Question Number 27102 by Tinkutara last updated on 02/Jan/18
Answered by ajfour last updated on 02/Jan/18
$${let}\:{reaction}\:{force}\:{be}\:{N}. \\ $$$${N}\mathrm{sin}\:\theta={mA}\:\:\:\left({A}:\:{acc}.{of}\:{shell}\right) \\ $$$${pseudo}\:{force}\:{on}\:{particle}\:{is}\:{mA}. \\ $$$${so}\:\:\frac{{N}}{{F}_{{pseudo}} }\:=\:\frac{{mA}\mathrm{cosec}\:\theta}{{mA}}\:=\mathrm{cosec}\:\theta\:. \\ $$
Commented by Tinkutara last updated on 02/Jan/18
$${Why}\:{N}\mathrm{sin}\:\theta={mA}?\:{Because}\:{N}\mathrm{sin}\:\theta \\ $$$${is}\:{not}\:{the}\:{net}\:{force}\:{on}\:{particle}\:{then} \\ $$$${why}? \\ $$
Commented by ajfour last updated on 02/Jan/18
$${N}\mathrm{sin}\:\theta\:{is}\:{the}\:{horizontal}\:{component} \\ $$$${of}\:{force}\:{of}\:{particle}\:{on}\:{shell}. \\ $$$${A}\:{is}\:{the}\:{acceleration}\:{of}\:{shell}. \\ $$
Commented by Tinkutara last updated on 03/Jan/18
Yes that is horizontal component of normal reaction then why it is equal to mA?
Commented by ajfour last updated on 03/Jan/18
$${the}\:{only}\:{horizontal}\:{force}\:{on}\:{shell} \\ $$$${is}\:{N}\mathrm{sin}\:\theta\:{and}\:{acc}.\:{of}\:{shell}\:{is}\:{thus} \\ $$$${given}\:{by}\:{mA}={N}\mathrm{sin}\:\theta\:. \\ $$$${In}\:{psedo}\:{force}\:{of}\:{particle}\:{we}\:{use} \\ $$$${this}\:{acc}.\:\:{and}\:{mass}\:{of}\:{particle} \\ $$$$={mA}\:. \\ $$
Commented by Tinkutara last updated on 03/Jan/18
Thank you very much Sir! I got the answer.