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Find the potential energy of the gravita...

Find the potential energy of the gravitational interaction of a point mass `m` and a rod of mass `m` and length `l`, if they are along a straight line. Point mass is at a distance of a from the end of the rod.

Find the potential energy of the gravitational

A

`U=(-Gm^(2))/l ln((2a+l)/(2a-l))`

B

`U=(-Gm^(2))/(l^(2)) ln((2a+l)/(2a-l))`

C

`U=(-Gm^(2))/(l^(2)) ln((2a-l)/(2a+l))`

D

`U=(-Gm^(2))/(l^(2)) ln((2a-l)/(2a+l))^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`dm=m/l.dx`
`dU=-(Gm.(dm))/((a-x))=(-Gm^(2)dx)/(l(a-x))`
`U=int_(-1//2)^(1//2)dU=-(Gm^(2))/l int_(-l//2)^(l//2)(dx)/(a-x)`
`=-(Gm^(2))/l ln((a+l//2)/(a-l//2))=-(Gm^(2))/lln((2a+l)/(2a-l))`
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Knowledge Check

  • Choose the most appropriate option. The potential energy of gravitational interaction of a point mass m and a thin uniform rod of mass M and length L, if they are located along a straight line at a distance a from each other, is

    A
    ` U = ( G M m )/(a) ln (( a + L )/( alpha ))`
    B
    `U = GM m ((1)/(a) - (1)/( a + L))`
    C
    `U = - ( GM m)/( a )`
    D
    `U =- ( GMm)/(L) ln (( a + L )/(a))`

  • A point mass is placed at a distance a from one end of a rod of mass M and length l on the line which passes through the central axis of the rod. The potential energy of the system.

    A
    `U =- ( GM m )/( a) log _(e) ((a + 1 )/(a))`
    B
    ` U =- G M m ((1)/(a) - (1)/(a +1))`
    C
    `U=- ( GMm )/(1) log _(e) ((a + 1)/( a))`
    D
    `U =- ( GMm )/(a)`

  • Moment of inertia of a uniform rod of mass m and length l is 7/12ml^2 mla about a line perpendicular to the rod. Find the distance of this line from mid point of the rod :

    A
    `l/2`
    B
    `l/sqrt2`
    C
    `l/4`
    D
    `l/sqrt3`

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