/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A smooth chain \(A B\) of mass \... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 52

A smooth chain \(A B\) of mass \(m\) rests against a surface in the form of a quarter of a circle of radius \(R\), If it released from rest, what is the velocity of the chain after it comes over the horizontal part of the surface? Solution (T.M.E.) \(=\) (T.M.E.) \(\quad \Rightarrow \quad \mathrm{K} \cdot \mathrm{E}_{i}+\mathrm{P.E}_{i}=\mathrm{K} \cdot \mathrm{E}_{y}+\mathrm{P} \mathrm{L}_{f}\) \(\Rightarrow U_{1}-U_{f}-K_{f}-K_{i}-\frac{1}{2} m v^{2}-0\) \(d U=(d m) g h=(\lambda R d \theta) g(R-R \sin \theta)-\frac{m R^{2} g}{(\pi R / 2)}(1 \quad \sin \theta) d \theta\) \(\Rightarrow d U-\left(\frac{2 m R g}{\pi}\right)(1 \sin \theta) d \theta\) \(\therefore U_{i}-\int d U-\left(\frac{2 m R g}{\pi}\right) \int_{o=0}^{\pi / 2}(1 \quad \sin \theta) d \theta-\left(\frac{2 m g R}{\pi}\right)\left[\int_{0-0}^{\pi / 2} d \theta \int_{0=f}^{\pi / 2} \sin \theta d \theta\right]\) \(-\left(\frac{2 m g R}{\pi}\right)\left[\frac{\pi}{2}-(\cos \theta)_{0}^{\pi / 2}\right]-\left(\frac{2 m g R}{\pi}\right)\left[\frac{\pi}{2}-\left(-\cos \frac{\pi}{2}+\cos 0\right)\right]-\left(\frac{2 m R g}{\pi}\right)\left[\frac{\pi}{2}-1\right]\) and \(U_{f}-0 ; K_{i}-0\) \(\therefore\) From (1) \(K_{f}-U_{i}-U_{f}-\left(\frac{2 m R g}{\pi}\right)\left(\frac{\pi}{2}-1\right) \Rightarrow \frac{1}{2} m v^{2}-m R g\left(1-\frac{2}{\pi}\right)\) \(\therefore U-\sqrt{2 R g\left(1 \frac{2}{\pi}\right)}\)

Short Answer

Expert verified
The velocity of the chain is \( v = \sqrt{2Rg\left(1 - \frac{2}{\pi}\right)}\)."

Step by step solution

01

Understand Initial and Final Energy

Initially, the chain is at rest, so its initial kinetic energy (\(K.E_i\)) is 0, and it has a potential energy (\(P.E_i\)) due to its height on the quarter circle. When the chain leaves the horizontal surface, its potential energy will be zero (\(P.E_f = 0\)). We need to find the final kinetic energy to determine the velocity.
02

Set Up the Conservation of Mechanical Energy Equation

The total mechanical energy (T.M.E) is conserved. This means the initial potential energy plus initial kinetic energy equals the final kinetic energy plus final potential energy. In formula terms: \(P.E_i + K.E_i = P.E_f + K.E_f\).
03

Calculate Change in Potential Energy

To find \(P.E_i\), integrate small mass elements \(dU = (dm)gh = (\lambda R d\theta)g(R-R \sin \theta)\) for the chain over the arc. With \(\lambda = \frac{m}{\frac{\pi R}{2}}\), the result becomes \(\frac{2mgR}{\pi}\left[\frac{\pi}{2} - 1\right]\).
04

Express the Final Kinetic Energy

Using the relation \(K.E_f = P.E_i - 0\), calculate \(\frac{1}{2} m v^2 = mRg\left(1 - \frac{2}{\pi}\right)\). This will allow us to solve for the final velocity \(v\).
05

Solve for the Final Velocity

Rearrange the equation to express \(v\): \(\frac{1}{2} m v^2 = mRg\left(1 - \frac{2}{\pi}\right)\). So, \(v = \sqrt{2Rg\left(1 - \frac{2}{\pi}\right)}\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's calculated using the formula \( K.E = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity. In the scenario of the chain on a quarter circle, initially, the chain is stationary, meaning its initial kinetic energy is zero.
As the chain moves and reaches the horizontal surface, it gains velocity, converting potential energy into kinetic energy. This change in energy allows us to calculate the velocity when all potential energy is converted to kinetic energy and potential energy is zero.
Potential Energy
Potential energy is related to an object's position in a gravitational field. For the chain resting on the quarter-circle surface, the height contributes to its initial potential energy. We calculate it using \( P.E = mgh \), where \( h \) is the height above the reference point, which in this case is the horizontal surface at the bottom of the circle.
As the chain starts from rest, it begins with maximum potential energy. As it descends, this energy decreases until it reaches zero, where the full potential energy is now kinetic, facilitating the calculation of the chain's final speed.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energies in a system. It is a fundamental concept in physics, especially in closed systems where energy is conserved.
For the chain problem, the Total Mechanical Energy (T.M.E.) conservation equation \( P.E_i + K.E_i = P.E_f + K.E_f \) holds because no external forces do work on the system. Initially, all energy is potential, but as the chain moves, this energy is converted into kinetic energy as there is no friction to account for.
Velocity Calculation
Velocity calculation in this exercise is derived from conserving energy principles. Initially, the velocity of the chain is zero, as it is at rest. Applying the equation \( \frac{1}{2} m v^2 = mRg\left(1 - \frac{2}{\pi}\right) \), where \( R \) is the radius of the quarter circle and \( g \) is the acceleration due to gravity, allows us to solve for the velocity the chain attains at the edge of the horizontal surface.
Rearranging, we find that \( v = \sqrt{2Rg\left(1 - \frac{2}{\pi}\right)} \), giving us a tangible result of how fast the chain is moving once it transitions from potential to pure kinetic energy.
Energy Integration
Energy integration involves calculating the total potential energy based on the sum of infinitesimal changes in potential energy over the entire path of the chain. This can be challenging because it requires integrating small segments of the chain's mass across the quarter-circle path.
Using \( dU = (dm)gh = (\lambda R d\theta)g(R-R \sin \theta) \), where \( \lambda = \frac{m}{\frac{\pi R}{2}} \), we integrate to find the initial potential energy over the circular arc. This approach underpins the derivation of the chain's velocity by providing a comprehensive understanding of how energy transforms as the system evolves from start to finish.

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Most popular questions from this chapter

Figure shows a smooth track, a part of which is a circle of radius \(R\). A block of mass \(m\) is pushed against a spring of spring constant \(k\) fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force \(m g\) when it reaches the point \(P\), where the radius of the track is horizontal.

Two particles of masses \(m_{1}\) and \(m_{2}\) are connected by an inextensible string which passes through a smooth vertical tube. The particle \(m_{1}\) is in equilibrium by the tension in the string which revolves the particle \(m_{2}\) in a horizontal circular path. Find the (a) angle \(\theta\) and (b) angular speed \(\omega\).

When a body is moving up with constant velocity: (a) work done by lifting force is positive (b) work done by lifting force is negative (c) work done by force of gravity is positive (d) work done by force of gravity is negative

Find the acceleration of a small body \(A\) which slides without initial velocity down a helical groove with a pitch \(h\) and radius \(R\) at the end of the \(n^{\text {th }}\) turn Neglect friction. Solution In \(n\) turns the particle will fall through a vertical height \(n h\). So, by the conservation of energy $$ m g(n h)-\frac{1}{2} m v^{2} \Rightarrow v-\sqrt{2 n g h} $$ The particle will have two accelerations at any instant: the tangential acceleration \(a_{t}=g \sin \alpha\), where \(\alpha=\) inclination of the groove with the horizontal and the normal acceleration $$ a_{n}-\frac{(v \cos \alpha)^{2}}{R} $$To find \(\alpha\), we mentally flatten the surlace of the cylinder with the helical groove into a plane surface. The groove will become the hypotenuse of a right-angled triangle of height \(n h\) and base \(2 \pi n R\) and so $$ \begin{aligned} & \tan \alpha-\frac{n h}{2 \pi n R}-\frac{h}{2 \pi R} \\ \therefore & a-\sqrt{a_{t}^{2}+a_{n}^{2}}-\sqrt{g^{2} \sin ^{2} \alpha+\frac{v^{4} \cos ^{4} \alpha}{R^{2}}} \\ \Rightarrow & a-\sqrt{g^{2} \frac{h^{2}}{h^{2}+4 \pi^{2} R^{2}}+\frac{4 n^{2} g^{2} h^{2}}{R^{2}} \frac{16 \pi^{4} R^{4}}{\left(h^{2}+4 \pi^{2} R^{2}\right)^{2}}} \\ \Rightarrow & a-\frac{g h \sqrt{h^{2} \mid 4 \pi^{2} R^{2}+64 \pi^{4} n^{2} R^{2}}}{h^{2}+4 \pi^{2} R^{2}} \end{aligned} $$

\(\Lambda\) string with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of \(2 \mathrm{~m}\) from the wall, has a point mass \(M=2 \mathrm{~kg}\) attached to it at a distance of \(1 \mathrm{~m}\) from the wall. \(\Lambda\) mass \(m=0.5 \mathrm{~kg}\) attached at the frec end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pullcy. What will be the speed with which the mass \(M\) will hit the wall when the mass \(m\) is relcased? Solution When mass \(m\) is released from rest, the heavier mass \(M\) loses potential energy and it strikes the wall with velocity \(v\) and the string remains stretched. It increases the kinetic energy of mass \(M\) and increases the kinetic and potential energies of mass \(m\), Decrease in potential energy of mass \(M\) \(=M_{g} \times 1=2.0 \times g \times 1=2 g \quad\) where \(g=9.8 \mathrm{~ms}^{-2}\) Increase in kinetic energy of mass \(M-\frac{1}{2} M v^{2}-\frac{1}{2} \times 2.0 \times v^{2}-v^{2}\) Increase in potential cnergy of mass \(m-m g\left(\sqrt{5} \quad\right.\) 1) \(-\frac{g}{2}(\sqrt{5} \quad 1)\) \(\\{\because\) Distance through which mass \(m\) is raised \(-A B+B C-A C-1+\sqrt{5}-2-(\sqrt{5}-1) \mathrm{m}\\}\) Increase in kinctic energy of mass \(m-\frac{1}{2} m v^{2}-\frac{1}{2} \times 0.5 \times v^{2}-\frac{v^{2}}{4}\) \Lambdas the total energy must be conserved, the decrease in the potential energy of the mass \(M\) cquals the increase in the potential and kinetic energies of both the masses.\(\therefore \quad M g \times 1-\frac{1}{2} M v^{2} \mid m g\left(\sqrt{5} \quad\right.\) 1) \(1 \frac{1}{2} m v^{2}\) \(\therefore 2 g-v^{2}+\frac{1}{2} g(\sqrt{5}-1)+\frac{1}{4} v^{2} \quad \Rightarrow \quad g\left[2-\frac{\sqrt{5}}{2}+\frac{1}{2}\right]-\frac{5}{4} v^{2}-g(3.6)\) or \(v^{2}-3.6 \times \frac{9.8 \times 4}{5} \quad \therefore \quad v-5.31 \mathrm{~m} / \mathrm{s}\)
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