A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is :
Solution
K.E = ${1 \over 2}m{V^2} + {1 \over 2}{I_{cm}}{\omega ^2}$
<br><br>= $${1 \over 2}m{V^2} + {1 \over 2} \times {2 \over 5}m{R^2} \times {{{V^2}} \over {{R^2}}}$$
<br><br>= ${1 \over 2}m{V^2} + {1 \over 5}m{V^2}$
<br><br>= ${7 \over {10}}m{V^2}$
<br><br>= ${7 \over {10}} \times 0.5 \times 25 \times {10^{ - 4}}$
<br><br>= ${{35} \over 4} \times {10^{ - 4}}$
<br><br>= 8.75 × 10<sup>–4</sup> J
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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