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What is the gravitational potential ener...

What is the gravitational potential energy of a particle of mass `m` kept at a distance `x` from the centre of a disc of mass `M` on its axis ? The radius of the disc is `R`.

Text Solution

Verified by Experts

The correct Answer is:
`-(2GMm)/(R^(2))(sqrt(R^(2)+x^(2))-x)`
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Knowledge Check

  • The gravitational potential energy at a body of mass m at a distance r from the centre of the earth is U. What is the weight of the body at this distance ?

    A
    `U`
    B
    `Ur`
    C
    `(U)/(r)`
    D
    `(U)/(2r)`

  • A disc of mass M and radius R can rotate freely in a vertical plane about a horizontal axis at O distance r from the centre of the disc as shown in Fig. The disc is released from rest in the shown position. Answer the following questions based on the above information The angular velocity of the disc in the above described case is

    A
    `sqrt((8gr)/(5{R^(2)+2r^(2)]))`
    B
    `sqrt((6gr)/(5[R^(2)+2r^(2)]))`
    C
    `sqrt((12gr)/(5[R^(2)+2r^(2)]))`
    D
    `sqrt((12gr)/(5R^(2)))`

  • A disc of mass M and radius R can rotate freely in a vertical plane about a horizontal axis at O distance r from the centre of the disc as shown in Fig. The disc is released from rest in the shown position. Answer the following questions based on the above information Reaction force exerted by the hinge on the disc at this instant is

    A
    `(Mg)/(5(R^(2)+2r^(2)))xxsqrt(9(R^(2)+6r^(2))^(2)+(4R^(2))^(2))`
    B
    `(Mg)/(5(R^(2)+2r^(2)))xx3(R^(2)+6r^(2))`
    C
    `(4Mg)/(5(R^(2)+2r^(2)))xxR^(2)`
    D
    `(Mg)/(5(R^(2)+2r^(2)))xx4R^(2)`