A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :
Solution
${1 \over 2}I{\omega ^2} = {1 \over 2} \times {1 \over 2}m{v^2}$<br><br>$I = {1 \over 2}m{R^2}$<br><br>Body is solid cylinder
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
This question is part of PrepWiser's free JEE Main question bank. 158 more solved questions on Rotational Motion are available — start with the harder ones if your accuracy is >70%.