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  • A uniform circular disc of mass '' and radius '<img alt="R" src="https://entrancecorner.oncodecogs.com/gif.latex?R

A uniform circular disc of mass 'M' and radius 'R' is rotating with an angular velocity of ' \omega_0 ' about its own axis, perpendicular to the plane. Two uniform circular discs, each of mass ' m ' and radius ' r '  are gently placed on the disc such that they touch each other along the axis of the larger disc and the rim of the disc. What will be the final angular velocity of the system. [Given, all discs have same mass density]

Option: 1

\omega=\frac{8}{11} \omega_0


Option: 2

\omega=\frac{11}{8} \omega_0


Option: 3

\omega=\frac{16}{11} \omega_0


Option: 4

\omega=\frac{11}{16} \omega_0


Answers (1)

best_answer

Applying conservation of angular momentum:-

\begin{aligned} & I_i \omega_i=I_f \omega_f \\ & I_i=\frac{1}{2} M R^2, \omega_i=\omega_0 \\ & I_f=\frac{1}{2} M R^2+2\left[\frac{1}{2} m r^2+m r^2\right] \\ & I_f=\frac{1}{2} M R^2+2 \times \frac{3}{2} m r^2 \\ & r=R / 2 \\ & I_f=\frac{1}{2} M R^2+3 \frac{m R^2}{4} \\ & \frac{M}{m}=\frac{K R^2}{K r^2}=\left(\frac{R}{r}\right)^2=\left({\frac{2R}{R}}\right)^2=\frac{4}{1} \end{aligned}

\begin{aligned} & m=\frac{M}{4} \\ & I_f=\frac{1}{2} M R^2+\frac{3}{4} \times \frac{M}{4} R^2 \\ & I_f=\left(\frac{8+3}{16}\right) M R^2=\frac{11}{16} M R^2 \\ & \left(\frac{1}{2} M R^2\right) \omega_0=\left(\frac{11}{16}\right) M R^2 \omega \\ & \omega=\frac{8}{11} \omega_0 \end{aligned}

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