A student on a plane stool rotates freely with an angular speed of 3.05 rev/s Th

Question

A student on a plane stool rotates freely with an angular speed of 3.05 rev/s The student tinkle 1.30 kg mass in each outstretched atm 0.769 m from the axis of rotation. The contained moment of inertia of the student and the stool Ignoring the two masses, is 5.23 kg middot m^2. a value that remains constant. As the student pulls his arms Inward, his angular speed increases to 3.73 rev/s How far are the masses from the axis of rotation at this time. considering the masses to be points? Calculate the initial kinetic energy of the system Calculate the initial kinetic energy of the system

Explanation / Answer

initiall moment of inertia of system (stool + student + masses in hand)

Ii = (5.23) + (2 x 1.30 x 0.769^2)

Ii = 6.77 kg m^2

wi = 3.05 rev/s

finally:

wf = 3.73 rev/s

Applying angular momentum conservation,

Ii wi = If wf

6.77 x 3.05 = If x 3.73

If = 5.53 kg / m^2

5.53 = 5.23 + (2 x 1.30 x d^2)

d = 0.343 m

(B) wi = 3.05 rev/s = 3.05 x 2pi rad/s = 19.16 rad/s

wf = 3.73 x 2 pi rad/s = 23.44 rad/s

Ki = Ii wi^2 /2

= 6.77 x 19.16^2 /2

= 1242.65 J

(C) Kf = If wf^2 /2

= 5.53 x 23.44^2 /2

= 1519.18 J

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