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A disc of mass M and radius R is rolling...

A disc of mass M and radius R is rolling with angular speed `omega` on a horizontal plane as shown in figure. The magnitude of angular momentum of the disc about the origin O is

A

`(1/2) MR^2omega`

B

`MR^2omega`

C

`(3/2)MR^2 omega`

D

`2 MR^2 omega`

Text Solution

Verified by Experts

The correct Answer is:
C
(c ) The disc has two types of motion namely translational
and rotational. Therfore there are two types of angular
momentum and the total angular momentum is the
vector sum of these two.
in this case both the angular momentum have
the same direction
(perpendicular to the
plane of paper and away from the reader).

`vecL = vecL_T + vecLr`
`L_T` angular momentum due to translational motion.
`L_R` = angular momentum due to rotaional motion
about C.M.
`L = MVxxR+I_(cm)omega`
`I_(cm)` = M.I. about centre of mass C.
`=M(R omega)R+(1)/(2)MR^2 omega`
(v = Romega` in case of rolling motion and surface at rest)
`=(3)/(2) MR^2 omega`
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Knowledge Check

  • A disc of mass M and radius R is rolling with angular speed omega on a horizontal plane. The magnitude of angular momentum of the disc about a point on ground along the line of motion of disc is :

    A
    `(1//2)MR^(2)omega`
    B
    `MR^(2)omega`
    C
    `(3//2)MR^(2)omega`
    D
    `2MR^(2)omega`

  • A disc of mass m and radius R moves in the x-y plane as shown in Fig. The angular momentum of the disc about the origin O at the instant shown is

    A
    `-5/2mR^(2)omegahatk`
    B
    `7/3mR^(2)omegahatk`
    C
    `-9/2mR^(2)omegahatk`
    D
    `5/2mR^(2)omegahatk`

  • A uniform circular disc of mass M and radius R rolls without slipping on a horizontal surface. If the velocity of its centre is v_0 , then the total angular momentum of the disc about a fixed point P at a height (3R)//2 above the centre C .

    A
    increases continuously as the disc moves away
    B
    decreases continuously as the disc moves away
    C
    is always equal to `2MRv_(0)`
    D
    is always equal to `MRv_(0)`
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