A crew of scientists has built a new space st | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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 $B.$

Similar Questions

$A$ constant torque of $31.4 \, Nm$ is exerted on a pivoted wheel. If the angular acceleration of the wheel is $4\pi \, rad/s^2$,then the moment of inertia will be ....... $kg \cdot m^2$.

$A$ uniform disc of mass $5\,g$ and radius $1\,cm$ is fixed to a thin stick $AB$ of negligible mass as shown in the figure. The system is initially at rest. The constant torque,that will make the system rotate about $AB$ at $25$ rotations per second in $5\,s$,is close to:

$A$ constant torque acting on a uniform circular wheel changes its angular momentum from $L_0$ to $4L_0$ in $4 \ s$. The magnitude of this torque is

Vedclass Test Series

Exam Paper Generator

Online Exam Module