Question

Two disks are mounted (like a merry-go-round) on low-friction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia 3.00 kg m about its central axis, is set spinning counterclockwise at 550 rev/min. The second disk, with rotational inertia 6.40 kg m about its central axis, is set spinning counterclockwise at 800 rev/min. They then couple together. (a) What is their angular speed after coupling? rev/min (b) If instead the second disk is set spinning clockwise at 800 rev/min, what is their angular speed after they couple together? rev/min

Answer 1

Moment of Inertia of first disk,

Moment of Inertia of second disk, $I_2=6.40 \textrm { kg}\cdot\rm{m^2}$

Angular speed of first disk, $\omega _1=550 \textrm { rev/min}$

Angular speed of second disk, $\omega _2=800 \textrm { rev/min}$

Let us consider that $I \textrm{ and } \omega$ are the moment of inertia and angular speed of the coupled unit.

(a) According to conservation of angular momentum (L), the final angular momentum must be equal to the inital angular momentum of the system.

$L_{final}=L_{initial}\\ I\omega =I_1\omega_1+I_2\omega_2 \\ \Rightarrow \omega =\frac{I_1\omega_1+I_2\omega_2}{I}\\ \Rightarrow \omega =\frac{I_1\omega_1+I_2\omega_2}{I_1+I_2}\\ \Rightarrow \omega =\frac{(3.00 \textrm{ kg}\cdot\rm{m^2})(550 \textrm{ rev/min})+(6.40 \textrm{ kg}\cdot\rm{m^2})(800 \textrm{ rev/min})}{(3.00 \textrm{ kg}\cdot\rm{m^2})+(6.40 \textrm{ kg}\cdot\rm{m^2})}=720.21 \textrm{ rev/min}\\$

Hence, the coupled unit rotates in the counterclockwise direction with the angular speed of 720.21 rev/min.

(b) If the second disk was set to spin in the opposite direction, then the final angular speed becomes,

$\omega =\frac{(3.00 \textrm{ kg}\cdot\rm{m^2})(550 \textrm{ rev/min})+(6.40 \textrm{ kg}\cdot\rm{m^2})(-800 \textrm{ rev/min})}{(3.00 \textrm{ kg}\cdot\rm{m^2})+(6.40 \textrm{ kg}\cdot\rm{m^2})}\\ \omega =\frac{(3.00 \textrm{ kg}\cdot\rm{m^2})(550 \textrm{ rev/min})-(6.40 \textrm{ kg}\cdot\rm{m^2})(800 \textrm{ rev/min})}{(3.00 \textrm{ kg}\cdot\rm{m^2})+(6.40 \textrm{ kg}\cdot\rm{m^2})}\\ \omega =-369.15 \textrm{ rev/min}$

-Ve sign indicates that the coupled unit rotates in the clockwise direction.

Hence, the coupled unit rotates in the clockwise direction with the angular speed of 369.15 rev/min.