Question

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstretched arm, 0.789 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.53 kg⋅m2 , a value that remains constant.As the student pulls his arms inward, his angular speed increases to 3.60 rev/s . How far are the masses from the axis of rotation at this time, considering the masses to be points

Answer 1

Answer:r'=0.327 m

Explanation:

Given

`N=2.85 rev/s`

angular velocity
`\omega =2\pi N=17.90 rad/s`

mass of objects
`m=1.5 kg`

distance of objects from stool
`r_1=0.789 m`

Combined moment of inertia of stool and student
`=5.53 kg.m^2`

Now student pull off his hands so as to increase its speed to 3.60 rev/s

`\omega _2=2\pi N_2`

`\omega _2=2\pi 3.6=22.62 rad/s`

Initial moment of inertia of two masses
`I_0=2mr_^2`

`I_0=2* 1.5* (0.789)^2=1.867`

After Pulling off hands so that r' is the distance of masses from stool

`I_0'=2* 1.5* (r')^2`

Conserving angular momentum

`I_1\omega =I_2\omega _2`

`(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62`

`I_0'=1.397* 0.791`

`I_0'=5.851`

`5.53+2* 1.5* (r')^2=5.851`

`2* 1.5* (r')^2=0.321`

`r'^2=0.107009`

`r'=0.327 m`