Flywheel angular acceleration ϵ=RmgJ+mR2=0.51(s−2)\epsilon = \frac{Rmg}{J+mR^2} = 0.51(s^{-2})ϵ=J+mR2Rmg=0.51(s−2)
which means the angular velocity is ω=ϵt=0.51⋅20=10.2(s−1)\omega = \epsilon t = 0.51\cdot20 = 10.2 (s^{-1})ω=ϵt=0.51⋅20=10.2(s−1)
and the linear velocity is v=ω(R−d)=10.2⋅0.73=7.4(m/s)v = \omega (R - d) = 10.2\cdot 0.73 = 7.4 (m/s)v=ω(R−d)=10.2⋅0.73=7.4(m/s)
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Sir please explain me #95750 question number detailly. Please explain the terms used.