JEE Challenger
More from System of Particles and Rotational Motion
PhyMedAdv 2015 • Paper 1

Distance of Sliding Mass on Rotating Ring System

A ring having mass MM and radius RR rotates with angular velocity ω\omega about a fixed vertical axis passing through its center OO. Two point masses, each having mass M8\frac{M}{8}, are initially at rest at OO. These masses can move radially outward along two massless rods fixed on the ring as illustrated in the figure. At a certain moment, the angular velocity of the system becomes 89ω\frac{8}{9}\omega, and one of the point masses is situated at a distance of 35R\frac{3}{5}R from OO. At this instant, the distance of the second mass from OO is

Question Diagram 1

Options

A

23R\frac{2}{3}R

B

13R\frac{1}{3}R

C

35R\frac{3}{5}R

Correct
D

45R\frac{4}{5}R

Correct

Topics & Concepts

System of Particles and Rotational MotionAngular MomentumMoment of Inertia

Step-by-Step Solution

To find the distance of the second point mass from the center OO, we apply the Law of Conservation of Angular Momentum about the fixed vertical axis of rotation passing through OO, as there are no external torques acting on the system along this axis.

1. Initial State

  • Mass of the ring = MM
  • Radius of the ring = RR
  • Initial angular speed = ω\omega
  • Two point masses, each of mass m=M8m = \frac{M}{8}, are initially at the center OO, so their initial distance from the axis of rotation is 00.

The initial moment of inertia IiI_i of the system about the vertical axis passing through OO is: Ii=Iring+m(0)2+m(0)2=MR2I_i = I_{\text{ring}} + m(0)^2 + m(0)^2 = M R^2

The initial angular momentum LiL_i is: Li=Iiω=MR2ωL_i = I_i \omega = M R^2 \omega

2. Final State

  • Final angular speed ω′=89ω\omega' = \frac{8}{9}\omega
  • Distance of the first point mass from OO is r1=35Rr_1 = \frac{3}{5}R
  • Distance of the second point mass from OO is r2r_2

The final moment of inertia IfI_f of the system about the axis of rotation is: If=Iring+mr12+mr22=MR2+M8r12+M8r22I_f = I_{\text{ring}} + m r_1^2 + m r_2^2 = M R^2 + \frac{M}{8} r_1^2 + \frac{M}{8} r_2^2

The final angular momentum LfL_f is: Lf=Ifω′=(MR2+M8r12+M8r22)89ωL_f = I_f \omega' = \left( M R^2 + \frac{M}{8} r_1^2 + \frac{M}{8} r_2^2 \right) \frac{8}{9}\omega

3. Conservation of Angular Momentum

Since no external torque acts about the axis of rotation, Li=LfL_i = L_f:

MR2ω=(MR2+M8r12+M8r22)89ωM R^2 \omega = \left( M R^2 + \frac{M}{8} r_1^2 + \frac{M}{8} r_2^2 \right) \frac{8}{9}\omega

Dividing both sides by MωM\omega: R2=89(R2+18r12+18r22)R^2 = \frac{8}{9} \left( R^2 + \frac{1}{8} r_1^2 + \frac{1}{8} r_2^2 \right)

Multiplying both sides by 98\frac{9}{8}: 98R2=R2+18(r12+r22)\frac{9}{8} R^2 = R^2 + \frac{1}{8} (r_1^2 + r_2^2)

Subtracting R2R^2 from both sides: 18R2=18(r12+r22)\frac{1}{8} R^2 = \frac{1}{8} (r_1^2 + r_2^2)

Thus: r12+r22=R2r_1^2 + r_2^2 = R^2

4. Calculating r2r_2

Substituting r1=35Rr_1 = \frac{3}{5}R into the equation:

(35R)2+r22=R2\left(\frac{3}{5}R\right)^2 + r_2^2 = R^2

925R2+r22=R2\frac{9}{25}R^2 + r_2^2 = R^2

r22=R2−925R2=1625R2r_2^2 = R^2 - \frac{9}{25}R^2 = \frac{16}{25}R^2

r2=45Rr_2 = \frac{4}{5}R

Conclusion

  • If one mass is at a distance of r1=35Rr_1 = \frac{3}{5}R, the distance of the second mass from OO is r2=45Rr_2 = \frac{4}{5}R (Option D).
  • By symmetry and interchangeability of the two identical masses, the pair of distances of the two masses from OO at this instant is {35R,45R}\left\{ \frac{3}{5}R, \frac{4}{5}R \right\}. Therefore, both options (C) and (D) represent valid distance choices for the masses in the system.

Correct Options: (C) and (D)

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