A student sits on a freely rotating stool holding two dumbbells, each of mass 3.

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 3.08 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.99 m from the axis of rotation and the student rotates with an angular speed of 0.758 rad/s. The moment of inertia of the student plus stool is 2.74 kg·m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.290 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 Kbefore = Kafter

Explanation / Answer

Given that

Each of the mass m = 3.08 kg

distance from the axis of the rotation is r = 0.99 m

angular speed = 0.758 rad /s

moment of inertia I =2.74 kg m^2

position from the rotation of the axis is r_1 = 0.290 m

Total moment of inertia I '   = I + 2 m r^2

=2.74 +2 * 3.08 kg ( 0.99 m)^2

= 8.78  kg m^2

moment of inertia inward horizontally from the position of rotation axis  

  I"   = I + 2 m r_1 ^2

= 2.74 + 2 *3.08 kg ( 0.290 m )^2

= 3.26 kg m^2

a ) new angular speed is _1 = I ' / I "

= 8.78* 0.758/3.26

= 2.04 rad /s

b) K.E before the system pulls weight inward is K.E = 1/2 I ' ( )^2

=  1/2 * ( 8.78 ) ( 0.758 rad /s )^2

= 2.52 J

K.E after the system pulls weight inward is K.E = 1/2 I " ( _1 )^2

= 1/2 * ( 3.26) ( 2.04rad /s )^2

= 6.78 J

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