Question

A student sits on a freely rotating stool holding two weights, each of mass 3.10 kg. When his arms are extended horizontally, the weights are 1.00 m from the axis of rotation and he rotates with an angular speed of 0.754 rad/s. The moment of inertia of the student plus stool is 3.10 kg·m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.290 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 weights inward.
in J (before)
in J (after)

Answer 1

Answer

Give that,

mass of two student m = 3.10 kg  

distance from the axis of the rotation is r = 1 m

angular speed ω = 0.754 rad /s  

moment of inertia I = 3.10 kg m²

position from the rotation of the axis is r_1 = 0.29 m

Total moment of inertia I '   = I + 2 m r²

                                          = 3.1 +2 x 3.1 x( 1)^2  

                                           = 9.3 kg m²

moment of inertia inward horizontally from the position of rotation axis is  

                                   I"   = I + 2 m r^2  

                                         = 3.1 + 2 * 3.1 kg ( 0.29)^2  

                                         = 3.62 kg m^2  

a ) new angular speed is ω_1 =$\dfrac{I'\omega}{I''}$

                                               = $\dfrac{9.3\times 0.754}{3.62}$  

                                               = 1.94 rad /s

b ) K.E before the system pulls weight inward is

$K.E = \dfrac{1}{2} I' \omega^2$

$K.E = \dfrac{1}{2}\times 9.3 \times 0.754^2$

$K.E =2.64\ J$

c )K.E after the system pulls weight inward is

$K.E = \dfrac{1}{2} I'' \omega_1^2$

$K.E = \dfrac{1}{2}\times 3.62 \times 1.92^2$

$K.E =6.67\ J$