Question
An annular ring with inner and outer radii $${R_1}$$ and $${R_2}$$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, $$\frac{{{F_1}}}{{{F_2}}}$$ is
A.
$${\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^2}$$
B.
$$\frac{{{R_2}}}{{{R_1}}}$$
C.
$$\frac{{{R_1}}}{{{R_2}}}$$
D.
$$1$$
Answer :
$$\frac{{{R_1}}}{{{R_2}}}$$
Solution :
$$\eqalign{ & {a_1} = \frac{{v_1^2}}{{{R_1}}} = \frac{{{\omega ^2}R_1^2}}{{{R_1}}} = {\omega ^2}{R_1} \cr & {a_2} = \frac{{v_2^2}}{{{R_2}}} = {\omega ^2}{R_2} \cr} $$
Taking particle masses equal
$$\frac{{{F_1}}}{{{F_2}}} = \frac{{m{a_1}}}{{m{a_2}}} = \frac{{{a_1}}}{{{a_2}}} = \frac{{{R_1}}}{{{R_2}}}$$
$$\eqalign{ & {a_1} = \frac{{v_1^2}}{{{R_1}}} = \frac{{{\omega ^2}R_1^2}}{{{R_1}}} = {\omega ^2}{R_1} \cr & {a_2} = \frac{{v_2^2}}{{{R_2}}} = {\omega ^2}{R_2} \cr} $$
Taking particle masses equal
$$\frac{{{F_1}}}{{{F_2}}} = \frac{{m{a_1}}}{{m{a_2}}} = \frac{{{a_1}}}{{{a_2}}} = \frac{{{R_1}}}{{{R_2}}}$$
