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(B) The moment of inertia of the space station about its central axis is $I = MR^2.$ Two rockets,each providing a thrust $T$ at a distance $R$ from th

Vedclass · Accepted Answer

$t=\frac{\sqrt{gR}M}{2T}$

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(B) The moment of inertia of the space station about its central axis is $I = MR^2.$ Two rockets,each providing a thrust $T$ at a distance $R$ from the center,create a total torque $	au = 2TR.$ According to the angular impulse-momentum theorem,the angular impulse equals the change in angular momentum: $\int 	au dt = L_f - L_i.$ Assuming the station starts from rest $(L_i = 0)$,we have $2TRt = I\omega = MR^2\omega.$ To simulate Earth's gravity,the radial acceleration at the rim must be $g,$ so $a_r = \omega^2R = g.$ Solving for angular velocity,we get $\omega = \sqrt{\frac{g}{R}}.$ Substituting $\omega$ into the impulse equation: $2TRt = MR^2 \sqrt{\frac{g}{R}} = M\sqrt{gR^3}.$ Solving for time $t,$ we get $t = \frac{M\sqrt{gR^3}}{2TR} = \frac{M\sqrt{gR}}{2T}.$ Thus,the correct option is $B.$

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