4.) A woman with mass 50kg is standing on the rim of a large disc that is rotati
ID: 1391141 • Letter: 4
Question
4.) A woman with mass 50kg is standing on the rim of a large disc that is rotating at .5 rev/s about an axis perpendicular to it through its center. The disc has a mass of 110kg and a radius of 4m.
a.) Calculate the magnitude of the total angular momentum and the kinetic energy of the woman-plus-disc-system, assuming that you can treat the woman as a point.
b.) The woman starts walking towards the center of the disc. Calculate the angular velocity of the system when she is at 1m from the center of the disc.
c.) What is the change in energy for the system? The moment of inertia for a disc is I=(1/2)MR^2
Explanation / Answer
mass of a women m= 50 kg, w= 5 rev/s = 5 x 2 pi = 10 pi radian/sec
mass of disc M = 110 kg, radius of disc = 4 m
a) Total angular momentum of women plus disc
L = ( M.I of disc + M.I of women) x w
= (MR2/2 + mR2) w
= ( 110 x 42/2 + 50 x 42) x 10 pi
= (880 + 800 ) x 10 pi
= 16800 pi = 16800 x3.14
L= 52752 kgm2 rad/s
Kinetic energy K.E = I w2/2
I = (M + m) R2/2
K.E =(( M + m)/2) R2 w2 /2
= (( 110 + 50)/2) x 42 x (10pi)2/2
= 160 x 4 x4 x 100 x pi2/ 4
= 64000 pi2 = 64000 x 3.142
K.E = 6.31x105 J
b) The angular velocity of system women at adistance from centre, theeta = s/R
s = total distance covered in a second = 5 rev/s
= 5 x 2 pi r
theeta = 5 x 2 pi r/R
= 5 x 2 x 3.14 x 1 / 4
= 31.4 /4
angular velocity, theeta = 7.85 radian / sec
c) Energy of the disc T= Iw2/2
= (MR2/2) x (5 x 2 pi)2/2
= 110 x 42 x 100 x pi2/ 4
= 44000 pi2 = 44000 x 3.142
T = 4.34x104 Joule
Change in energy = K.E of disc with womem - K.E of disc
= 6.31 x105 - 4.34x105
Change in energy of system=1.97x105 joule
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.