A solid metallic sphere is rotating about its diameter with constant angular speed omega if it's summer
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Answer:
Given that a solid metal sphere is rotating about its diameter with a constant angular speed. It is heated and its temperature increases appreciably. We assume that all heat energy has been used for expansion of the sphere.
Due to coefficient of volume expansion of the metal of the sphere, there is increase in the volume of the sphere. It is same as saying that radius of the sphere increases.
Now rotational kinetic energy of sphere is given as
Also, moment of inertia is given as
In the absence of any external turning moment/torque, rotational kinetic energy must be conserved.
From equation (2) we see that with the increase of radius, moment of inertia of the sphere increases. From (1) we infer that in order to conserve rotational kinetic energy, must decrease.
.-.-.-
Alternatively, angular momentum of rotating sphere
Since there is no external torque, angular momentum needs to be conserved. With the help of (2) and (3) we arrive at the same result.
Answer:
Given that a solid metal sphere is rotating about its diameter with a constant angular speed. It is heated and its temperature increases appreciably. We assume that all heat energy has been used for expansion of the sphere.
Due to coefficient of volume expansion of the metal of the sphere, there is increase in the volume of the sphere. It is same as saying that radius of the sphere increases.
Now rotational kinetic energy of sphere is given as
KE_{rot}=\frac{1}{2}I\omega^2 ........(1)KE
rot
=
2
1
Iω
2
........(1)
Also, moment of inertia is given as
I = \frac {2}{5}MR^2 .........(2)I=
5
2
MR
2
.........(2)
In the absence of any external turning moment/torque, rotational kinetic energy must be conserved.
From equation (2) we see that with the increase of radius, moment of inertia of the sphere increases. From (1) we infer that in order to conserve rotational kinetic energy, \omegaω must decrease.
.-.-.-
Alternatively, angular momentum of rotating sphere
I = I \omega......(3)I=Iω......(3)
Since there is no external torque, angular momentum needs to be conserved. With the help of (2) and (3) we arrive at the same result.