An LPG cylinder [consider it to be perfectly cylindrical] of mass 15 kg containing 10 kg of LPG is performing pure rolling on horizontal surface at speed of 0.5 m/s. Kinetic energy (in J) of cylinder with gas is (Ignore the internal energy of the gas)

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Solution

The cylinder perform translational as well as rotational motion, whereas the LPG can only perform translational motion as gas cannot be a rigid body.
Therefore the total kinetic energy of the system can be given by,
$K E_{T o t a l} = \frac{1}{2} I_{c y} w^{2} + \frac{1}{2} m_{c y} v^{2} + \frac{1}{2} m_{g a s} v^{2}$
$K E_{T o t a l} = \frac{1}{2} I_{c y} \frac{v^{2}}{R^{2}} + \frac{1}{2} m_{c y} v^{2} + \frac{1}{2} m_{g a s} v^{2}$
$K E_{T o t a l} = \frac{1}{2} m_{c y} R_{c y}^{2} \frac{v^{2}}{R_{c y}^{2}} + \frac{1}{2} m_{c y} v^{2} + \frac{1}{2} m_{g a s} v^{2}$
$K E_{T o t a l} = m_{c y} v^{2} + \frac{1}{2} m_{g a s} v^{2}$
$K E = 15 \times {(\frac{1}{2})}^{2} + \frac{1}{2} (10) {(\frac{1}{2})}^{2} = 5 J$

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