Question

A student sits on a rotating stool holding two 2.8-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg · m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.34 m from the rotation axis. (a) Find the new angular speed of the student.

Answer 1

Step-by-step explanation:

Given

mass of objects m is 2.8 kg

objects is
`r=1 m` away from axis of rotation

angular velocity
`\omega _1=0.75 rad/s`

moment of inertia of stool and student is
`=3 kg-m^2`

New distance of objects
`r'=0.34 m`

conserving Angular momentum

Initial Angular momentum
`L_1=I\omega _1`

`I_1=3+2* 2.8* 1^2`

`I_1=3+5.6=8.6 kg-m^2`

`L_1=8.6* 0.75=6.45`

For Second case

`I_2=3+2* 2.8* 0.34^2=3.32 kg-m^2`

`L_2=3.32* \omega _2`

`L_1=L_2`

`6.45=3.32* \omega _2`

`\omega _2=1.94 rad/s`