A lawn roller of mass $80 \mathrm{~kg}$, radius $0.3 \mathrm{~m}$ and moment of inertia $3.6 \mathrm{~kg} \cdot \mathrm{m}^2$, is drawn along a level surface at a constant speed of $1.8 \mathrm{~m} / \mathrm{s}$. Find
(i) the translational kinetic energy
(ii) the rotational kinetic energy
(iii) the total kinetic energy of the roller.
(i) the translational kinetic energy
(ii) the rotational kinetic energy
(iii) the total kinetic energy of the roller.
Q 128.1
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Data : $M=80 \mathrm{~kg}, \mathrm{R}=0.3 \mathrm{~m}, \mathrm{I}=3.6 \mathrm{~kg} \cdot \mathrm{m}^2, \mathrm{v}=1.8 \mathrm{~m} / \mathrm{s}$
(i) The translational kinetic energy of the centre of mass of the roller,
$
E_{\text {tran }}=\frac{1}{2} M v^2=\frac{1}{2} \times 80 \times(1.8)^2=40 \times 3.24=129.6 \mathrm{~J}
$
(ii) The rotational kinetic energy about the roller's axle,
$
\begin{aligned}
E_{\text {rot }}=\frac{1}{2} I \omega^2=\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =\frac{1}{2} \times 3.6 \times\left(\frac{1.8}{0.3}\right)^2 \\ \\
& =1.8 \times 36=64.8 \mathrm{~J}
\end{aligned}
$
(iii) The total kinetic energy of the roller,
$
E=E_{\operatorname{tran}}+E_{r o t}=129.6+64.8=194.4 \mathrm{~J}
$
(i) The translational kinetic energy of the centre of mass of the roller,
$
E_{\text {tran }}=\frac{1}{2} M v^2=\frac{1}{2} \times 80 \times(1.8)^2=40 \times 3.24=129.6 \mathrm{~J}
$
(ii) The rotational kinetic energy about the roller's axle,
$
\begin{aligned}
E_{\text {rot }}=\frac{1}{2} I \omega^2=\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =\frac{1}{2} \times 3.6 \times\left(\frac{1.8}{0.3}\right)^2 \\ \\
& =1.8 \times 36=64.8 \mathrm{~J}
\end{aligned}
$
(iii) The total kinetic energy of the roller,
$
E=E_{\operatorname{tran}}+E_{r o t}=129.6+64.8=194.4 \mathrm{~J}
$
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