A 3 kg sphere dropped through air has a term | vedclass

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Question

(B) At terminal velocity,the drag force equals the gravitational force: $mg = bv_{t}$.Given $m = 3 \ kg$,$v_{t} = 25 \ m/s$,and taking $g = 10 \ m/s^2

Vedclass · Accepted Answer

$9.998$

Vedclass · Answer

(B) At terminal velocity,the drag force equals the gravitational force: $mg = bv_{t}$.Given $m = 3 \ kg$,$v_{t} = 25 \ m/s$,and taking $g = 10 \ m/s^2$,we have $3 	imes 10 = b 	imes 25$,which gives $b = \frac{30}{25} = 1.2 \ kg/s$.The damping constant $\gamma = \frac{b}{2m} = \frac{1.2}{2 	imes 3} = 0.2 \ s^{-1}$.The angular frequency of damped $SHM$ is given by $\omega' = \sqrt{\omega_0^2 - \gamma^2}$,where $\omega_0 = \sqrt{\frac{k}{m}}$.$\omega_0 = \sqrt{\frac{300}{3}} = \sqrt{100} = 10 \ rad/s$.$\omega' = \sqrt{10^2 - (0.2)^2} = \sqrt{100 - 0.04} = \sqrt{99.96}$.Using the binomial approximation $\sqrt{x^2 - a} \approx x - \frac{a}{2x}$ for small $a$,we get $\omega' \approx 10 - \frac{0.04}{20} = 10 - 0.002 = 9.998 \ rad/s$.

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