Question

Two rigid bodies have same angular momentum about their axes of symmetry

Answer 1

Answer:

we know, rotational kinetic energy is given by

K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)

where I denotes moment of inertia and \omega denotes angular velocity.

we know, L=I\omega, where L denotes angular momentum.

or, \omega=\frac{L}{I}, put it in equation (1),

so, K.E_{\textbf{rotational}}=\frac{1}{2}\frac{L^2}{I}

here external torque, \tau_{\textbf{external}} = 0,

so, angular momentum remains constant.

or, K.E_{\textbf{rotational}}\propto\frac{1}{I}

hence,The rigid body having less moment of inertia will have greater kinetic energy.

Answer 2

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we know, rotational kinetic energy is given by

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)where I denotes moment of inertia and \omega denotes angular velocity.

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)where I denotes moment of inertia and \omega denotes angular velocity.we know, L=I\omega, where L denotes angular momentum.

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)where I denotes moment of inertia and \omega denotes angular velocity.we know, L=I\omega, where L denotes angular momentum.or, \omega=\frac{L}{I}, put it in equation (1),

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)where I denotes moment of inertia and \omega denotes angular velocity.we know, L=I\omega, where L denotes angular momentum.or, \omega=\frac{L}{I}, put it in equation (1), so, K.E_{\textbf{rotational}}=\frac{1}{2}\frac{L^2}{I}

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)where I denotes moment of inertia and \omega denotes angular velocity.we know, L=I\omega, where L denotes angular momentum.or, \omega=\frac{L}{I}, put it in equation (1), so, K.E_{\textbf{rotational}}=\frac{1}{2}\frac{L^2}{I}here external torque, \tau_{\textbf{external}} = 0,

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)where I denotes moment of inertia and \omega denotes angular velocity.we know, L=I\omega, where L denotes angular momentum.or, \omega=\frac{L}{I}, put it in equation (1), so, K.E_{\textbf{rotational}}=\frac{1}{2}\frac{L^2}{I}here external torque, \tau_{\textbf{external}} = 0, so, angular momentum remains constant.

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)where I denotes moment of inertia and \omega denotes angular velocity.we know, L=I\omega, where L denotes angular momentum.or, \omega=\frac{L}{I}, put it in equation (1), so, K.E_{\textbf{rotational}}=\frac{1}{2}\frac{L^2}{I}here external torque, \tau_{\textbf{external}} = 0, so, angular momentum remains constant. or, K.E_{\textbf{rotational}}\propto\frac{1}{I}

we know, rotational kinetic energy is given by K.E_{\textbf{rotational}}=\frac{1}{2}I\omega^2........(1)where I denotes moment of inertia and \omega denotes angular velocity.we know, L=I\omega, where L denotes angular momentum.or, \omega=\frac{L}{I}, put it in equation (1), so, K.E_{\textbf{rotational}}=\frac{1}{2}\frac{L^2}{I}here external torque, \tau_{\textbf{external}} = 0, so, angular momentum remains constant. or, K.E_{\textbf{rotational}}\propto\frac{1}{I}hence,The rigid body having less moment of inertia will have greater kinetic energy.