A solid disk rotates in the horizontal plane at an angular velocityof 0.046 rad/

Question

A solid disk rotates in the horizontal plane at an angular velocityof 0.046 rad/s with respect to an axisperpendicular to the disk at its center. The moment of inertia ofthe disk is 0.12 kg ·m2. From above, sand is dropped straight down onto thisrotating disk, so that a thin uniform ring of sand is formed at adistance of 0.40 m from the axis. The sand in the ring has a massof 0.50 kg. After all the sand is in place, what is the angularvelocity of the disk?

I just don't get it. How can I find the answer without knowingeither the radius of the disk or its mass. If there is a connectionsomewhere, I don't see it.

Explanation / Answer

L=I accordingto definition of angular momentum. the principle of conservation of angular momentum saysLf=L0. SoIff=I00. Moments of Inertia can be added like masses can be addedmr2=I. If=(0.5kg x 0.42m) + 0.12kgm2= newinertia is 0.2kgm2 f=(I00)/If=(0.12x 0.046)/0.2=0.0276 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.