Question

A student holds two lead weights, each of mass 6.4 kg. When the students’ arms are extended horizontally, the lead weights are 0.84 m from the axis of rotation and the student rotates with an angular speed of 2.4 rad/sec. The moment of inertia of student plus stool is 9.4 kg m2 and is assumed to be constant; i.e., the student’s arms are massless! Then the student pulls the lead weights horizontally to a radius 0.23 m from the axis of rotation.

1). Find the new angular speed of the student?
2). Find the kinetic energy of the rotating system (comprised of student, stool, and weights) before and after he pulls the weights inward?

Answer 1

(1) ω = 4.3 rad/sec

(2) 52.99 j and 93.4 j

Step-by-step explanation:

from the question we are given the following:

mass of weight (m) = 6.4 kg

initial radius (ri) = 0.84 m

final radius (rf) = 0.23 m

angular speed (ω) = 2.4 rad/sec

moment of inertia of the student and stool (I) = 9.4 kgm^{2}

find the new angular speed and the kinetic energy of the rotating system before and after pulling the weight inwards.

(1) We can find the new angular momentum from the equation

initial angular momentum = final angular momentum

where

• angular momentum = (total inertia) x angular speed

• total inertia = inertia of student and stool + inertia of the masses

• inertia of the masses = mr^{2}

the equation now becomes

( I + 2m(ri)^{2}) x ω = ( I + m(rf)^{2}) x ω

(9.4 + (2 X 6.4 X 0.84^{2})) X 2.4 = ( 9.4 + (2 X 6.4 X 0.23^{2})) X ω

44.2 = 10.1 X ω

ω = 4.3 rad/sec

(2) kinetic energy = 0.5 x I x ω^{2}

kinetic energy before = 0.5 x 18.4 x 2.4^{2} = 52.99 j

kinetic energy after = 0.5 x 10.1 x 4.3^{2} = 93.4 j

Answer 2

1)
`\omega_f=10.5354\,rad.s^(-1)`

2)
`KE_i=53.0833\,J` &
`KE_f=97.0946\,J`

Step-by-step explanation:

Given:

Mass of lead weight in each hand,
`m=6.4\,kg`

distance of weights while arms extend,
`r_i=0.84\,m`

initial angular speed of student holding the masses in extended arms,
`\omega_i=2.4\,rad.s^(-1)`

moment of inertial of student and stool,
`I_s=9.4\,kg.m^2`

final radius of lead weights,
`r_f=0.23\,m`

1.

Moment of inertia due to lead weights in the extended arms:

`I_w_i=2* m.r_i\,^2`

`I_w_i=2* 6.4* 0.84^2`

`I_w_i=9.0317\,kg.m^2`

∴Total moment of inertia initially

`I_i=I_s+I_w_i`

`I_i=9.4+9.0317`

`I_i=18.4317\,kg.m^2`

Moment of inertia due to lead weights in the pulled-in arms:

`I_w_f=2* m.r_f\,^2`

`I_w_f=2* 6.4* 0.23^2`

`I_w_f=0.6771\,kg.m^2`

∴Total moment of inertia in final condition:

`I_f=I_s+I_w_f`

`I_f=9.4+0.6771`

`I_f=10.0771\,kg.m^2`

• According to the law of conservation of angular momentum:

`I_i.\omega_i=I_f.\omega_f`

`18.4317* 2.4=10.0771* \omega_f`

`\omega_f=4.3898\,rad.s^(-1)`

2.

Total Kinetic Energy before the student pulls his arm:

`KE=(1)/(2) I_i.\omega_i\,^2`

`KE=(1)/(2) * 18.4317* 2.4^2`

`KE_i=53.0833\,J`

Total Kinetic Energy after the student pulls his arm:

`KE=(1)/(2) I_f.\omega_f\,^2`

`KE=(1)/(2) * 10.0771* 4.3898^2`

`KE_f=97.0946\,J`