JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 11)
A particle is moving with uniform speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to R3. Its time period of revolution will be given by :
$$T \propto {R^{{4 \over 3}}}$$
$$T \propto {R^{{5 \over 2}}}$$
$$T \propto {R^{{3 \over 2}}}$$
$$T \propto {R^2}$$
설명
$$F \propto {1 \over {{R^3}}}$$
$$F = {K \over {{R^3}}}$$
$${{m{v^2}} \over R} = {K \over {{R^3}}}$$
$$m{(\omega R)^2} = {K \over {{R^2}}}$$
$$m{\omega ^2}{R^2} = {K \over {{R^2}}}$$
$${\omega ^2} = {K \over m}\left( {{1 \over {{R^4}}}} \right)$$
$${\left( {{{2\pi } \over T}} \right)^2} \propto {1 \over {{R^4}}}$$
$${{4{\pi ^2}} \over {{T^2}}} \propto {1 \over {{R^4}}}$$
$$T \propto {R^2}$$
$$F = {K \over {{R^3}}}$$
$${{m{v^2}} \over R} = {K \over {{R^3}}}$$
$$m{(\omega R)^2} = {K \over {{R^2}}}$$
$$m{\omega ^2}{R^2} = {K \over {{R^2}}}$$
$${\omega ^2} = {K \over m}\left( {{1 \over {{R^4}}}} \right)$$
$${\left( {{{2\pi } \over T}} \right)^2} \propto {1 \over {{R^4}}}$$
$${{4{\pi ^2}} \over {{T^2}}} \propto {1 \over {{R^4}}}$$
$$T \propto {R^2}$$