Answer in Mechanics | Relativity for A #108105
March 12th, 2023
Given data,
Angular velocity =43rpm
Time = 6.5min=390second
Mass of the satellite M=4000kg
Mass of the rocket m = 250kg
Radius of the satellite R =5m
Angular velocity “omega=43timesfrac{2pi}{60}=4.50 rad s^{-1}”
Angular acceleration “alpha”,
“omega=omega_0+alphatimes t”
“alpha=frac{4.5}{390}=0.0115rads^{-2}”
Inertia is given by
“I=0.5MR^2+4mR^2”
And the net torque,
“tau_{net}=Itimesalpha=4Ftimes R”
“F=frac{Itimesalpha}{4R}”
“F=frac{R(0.5M+4m)timesalpha}{4}”
F=43.125N