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

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nuuk nuuk · Answer 1 · 2019-11-11T08:32:35+01:00

Answer:r'=0.327 m

Step-by-step explanation:

Given

[tex]N=2.85 rev/s[/tex]

angular velocity [tex]\omega =2\pi N=17.90 rad/s[/tex]

mass of objects [tex]m=1.5 kg[/tex]

distance of objects from stool [tex]r_1=0.789 m[/tex]

Combined moment of inertia of stool and student [tex]=5.53 kg.m^2[/tex]

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

[tex]\omega _2=2\pi N_2[/tex]

[tex]\omega _2=2\pi 3.6=22.62 rad/s[/tex]

Initial moment of inertia of two masses [tex]I_0=2mr_^2[/tex]

[tex]I_0=2\times 1.5\times (0.789)^2=1.867[/tex]

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

[tex]I_0'=2\times 1.5\times (r')^2[/tex]

Conserving angular momentum

[tex]I_1\omega =I_2\omega _2[/tex]

[tex](5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62[/tex]

[tex]I_0'=1.397\times 0.791[/tex]

[tex]I_0'=5.851[/tex]

[tex]5.53+2\times 1.5\times (r')^2=5.851[/tex]

[tex]2\times 1.5\times (r')^2=0.321[/tex]

[tex]r'^2=0.107009[/tex]

[tex]r'=0.327 m[/tex]