Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 11, Problem 103P

(a)

To determine

ToShow: That the potential energy shared by an element of the rod of mass dm and the point particle of mass m0 located at x0 12L is given by:

  dU=Gm0dmx0xsdU=GMm0L(x0xs)dxs

(a)

Expert Solution
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Explanation of Solution

Given information :

Mass of point particle =m0

Formula used :

Gravitational potential energy

  U=Gmomr

U is gravitational potential energy, G is the gravitational constant, m is the mass of large body and m0 is the mass of the other body.

Calculation:

Let U = 0 at x =  .

The potential energy of an element of the stick dm and the point mass m0 is given by the definition of gravitational potential energy.

  U=Gmomr

Where r is the separation of dm and m0 .

  dU=Gm0dmx0xs

  dm=λdxo

  λ=ML

  dU=Gm0λdx0x0xsdU=GMm0dx0L(x0xs)

Conclusion:

The potential energy shared by an element of the rod of mass dm and the point particle of mass m0 located at x0 12L is given by:

  dU=Gm0dmx0xsdU=GMm0L(x0xs)dxs

(b)

To determine

ToIntegrate:The result had in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 103P

  U=GMm0Lln(x0+L/2x0L/2)

Explanation of Solution

Given information:

Mass of point particle =m0

Formula used:

Gravitational potential energy

  U=Gmomr

U is gravitational potential energy, G is the gravitational constant, m is the mass of large body and m0 is the mass of the other body.

Calculation:

Integrate dU to find the total potential energy of the system:

  U=GMm0LL/2L/2dxsx0xsU=GMm0L[ln(x0L2)ln(x0+L2)]U=GMm0Lln(x0+L/2x0L/2)

Conclusion:

The integration of dU=GMm0dxL(x0xs) is, U=GMm0Lln(x0+L/2x0L/2) .

(c)

To determine

To Calculate: The force on m0 at a general point x using Fx=dU/dx and compare the result with m0g .

(c)

Expert Solution
Check Mark

Answer to Problem 103P

  F(x0)=Gmm0x2L2/4

Explanation of Solution

Given information:

Mass of point particle =m0

Formula used:

Force on mass  m0

  F(x0)=dUdx0

Where, dU is the potential energy difference and dxo is the distance to mass  m0 .

Calculation:

  U=GMm0L[ln(x0L2)ln(x0+L2)]

Because x0 is a general point along the x axis:

  F(x0)=dUdx0F(x0)=GMm0L[1x0+L21x0L2]

  F(x0)=Gmm0xo2L2/4

This answer and the answer given in Example 11-8 are the same.

Conclusion:

The force on m0 at a general point x using Fx=dU/dx is, F(x0)=Gmm0x2L2/4 .

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Chapter 11 Solutions

Physics for Scientists and Engineers, Vol. 1

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