
A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass the same, which one of the following will not be affected?
a)Moment of inertia
b) Angular momentum
c) Angular velocity
d)Rotational kinetic energy
Answer
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Hint: In the question it is given to us that the sphere rotates about the axis passing through its center. First we will define all the terminologies provided to us in the options. Further we will verify whether on increasing the radius of the sphere the individual quantities get affected or not and accordingly come to a conclusion.
Formula used:
$I=\dfrac{2}{5}m{{R}^{2}}$
$L=I\omega $
$K.E(rot)=\dfrac{1}{2}I{{\omega }^{2}}J$
Complete step-by-step answer:
Let us say the above sphere of mass ‘m’ has a radius R and rotates with angular velocity $\omega $ about its axis of rotation passing through the centre of the sphere.
The moment of inertia(I) of the above sphere about its centre of mass is given by,
$I=\dfrac{2}{5}m{{R}^{2}}$
When the radius of the sphere is increased, from the above equation we can conclude that the moment of inertia of the sphere will change.
The angular momentum (L) of the above sphere can be written as $L=I\omega $.If we observe this equation we can see that the moment of inertia changes. As per the law of conservation of angular momentum, the initial momentum and the final momentum remains the same. Therefore the value of L will be a constant. From this we can imply that since the moment of inertia increases, the angular velocity has to decrease. As the body rotates it has a rotational kinetic energy $K.E(rot)$. This is numerically equal to,
$\begin{align}
& K.E(rot)=\dfrac{1}{2}I{{\omega }^{2}}J\text{, }\because L=I\omega \\
& \Rightarrow K.E(rot)=\dfrac{1}{2}L\omega \\
\end{align}$
We know that the angular momentum remains constant. But the angular velocity decreases when the radius of the sphere is increased. Hence we can conclude that the kinetic energy also changes.
So, the correct answer is “Option b”.
Note: It is to be noted that the sphere rotates on its own when the radius is increased. Hence we can say that the external torque acting on the sphere is zero. Since external torque is zero, by law of conservation of angular momentum we can imply that the momentum of the sphere should be conserved i.e. constant.
Formula used:
$I=\dfrac{2}{5}m{{R}^{2}}$
$L=I\omega $
$K.E(rot)=\dfrac{1}{2}I{{\omega }^{2}}J$
Complete step-by-step answer:
Let us say the above sphere of mass ‘m’ has a radius R and rotates with angular velocity $\omega $ about its axis of rotation passing through the centre of the sphere.
The moment of inertia(I) of the above sphere about its centre of mass is given by,
$I=\dfrac{2}{5}m{{R}^{2}}$
When the radius of the sphere is increased, from the above equation we can conclude that the moment of inertia of the sphere will change.
The angular momentum (L) of the above sphere can be written as $L=I\omega $.If we observe this equation we can see that the moment of inertia changes. As per the law of conservation of angular momentum, the initial momentum and the final momentum remains the same. Therefore the value of L will be a constant. From this we can imply that since the moment of inertia increases, the angular velocity has to decrease. As the body rotates it has a rotational kinetic energy $K.E(rot)$. This is numerically equal to,
$\begin{align}
& K.E(rot)=\dfrac{1}{2}I{{\omega }^{2}}J\text{, }\because L=I\omega \\
& \Rightarrow K.E(rot)=\dfrac{1}{2}L\omega \\
\end{align}$
We know that the angular momentum remains constant. But the angular velocity decreases when the radius of the sphere is increased. Hence we can conclude that the kinetic energy also changes.
So, the correct answer is “Option b”.
Note: It is to be noted that the sphere rotates on its own when the radius is increased. Hence we can say that the external torque acting on the sphere is zero. Since external torque is zero, by law of conservation of angular momentum we can imply that the momentum of the sphere should be conserved i.e. constant.
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