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 2.93 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.97 m from the axis of rotation and the student rotates with an angular speed of 0.741 rad/s. The moment of inertia of the student plus stool is 2.75 kg middot m^2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.291 m from the rotation axis (Figure b). (a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward. K_before = 2.32 K_after = J

Explanation / Answer

Initial Moment of inertia of the system is I1 = (Ix) + (2*m*r^2) = (2.75)+(2*2.93*0.97^2)= 8.26 kg-m^2

Final moment ofinertia of the system is I2 = (Ix)+(2*m*r1^2) = (2.75)+(2*2.93*0.291^2)= 3.25 Kg-m^2

w1 = 0.741 rad/sec

w2 = ?

Using law of conservation of angular momentum

I1*w1 = I2*w2

(8.26*0.741) = (3.25*w2)

w2 = 1.88 rad/sec is the answer for a)

b) Kbefore is (0.5*I1*w1^2) = (0.5*8.26*0.741^2) = 2.26 J

Kafter = 0.5*I2*w2^2 = (0.5*3.25*1.88^2) = 5.74 J

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