Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for the increase in kinetic energy?
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Hey mate,
◆ Answer- KEf = 5/2 KEi
◆ Explaination-
# Given-
ωi = 40 rpm
ωf = 100 rpm
If = 2/5 Ii
# Solution-
Let, KEi be K.E. of boy with outstretched hands and KEf be K.E. of boy with folded hands.
Initial kinetic energy,
KEi = (1/2)Iiωi^2
Final kinetic energy,
KEi = (1/2)Iiωi^2
Taking ratio,
KEf/KFi = (0.5×If×ωf^2) / 0.5×Ii×ωi^2
KEf/KFi = (2/5)Ii(100)^2 / Ii(40)^2
KEf/KFi = 5/2
Thus, final kinetic energy is greater than initial kinetic energy.
Increase in kinetic energy can be accounted for internal energy of the boy.
Hope, this helps you...
◆ Answer- KEf = 5/2 KEi
◆ Explaination-
# Given-
ωi = 40 rpm
ωf = 100 rpm
If = 2/5 Ii
# Solution-
Let, KEi be K.E. of boy with outstretched hands and KEf be K.E. of boy with folded hands.
Initial kinetic energy,
KEi = (1/2)Iiωi^2
Final kinetic energy,
KEi = (1/2)Iiωi^2
Taking ratio,
KEf/KFi = (0.5×If×ωf^2) / 0.5×Ii×ωi^2
KEf/KFi = (2/5)Ii(100)^2 / Ii(40)^2
KEf/KFi = 5/2
Thus, final kinetic energy is greater than initial kinetic energy.
Increase in kinetic energy can be accounted for internal energy of the boy.
Hope, this helps you...
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