A student sits on a freely rotating stool holding two dumbbells, each of mass 2.
ID: 1634998 • Letter: A
Question
A student sits on a freely rotating stool holding two dumbbells, each of mass 2.97 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 1.03 m from the axis of rotation and the student rotates with an angular speed of 0.752 rad/s. The moment of inertia of the student plus stool is 2.77 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.299 m from the rotation axis (Figure b).
(a) Find the new angular speed of the student.
rad/s
(b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.
Explanation / Answer
moment of inertia when arms are extended,
Ii = (2.77) + 2 ( 2.97 x 1.03^2) = 9.07 kg m^2
wi = 0.752 rad/s
when dumbell are pushed closer,
If = (2.77) + 2(2.97 x 0.299^2) = 3.30 kg m^2
(A) Applying angular momentum conservation,
Ii wi = If wf
(9.07) (0.752) = (3.30) wf
wf = 2.07 rad/s
(B) K_before = Ii wi^2 / 2
= 2.56 J
K_after = If wf^2 / 2
= 7.07 J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.