/*! 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 14 A particle \(P\) of mass \(m\) s... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 14

A particle \(P\) of mass \(m\) slides on a smooth horizontal table. \(P\) is connected to a second particle \(Q\) of mass \(M\) by a light inextensible string which passes through a small smooth hole \(O\) in the table, so that \(Q\) hangs below the table while \(P\) moves on top. Investigate motions of this system in which \(Q\) remains at rest vertically below \(O\), while \(P\) describes a circle with centre \(O\) and radius \(b\). Show that this is possible provided that \(P\) moves with constant speed \(u\), where \(u^{2}=M g b / m\).

Short Answer

Expert verified
For the two-body system described, the particle \(P\) can move in a circle with constant speed while the particle \(Q\) remains vertically stationary only if the speed \(u\) of the particle \(P\) is given by \(u^{2}=M g b / m\).

Step by step solution

01

Understanding the Physical Situation

From the problem, we are encouraged to assume that particle \(Q\) remains stationary under the table and that particle \(P\) is moving in a circular path with a uniform speed, \(u\), around the point where the string crosses the table. Both bodies are tied by a string and move under the effect of their respective weights. The system equilibrium will be determined by the balance between the centripetal force that keeps \(P\) in its circular path and the weight of \(Q\).
02

Calculation of the forces on both particles

The forces on the particles are their weights and the tension in the string. For \(P\), the tension \(T\) in the string is the only force acting on it. It provides the necessary centripetal force for circular motion of \(P\). This force is given by \(T=m u^{2} / b\). On the \(Q\), two forces act: its weight \(Mg\) and the tension of the string, equal to \(T\). Since \(Q\) is stationary, these forces must balance, i.e., \(T=Mg\).
03

Equating the expressions for Tension

Now equating the expressions for tension from above steps, we get \(m u^{2} / b = Mg\). This equation must hold true for \(P\) to move in circular path with constant speed while \(Q\) remains stationary.
04

Finding the Expression for Constant Speed

Rearranging the equation from step 3, we get the expression for constant speed \(u\) as \(u^{2}=M g b / m\). This equation confirms that for given masses \(m\) and \(M\), a specific radius \(b\), and the acceleration due to gravity \(g\), there is a specific constant speed \(u\) that must be maintained for the described behavior of the system.

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Most popular questions from this chapter

It is required to project a body from a point on level ground in such a way as to clear a thin vertical barrier of height \(h\) placed at distance \(a\) from the point of projection. Show that the body will just skim the top of the barrier if $$ \left(\frac{g a^{2}}{2 u^{2}}\right) \tan ^{2} \alpha-a \tan \alpha+\left(\frac{g a^{2}}{2 u^{2}}+h\right)=0 $$ where \(u\) is the speed of projection and \(\alpha\) is the angle of projection above the horizontal. Deduce that, if the above trajectory is to exist for some \(\alpha\), then \(u\) must satisfy $$ u^{4}-2 g h u^{2}-g^{2} a^{2} \geq 0 $$

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A particle of mass \(m\) can move on a rough horizontal table and is attached to a fixed point on the table by a light inextensible string of length \(b\). The resistance force exerted on the particle is \(-m K v\), where \(v\) is the velocity of the particle. Initially the string is taut and the particle is projected horizontally, at right angles to the string, with speed \(u\). Find the angle turned through by the string before the particle comes to rest. Find also the tension in the string at time \(t\).

A particle of mass \(m\) can move on a rough horizontal table and is attached to a fixed point on the table by a light inextensible string of length \(b\). The resistance force exerted on the particle is \(-m K v\), where \(v\) is the velocity of the particle. Initially the string is taut and the particle is projected horizontally, at right angles to the string, with speed \(u\). Find the angle turned through by the string before the particle comes to rest. Find also the tension in the string at time \(t\).

The radius of the Moon's approximately circular orbit is \(384,000 \mathrm{~km}\) and its period is \(27.3\) days. Estimate the mass of the Earth. \(\left[G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}\right.\).] The actual mass is \(5.97 \times 10^{24} \mathrm{~kg}\). What is the main reason for the error in your estimate? An artificial satellite is to be placed in a circular orbit around the Earth so as to be 'geostationary'. What must the radius of its orbit be? [The period of the Earth's rotation is \(23 \mathrm{~h} 56\) \(\mathrm{m}\), not \(24 \mathrm{~h}\). Why?]
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