Question Number 22703 by Sahib singh last updated on 22/Oct/17
Commented by Sahib singh last updated on 22/Oct/17
Answered by ajfour last updated on 22/Oct/17
$${F}−{f}={ma}\:\:\:\:\:\:\left({a}\:{is}\:{acc}.\:{of}\:{sphere}\right) \\ $$$${f}={mA}\:\:\:\:\:\:\:\:\:\:\:\:\left({A}\:{is}\:{acc}.\:{of}\:{plank}\right) \\ $$$${for}\:{pure}\:{rolling}\:{of}\:{sphere}\:{to}\: \\ $$$${begin}\:{on}\:{plank}\:\:\:\:\:\:{A}={a}−\alpha{R} \\ $$$$\boldsymbol{{and}}\:\:\:\:\:\:{fR}=\left(\frac{\mathrm{2}}{\mathrm{5}}{mR}^{\mathrm{2}} \right)\alpha \\ $$$${so}\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{f}=\mathrm{2}{m}\left({a}−{A}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{f}=\mathrm{2}{F}−\mathrm{2}{f}−\mathrm{2}{mA} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}{f}=\mathrm{2}{F}−\mathrm{2}{mA} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}{mA}=\mathrm{2}{F}−\mathrm{2}{mA} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\boldsymbol{{A}}=\frac{\mathrm{2}\boldsymbol{{F}}}{\mathrm{9}\boldsymbol{{m}}}\:.\:\: \\ $$
Commented by Sahib singh last updated on 22/Oct/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$
Commented by Sahib singh last updated on 22/Oct/17
$$\mathrm{I}\:\mathrm{just}\:\mathrm{made}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{condition}\:\mathrm{for}\:\mathrm{no}\: \\ $$$$\mathrm{slipping}. \\ $$$$ \\ $$