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Two equal masses m are attached by a str...

Two equal masses m are attached by a string. One mass lies at radial distance r from the centre of a horizontal turntable which rotates with constant angular velocity `omega=2` rad `s(1)`, while the second hangs from the string inside the tumtable's hollow spindle (see fig.) The coefficient of static friction between the turntable and the mass lying on it is `mu_(s)=0.5`. The maximum and minimum values of radius such that the mass lying on the turntable does not slide are `r_("max")+r_("min")` in meter is

Text Solution

Verified by Experts

T=mg
`m_("max")omega^(2)=T+mumg`
`mr_("min")omega^(2)=T-mumg`
`m(r_("max+r_("min"))omega^(2)=2mg`
`therefore r_("max")+r_("min")=(2g)/(omega^(2))=5m`
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Knowledge Check

  • Two men of equal masses stand at opposite ends of the diameter of a turntable disc of a certain mass. Moving with constant angular velocity. The two men make their way to the middle of the turntable at equal rates in doing so

    A
    kinetic energy of rotation has increased while angular momentum remains same
    B
    kinetic energy of rotation has decreased while angular momentum remains same
    C
    kinetic energy of rotation has decreased but angular momentum has increased
    D
    both, kinetic energy of rotation and angular momentum have decreased

  • Stone of mass 1 kg tied to the end of a string of length 1m , is whirled in horizontal circle with a uniform angular velocity 2 rad s^(-1) . The tension of the string is (in newton)

    A
    2
    B
    `1/3`
    C
    `4`
    D
    `1/4`

  • The coefficient of static friction between a small coin and the surface of a turntable is 0.30 . The turntable rotates at 2 rad s^(-1) . What is maximum distance from the centre of the turntable at which the coin will not slide ?

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    B
    1.5 m
    C
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    D
    0.75 m

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