A student sitting on a frictionless rotating stool has rotational inertia 0.95 k
ID: 1371890 • Letter: A
Question
A student sitting on a frictionless rotating stool has rotational inertia 0.95 kg # m2 about a vertical axis through her center of mass when her arms are tight to her chest. The stool rotates at 6.80 rad>s and has negligible mass. The student extends her arms until her hands, each holding a 5.0-kg mass, are 0.75 m from the rotation axis. (a) Ignoring her arm mass, what s her new rotational velocity? (b) Repeat if each arm is modeled as a 0.75-m-long uni- form rod of mass of 5.0 kg and her total body mass is 65 kg.
Explanation / Answer
From the conservation of momentum
I11 = I22
2= I11 /I2
= 0.95 kg m^2 ( 6.8 rad/s)/ 0.96 kg m^2 + 10 kg ( 0.75 m)^2
=0.981 rad/s
if her arms are also contributing in inertia new inertia
I new = 0.96 kg m^2 + 10 kg ( 0.75 m)^2+2ml^/3=0.96 kg m^2 + 10 kg ( 0.75 m)^2+2(5 kg )(0.75 m)^2/3
=6.585 kg m^2 +1.875 kg m^2
=8.464 kg m^2
her new inertia
w2 = 0.95 kg m^2 ( 6.8 rad/s)/8.463 kg m^2= 0.763 rad/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.