Question Number 31767 by NECx last updated on 14/Mar/18 $${calculate}\:{the}\:{angular}\:{velocity}\:{of} \\ $$$${the}\:{earth}\:{about}\:{its}\:{axis}\:{and}\:{its} \\ $$$${angular}\:{velocity}\:{about}\:{the}\:{axis}\: \\ $$$${and}\:{the}\:{sun}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31768 by NECx last updated on 14/Mar/18 $${Please}\:{help} \\ $$$$ \\ $$$${What}\:{is}\:{the}\:{kinetic}\:{energy}\:{of}\:{the} \\ $$$${earth}?{Please}\:{prove}\:{the}\:{result}. \\ $$$$ \\ $$$${Thanks}\:{in}\:{advance}! \\ $$ Terms of Service…
Question Number 31766 by NECx last updated on 14/Mar/18 $${Calculate}\:{the}\:{moment}\:{of}\:{inertia} \\ $$$${of}\:{the}\:{earth}\:{abouth}\:{its}\:{axis}.{If} \\ $$$${the}\:{mass}\:{of}\:{the}\:{earth}\:{is}\:\mathrm{6}.\mathrm{0}×\mathrm{10}^{\mathrm{34}} {kg} \\ $$$${and}\:{radius}\:\mathrm{6}.\mathrm{4}×\mathrm{10}^{\mathrm{6}} {m}. \\ $$$$\left({Assume}\:{the}\:{earth}\:{to}\:{be}\:{a}\:{perfect}\right. \\ $$$$\left.{sphere}\right) \\ $$ Commented…
Question Number 31628 by Tinkutara last updated on 11/Mar/18 $${Systems}\:{of}\:{particles}\:{doubt} \\ $$ Commented by Tinkutara last updated on 11/Mar/18 If a rigid object has only rotational motion about a fixed axis are there any points within the object that always have the same velocity as the centre of mass? If so which ones? Commented by mrW2 last updated…
Question Number 31584 by Tinkutara last updated on 10/Mar/18 Answered by mrW2 last updated on 10/Mar/18 $${x}_{{A}} ={l}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$${v}_{{A}} =\frac{{dx}_{{A}} }{{dt}}=−{l}\:\mathrm{sin}\:\theta\:\omega \\…
Question Number 31566 by ajfour last updated on 10/Mar/18 Commented by ajfour last updated on 10/Mar/18 $${Frame}\:{XYZ}\:{is}\:{fixed}.\:{Frame}\:{xyz} \\ $$$${rotates}\:{with}\:{angular}\:{velocity}\:\Omega. \\ $$$${Relate}\:{acceleration}\:{of}\:{particle}\: \\ $$$${in}\:{fixed}\:{frame}\:{to}\:{that}\:{in}\:{rotating} \\ $$$${frame}.…
Question Number 31476 by NECx last updated on 09/Mar/18 $${From}\:{a}\:{circular}\:{disc}\:{of}\:{radius}\:{R} \\ $$$${a}\:{circular}\:{hole}\:{of}\:{radius}\:{R}/\mathrm{2}\:{is} \\ $$$${cut}\:{out}.{The}\:{centre}\:{of}\:{the}\:{circular} \\ $$$${hole}\:{is}\:{at}\:{R}/\mathrm{2}\:{from}\:{the}\:{centre}\:{of} \\ $$$${the}\:{original}\:{disc}.{Locate}\:{the}\:{centre} \\ $$$${of}\:{gravity}\:{of}\:{the}\:{resulting}\:{flat} \\ $$$${body}. \\ $$$$ \\…
Question Number 31382 by Tinkutara last updated on 07/Mar/18 Answered by mrW2 last updated on 07/Mar/18 Commented by mrW2 last updated on 07/Mar/18 $${x}_{{A}} ={s}…
Question Number 31350 by NECx last updated on 06/Mar/18 $${Are}\:{centripetal}\:{and}\:{centrifugal} \\ $$$${forces}\:{always}\:{equal}?{If}\:{No},{then} \\ $$$${in}\:{what}\:{conditions}\:{are}\:{they}\:{equal} \\ $$$${or}\:{not}\:{equal}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31329 by Tinkutara last updated on 06/Mar/18 Commented by Tinkutara last updated on 06/Mar/18 Commented by Tinkutara last updated on 06/Mar/18 Answer given for (iii) part is 4.5 cm/s^2. Is it correct? Commented…