Question

A uniform disk turns at 4.1 rev/s around a frictionless central axis. A nonrotating rod, of the same mass as the disk and length equal to the disk’s diameter, is dropped onto the freely spinning disk, Fig. 11–32. They then turn together around the spindle with their centers superposed. What is the angular frequency in rev/s of the combination?

Accepted Answer

Start by identifying the principle of conservation of angular momentum. Since there are no external torques acting on the system, the total angular momentum before and after the rod is dropped must remain constant. Write the expression for the initial angular momentum of the system. The disk is rotating with an angular velocity of $$ \omega_i = 4.1 \; 	ext{rev/s} . $$The moment of inertia of a uniform disk is $$ I_{	ext{disk}} = \frac{1}{2} M R^2 $$, where $$ M  is $$the mass of the disk and $$ R  is $$its radius. Thus, the initial angular momentum is $$ L_i = I_{	ext{disk}} \omega_i . $$Next, calculate the moment of inertia of the rod. The rod is dropped onto the disk and rotates about its center. The moment of inertia of a uniform rod about its center is $$ I_{	ext{rod}} = \frac{1}{12} M L^2 $$, where $$ L  is $$the length of the rod. Since the length of the rod equals the diameter of the disk, $$ L = 2R . $$Substitute $$ L = 2R $$ into the formula for $$ I_{	ext{rod}} . $$After the rod is dropped, the disk and rod rotate together as a single system. The total moment of inertia of the system is the sum of the individual moments of inertia: $$ I_{	ext{total}} = I_{	ext{disk}} + I_{	ext{rod}} . $$Apply the conservation of angular momentum: $$ L_i = L_f $$, where $$ L_f = I_{	ext{total}} \omega_f . $$Solve for the final angular velocity $$ \omega_f  in $$terms of the given quantities: $$ \omega_f = \frac{I_{	ext{disk}} \omega_i}{I_{	ext{total}}} . $$Convert $$ \omega_f $$ back to revolutions per second (rev/s) if necessary.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Angular Momentum Conservation

Moment of Inertia

Angular Frequency