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Question Number 111774 by I want to learn more last updated on 04/Sep/20

Answered by mr W last updated on 04/Sep/20

$$\alpha =\frac{10}{\frac{10\times 0.{02}^2}{2}}=5000rad/s^2$$$$t=\sqrt{\frac{2\times 10\times 2\pi }{5000}}=\sqrt{\frac{\pi }{125}}$$$$\omega =\alpha t=5000\sqrt{\frac{\pi }{125}}\approx 793rad/s$$$$\Rightarrow Biscorrect$$

Commented by mr W last updated on 05/Sep/20

$$and$$$$\frac{1}{2}I{\omega }^2=\mathrm{\Gamma }\mathrm{\Delta }\theta$$

Commented by Dwaipayan Shikari last updated on 04/Sep/20

$$Yessir!$$

Commented by mr W last updated on 04/Sep/20

$$or$$$$\frac{1}{2}\times \frac{10\times 0.{02}^2}{2}\times {\omega }^2=10\times 10\times 2\pi$$$$\Rightarrow \omega =200\sqrt{5\pi }\approx 793rad/s$$

Commented by I want to learn more last updated on 04/Sep/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{really}\mathrm{appreciate}.$$$$\mathrm{sir},\mathrm{please}\mathrm{what}\mathrm{formular}\mathrm{do}\mathrm{you}\mathrm{use}\mathrm{to}\mathrm{find}\alpha \mathrm{and}\mathrm{t}$$

Commented by Dwaipayan Shikari last updated on 04/Sep/20

$${\mathrm{\Gamma }}_{torque}=\mid I\overrightarrow{\alpha }\mid I=momentofinertia$$$$Heremomentofinertiais\frac{1}{2}{MR}^2$$$$\alpha =\frac{2{\mathrm{\Gamma }}_{torque}}{{MR}^2}(\mathrm{\Gamma }=\mid \overrightarrow{F}\times \overrightarrow{R}\mid =FRsin\theta =MaR=MR.\alpha .R=I.\alpha )$$$$a=R\alpha \left(Rotationalanalogue\right)(\theta =\frac{\pi }{2}here)$$$${\theta }_1-{\theta }_0=w_{0}t+\frac{1}{2}\alpha t^2(w_0=0)$$$$△\theta =\frac{1}{2}\alpha t^2$$$$t=\sqrt{\frac{2△\theta }{\alpha }}$$

Commented by I want to learn more last updated on 05/Sep/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}$$

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