
A uniform metallic rod rotates about its perpendicular bisector with a constant angular speed. If it is heated uniformly to raise its temperature slightly, then
(A) Its speed of rotation increases
(B) Its speed of rotation decreases
(C) Its speed of rotation remains the same
(D) Its speed increases because its moment of inertia increases
Answer
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Hint: In the given question, we have been provided with a metallic rod and the only variable factor is the temperature of the rod as heat is uniformly applied to it. We have to establish a relation between this variable factor and the speed of rotation or the angular velocity.
Formula Used: \[L=I\omega \] , \[I=m{{L}^{2}}\]
Complete step by step solution:
The moment of inertia of a regular solid rod about its perpendicular bisector is given as \[(I)=m{{L}^{2}}\] where \[m\] is the mass of the rod and \[L\] is the length of the rod.
The angular momentum of a rotating object is given as \[L=I\omega \] where \[\omega \] is the angular velocity of the object.
In absence of any external torque on the body, as in case of the given question, the angular momentum remains constant which means if the moment of inertia increases, the angular velocity must decrease.
Now we have been told that the heat is being supplied at a constant rate to the rod, which means the temperature of the rod will increase and the rod will undergo linear expansion. As the length of the rod increases, the moment of inertia of the rod will increase as we have established a relation between the two quantities above. So the moment of inertia of the rod will increase when the heat is supplied to it and a higher moment of inertia means a lesser angular velocity. Hence we can say that the speed of the rotation would decrease.
Therefore, option (B) is the correct answer to the given question.
Note:
Here we have given our result based on the assumption that the melting point of the metal is very high as compared to the temperature rise after the supply of heat to the metal. This is very important because if the temperature rise approaches the melting point, some parts of the metal may tend to melt off and hence the mass would reduce and the moment of inertia would decrease and the angular speed would increase.
Formula Used: \[L=I\omega \] , \[I=m{{L}^{2}}\]
Complete step by step solution:
The moment of inertia of a regular solid rod about its perpendicular bisector is given as \[(I)=m{{L}^{2}}\] where \[m\] is the mass of the rod and \[L\] is the length of the rod.
The angular momentum of a rotating object is given as \[L=I\omega \] where \[\omega \] is the angular velocity of the object.
In absence of any external torque on the body, as in case of the given question, the angular momentum remains constant which means if the moment of inertia increases, the angular velocity must decrease.
Now we have been told that the heat is being supplied at a constant rate to the rod, which means the temperature of the rod will increase and the rod will undergo linear expansion. As the length of the rod increases, the moment of inertia of the rod will increase as we have established a relation between the two quantities above. So the moment of inertia of the rod will increase when the heat is supplied to it and a higher moment of inertia means a lesser angular velocity. Hence we can say that the speed of the rotation would decrease.
Therefore, option (B) is the correct answer to the given question.
Note:
Here we have given our result based on the assumption that the melting point of the metal is very high as compared to the temperature rise after the supply of heat to the metal. This is very important because if the temperature rise approaches the melting point, some parts of the metal may tend to melt off and hence the mass would reduce and the moment of inertia would decrease and the angular speed would increase.
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