A thin rod (mass M =40grams, length 20cm) is originally at rest and free to rota

Question

A thin rod (mass M =40grams, length 20cm) is originally at rest and free to rotate about one of its ends in a horizontal plane (ignore gravity). A person pushes with a force of F =0.2N for 0.08seconds at the lose end of the rod, perpendicular to the rod, see figure below. A bug (approximated by a pointlike mass of m_bug = 10grams) originally sits at the fixed end of the rod (that is, in the center of the rotation). What is the moment of inertia of the rotating rod? How large is the torque exerted by the person on the rod, and what is the rod's resulting angular speed? Now the bug crawls from the center to the lose end of the rod. With no external torque acting, what is the tangential speed of the bug once it arrives at the end?

Explanation / Answer

Here, Initial angular momentum = 0.2 * 0.08 * 0.2 = 3.2 * 10-3 kg*m2/s

=> Final angular momentum = (1/3 * 0.04 * 0.2 * 0.2 + 0.01 * 0.2 * 0.2) * w

                                                = 9.33 * 10-4 * w

=>     Initial angular momentum = Final angular momentum

=>     3.2 * 10-3 =    9.33 * 10-4 * w

=>    w = 3.429 rad/sec

=>    angular speed of bug once it arrives at end = 3.429 rad/sec

                                            

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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