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Question

Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length l=√24a through their centres. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The angular momentum of the entire assembly about the point 'O' is →L (see the figure). Which of the following statement(s) is(are) true?

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Solution

The correct options are

**C** The centre of mass of the assembly rotates about the z-axis with an angular speed of ω/5

**D** The magnitude of angular momentum of the assembly about its centre of mass is 17ma2ω/2

Let the angle of axis with horizontal be θ.

cosθ=l√l2+a2=√245 (∵l=√24a)

Distance of centre of mass of the system from O,

lcm=ml+4m(2l)m+4m=95l

Let the angle of axis with horizontal be θ.

cosθ=l√l2+a2=√245 (∵l=√24a)

Distance of centre of mass of the system from O,

lcm=ml+4m(2l)m+4m=95l

Velocity of centre of mass, Vcm=(m×ωa)+(4m.ω.2a)m+4m=95ωa

Magnitude of angular momentum of centre of mass about point O, =(m+4m)Vcmlcm

=5m×95ωa×95l=81√245mωa2

Hence option B is incorrect.

Now angular momentum →L of entire assembly = Angular momentum of centre of mass system about point O + Angular momentum about the rod connecting two discs

z- component of angular momentum of centre of mass of entire assembly about point O, →L1=81√245mωa2cosθ

Angular momentum of assembly about the axis passing along rod, →L2=[12ma2+12(4m)(2a)2]ω

→L2=172mωa2

z-component of →L2=172mωa2(−sinθ)

Thus angular momentum of entire assembly about point O, →L=→L1+→L2=81√245mωa2cosθ+172mωa2(−sinθ)

→L=380350mωa2 (∵cosθ=√245 and sinθ=15)

Hence option A is incorrect.

Velocity of point P (center of lower disk) aω=lΩ

Where Ω is the angular velocity of centre of mass (C.M) about axis perpendicular to the extension of rod connecting two discs.

Then Ω=aωl

Thus angular velocity of C.M. w.r.t z axis =Ωcosθ

⇒ωCM−z=aωl√245=ω5

Hence option C is correct.

LD−CM=I1ω+I2ω

LD−CM=ma22ω+4m(2a)22ω=17ma2ω2

Hence option D is correct.

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