A student sits on a freely rotating stool holding two dumbbells, each of mass 2.

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 2.92 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.96 m from the axis of rotation and the student rotates with an angular speed of 0.743 rad/s. The moment of inertia of the student plus stool is 2.71 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.304 m from the rotation axis

(a) Find the new angular speed of the student. in (rad/s)

(b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward. in (J)

Explanation / Answer

nitial I = 2.71 + 2*2.92*(0.96)^2 = 8.092 kg·m^2

final I = 2.71 + 2*2.92*(0.304)^2 = 3.2497 kg·m^2

(a)

Conserve angular momentum: initial I* = final I*

8.092 * 0.743 = 3.2497 *

= 1.85 rad/s

(b)

initial KE = 1/2I^2 = 0.5 * 8.092 * (0.743)^2 = 2.2335 J

final KE = 1/2 * 3.2497 * (1.85)^2 = 5.5610 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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