Question Number 188392 by yawa7373 last updated on 28/Feb/23
Commented by Rasheed.Sindhi last updated on 28/Feb/23
$$\boldsymbol{{sir}}\:{mr}\:{W}\:,\:{please}\:{see}\:{my}\:{third}\:{answer} \\ $$$${to}\:{Q}#\mathrm{188327}\:{in}\:{which}\:{I}'{ve}\:{used} \\ $$$${derivative}. \\ $$
Commented by mr W last updated on 01/Mar/23
$${very}\:{nice}\:{approach}\:{sir}! \\ $$
Commented by Rasheed.Sindhi last updated on 01/Mar/23
$$\mathcal{T}{han}\mathcal{X}\:{a}\:\mathcal{LOT}\:\:\boldsymbol{{sir}}! \\ $$
Answered by mr W last updated on 28/Feb/23
$${Q}\mathrm{1} \\ $$$$\left({a}\right) \\ $$$$\theta=\frac{\alpha{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{t}=\sqrt{\frac{\mathrm{2}\theta}{\alpha}}=\sqrt{\frac{\mathrm{2}×\mathrm{2}\pi}{\mathrm{4}.\mathrm{5}×\mathrm{10}^{−\mathrm{3}} }}=\mathrm{52}.\mathrm{844}\:{s} \\ $$$$\left({b}\right) \\ $$$$\omega=\alpha{t}=\mathrm{4}.\mathrm{5}×\mathrm{10}^{−\mathrm{3}} ×\mathrm{52}.\mathrm{844}=\mathrm{0}.\mathrm{238}\:{rad}/{s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{or}\:\mathrm{13}.\mathrm{625}°/{s} \\ $$
Answered by mr W last updated on 28/Feb/23
$${Q}\mathrm{2} \\ $$$${r}=\frac{\mathrm{0}.\mathrm{5}}{\mathrm{2}}=\mathrm{0}.\mathrm{25}\:{m} \\ $$$$\left({a}\right) \\ $$$$\omega=\mathrm{60}\:{rpm}=\frac{\mathrm{60}×\mathrm{2}\pi}{\mathrm{60}}=\mathrm{6}.\mathrm{28}\:{rad}/{s} \\ $$$$\left({b}\right) \\ $$$${v}=\omega{r}=\mathrm{6}.\mathrm{28}×\mathrm{0}.\mathrm{25}=\mathrm{3}.\mathrm{41}\:{m}/{s} \\ $$