Question

Two discs having masses in the ratio 1:2 and radii in the ratio 1: 8 roll down wi

Answer 1

Answer:

here the potential energy is converted to kinetic energy

for disc 1

$m_1gh=\frac{1}{2}m_1v_1^2+\frac{1}{2}I\omega^2\\ since\ \omega=\frac{v}{r_1}and \ I=\frac{m_1r_1^2}{2}\\m_1gh=\frac{1}{2}m_1v_1^2+\frac{1}{2}\frac{m_1v_1^2}{2}=\frac{3m_1v_1^2}{4}\\so\ v_1=\sqrt{\frac{4gh}{3}}\\ simillarly\ v_2=\sqrt{\frac{4gh}{3}}\\therefor\ \frac{v_1}{v_2}=\frac{1}{1}$