Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 22059 by tawa tawa last updated on 10/Oct/17

$$\mathrm{A}\mathrm{flywheel}\mathrm{whose}\mathrm{diameter}\mathrm{is}1.5\mathrm{m}\mathrm{decrease}\mathrm{uniformly}\mathrm{from}240\mathrm{rad}/\min$$$$\mathrm{until}\mathrm{it}\mathrm{came}\mathrm{to}\mathrm{rest}10\mathrm{s}.\mathrm{Find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{revolution}\mathrm{made}.$$

Answered by Tinkutara last updated on 10/Oct/17

$$r=\frac{3}{4}m;{\omega }_1=2\pi \left(4rps\right)=8\pi rad/s$$$$0={\omega }_1-\alpha t$$$$\alpha =\frac{{\omega }_1}{t}=\frac{4\pi }{5}rad/s^2$$$$\theta ={\omega }_{1}t-\frac{\alpha }{2}{t}^2$$$$\theta =80\pi -\frac{2\pi }{5}\times 100=40\pi$$$$Numberofrevolutions=\frac{\theta }{2\pi }=20$$

Commented by tawa tawa last updated on 10/Oct/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com