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