacement, the potential energy is converted into kinetic energy at the equilibrium position. For small angular amplitudes \( \Theta \), the maximum height is approximated by \[ h \rightarrow \frac{R(1-\text{cos} \theta)}{2} \ \rightarrow \frac{R \theta^2}{2} \]Using energy conservation, where all the gravitational potential energy is converted to kinetic energy at the bottom:\[ \frac{1}{2} m v^2 = m g h \ \rightarrow v^2 = 2gR \theta^2 \]
03

- Determine the radial component of tension

In circular motion, the tension \( T_r\) in the string provides the necessary centripetal force. Thus, \[ T_r - mg = \frac{m v^2}{R} \-> T_r = mg + \frac{m v^2}{R} \so Using velocity (simplified for small angle) v^ written previously\Rightarrow T_r = mg + m(2gR \theta^2)/R \ \rightarrow T_r = mg + 2 mg \theta^2 \]
04

- Simplify to obtain the final expression for tension

Combining the terms from the previous step results in the simplified expression for the tension in the string at the equilibrium position: \[ T_r = mg (1 + 2 \theta^2 ) \]which is the required solution.

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Key Concepts

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

radial acceleration
Radial acceleration (also known as centripetal acceleration) is crucial in understanding the motion of a pendulum bob. When a pendulum moves through its equilibrium position, it behaves like an object in uniform circular motion. The formula for radial acceleration is given by: \[a_r = \frac{v^2}{R}\] Here, \(v\) represents the velocity of the pendulum bob at the equilibrium position, while \(R\) stands for the length of the pendulum string. This acceleration keeps the bob moving along its curved path by pointing towards the center of the circle, which is the pivot point of the pendulum.
It is essential to know this formula because it helps us find other quantities like tension in the pendulum string, which is directly connected to the radial acceleration.
It provides a clearer understanding of forces acting perpendicularly to the motion of the pendulum bob, thereby helping you to grasp how energy and force are related in pendulum motion.
energy conservation
Energy conservation is a fundamental principle in physics. For a pendulum, it helps explain how energy transforms as the pendulum swings. When the pendulum bob is at its highest point, all its energy is potential energy due to its height. As it swings to the equilibrium position, potential energy converts into kinetic energy.
The energy conservation equation for a pendulum can be outlined as: \[ \frac{1}{2} m v^2 = m g h \] Here, \(m\) denotes the mass of the pendulum bob, \(v\) is its speed at equilibrium, \(g\) is the gravitational constant, and \(h\) signifies the height. For small angular amplitudes \( \theta \), the height \(h\) can be approximated by: \[ h \rightarrow \frac{R \theta^2}{2} \] Where \(R\) is the length of the pendulum.
By understanding these energy transformations, one can determine that at equilibrium, the entire potential energy is converted into kinetic energy. This helps you derive the speed of the pendulum bob, which further aids in calculating other dynamic quantities like radial acceleration and tension.
centripetal force
Centripetal force is what keeps an object moving in a circular path. For a pendulum, this force is provided by the tension in the string and the gravitational force acting on the bob. When the bob is at the equilibrium position, the centripetal force equation is: \[ T_r - mg = \frac{m v^2}{R} \]Here, \(T_r\) is the tension in the string, \(mg\) is the weight of the pendulum bob, and \( \frac{m v^2}{R} \) is the centripetal force itself.
This equation shows that the tension not only has to balance the weight of the bob but also provide the necessary centripetal force to keep the bob moving in a circular path. By calculating the tension, replacing the velocity \(v\) using the derived energy conservation equation, we get: \[ T_r = mg + 2mg \theta^2 \]Therefore, the final tension in the string at the equilibrium position becomes: \[ T_r = mg(1 + 2 \theta^2) \]Through understanding centripetal force, one gets to see the interplay of forces acting on the pendulum, which are essential in keeping it in motion.
small-angle approximation
The small-angle approximation is a helpful tool for simplifying the equations of motion for a pendulum. When the amplitude of the pendulum's swing is small (specifically, when \( \theta \) is small), the trigonometric functions can be approximated:
\[ \text{cos} \theta \rightarrow 1 - \frac{\theta^2}{2} \]This simplification is useful because it allows us to convert complex trigonometric expressions into easier-to-handle forms. For example, the height \(h\) of the pendulum bob, which would usually be calculated using another function, simplifies to: \[ h \rightarrow \frac{R \theta^2}{2} \]Where \(R\) is the length of the pendulum string and \( \theta\) is the angular amplitude.
This approximation is the foundation for deriving many other useful equations, including the velocity at equilibrium and the expression for tension. Remember though, the small-angle approximation only holds when \( \theta \) is sufficiently small (usually less than 15 degrees), making it crucial to the accurate analysis of pendulum motion within these limits.

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Most popular questions from this chapter

46\. Angular Amplitude For a simple pendulum, find the angular amplitude \(\Theta\) at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by \(1.0 \%\). (See "Trigonometric Expansions" in Appendix E.)

23\. Two Particles Oscillate Two particles oscillate in simple harmonic motion along a common straight-line segment of length \(A\). Each particle has a period of \(1.5 \mathrm{~s}\), but they differ in phase by \(\pi / 6\) rad. (a) How far apart are they (in terms of \(A\) ) \(0.50 \mathrm{~s}\) after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?

56\. Vertical Spring A vertical spring stretches \(9.6 \mathrm{~cm}\) when a \(1.3 \mathrm{~kg}\) block is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional \(5.0 \mathrm{~cm}\) downward and released from rest. Find (b) the period, (c) the frequency, (d) the amplitude, and (e) the maximum speed of the resulting SHM.

53\. Displacement in SHM When the displacement in SHM is onehalf the amplitude \(X\), what fraction of the total energy is (a) kinetic energy and (b) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

26\. Tuning Fork The end of one of the prongs of a tuning fork that executes simple harmonic motion of frequency \(1000 \mathrm{~Hz}\) has an amplitude of \(0.40 \mathrm{~mm}\). Find (a) the magnitude of the maximum acceleration and (b) the maximum speed of the end of the prong. Find (c) the magnitude of the acceleration and (d) the speed of the end of the prong when the end has a displacement of \(0.20 \mathrm{~mm}\).

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