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 by ajfour last updated on 06/Mar/18
$${a}_{{C}} =\omega^{\mathrm{2}} \left(\frac{{l}}{\mathrm{2}}\right)=\frac{{v}_{{C}} ^{\mathrm{2}} }{\left({l}/\mathrm{2}\right)}=\frac{\left(\mathrm{15}/\mathrm{2}\right)^{\mathrm{2}} {cm}^{\mathrm{2}} /{s}^{\mathrm{2}} }{\mathrm{25}{cm}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{225}}{\mathrm{100}}{cm}/{s}^{\mathrm{2}} \:=\:\mathrm{2}.\mathrm{25}\:{cm}/{s}^{\mathrm{2}} \:. \\ $$
Commented by Tinkutara last updated on 06/Mar/18
Is it wrong in book❓
Commented by MJS last updated on 06/Mar/18
$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure},\:\mathrm{but}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be} \\ $$$${l}_{{C}} =\mathrm{25}{cm}\:\Rightarrow\:\frac{{l}_{{C}} }{\mathrm{2}}=\mathrm{12}.\mathrm{5}{cm} \\ $$$$\Rightarrow\:{a}_{{C}} =\mathrm{4}.\mathrm{5}{cm}/{s}^{\mathrm{2}} \\ $$$$\mathrm{please}\:\mathrm{correct}\:\mathrm{me}\:\mathrm{if}\:\mathrm{I}'\mathrm{m}\:\mathrm{wrong} \\ $$