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Question Number 20916 by Tinkutara last updated on 07/Sep/17

$$\mathrm{A}\mathrm{body}\mathrm{starts}\mathrm{rotating}\mathrm{about}\mathrm{a}$$$$\mathrm{stationary}\mathrm{axis}\mathrm{with}\mathrm{an}\mathrm{angular}$$$$\mathrm{acceleration}b=2t\mathrm{rad}/{\mathrm{s}}^2.\mathrm{How}\mathrm{soon}$$$$\mathrm{after}\mathrm{the}\mathrm{beginning}\mathrm{of}\mathrm{rotation}\mathrm{will}\mathrm{the}$$$$\mathrm{total}\mathrm{acceleration}\mathrm{vector}\mathrm{of}\mathrm{an}\mathrm{arbitrary}$$$$\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{body}\mathrm{forms}\mathrm{an}\mathrm{angle}\mathrm{of}60°$$$$\mathrm{with}\mathrm{its}\mathrm{velocity}\mathrm{vector}?$$$$\left(1\right){\left(2\sqrt{3}\right)}^{1/3}\mathrm{s}$$$$\left(2\right){\left(2\sqrt{3}\right)}^{1/2}\mathrm{s}$$$$\left(3\right)\left(2\sqrt{3}\right)\mathrm{s}$$$$\left(4\right){\left(2\sqrt{3}\right)}^2\mathrm{s}$$

Answered by ajfour last updated on 08/Sep/17

$$tangentialaccelerationisinthe$$$$directionofvelocityvector.$$$$a_{\theta }=bR=2Rt$$$$\mid a_r\mid ={\omega }^{2}R(where\omega ={\int }_0^{t}2tdt)$$$$so{a}_r=t^{4}R$$$$\tan 60°=\sqrt{3}=\frac{\mid a_r\mid }{a_{\theta }}=\frac{t^{4}R}{2Rt}=\frac{t^3}{2}$$$$\Rightarrow t^3=2\sqrt{3}$$$$\mathit{t}={\left(2\sqrt{3}\right)}^{1/3}\mathit{s}.$$

Commented by Tinkutara last updated on 08/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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