Question Number 30613 by NECx last updated on 23/Feb/18
$${A}\:{car}\:{negotiates}\:{a}\:{bend}\:{of}\:{radius} \\ $$$$\mathrm{20}{m}\:{with}\:{an}\:{acceleration}\:{of}\: \\ $$$$\mathrm{12}{m}/{s}^{\mathrm{2}} .{What}\:{is}\:{the}\:{maximum} \\ $$$${speed}\:{the}\:{car}\:{can}\:{attain}\:{without} \\ $$$${skidding}? \\ $$
Commented by NECx last updated on 24/Feb/18
$${please}\:{help} \\ $$
Answered by mrW2 last updated on 25/Feb/18
$${v}_{{max}} =\sqrt{{ar}}=\sqrt{\mathrm{12}×\mathrm{20}}=\mathrm{15}.\mathrm{5}{m}/{s}=\mathrm{56}\:{km}/{h} \\ $$
Commented by NECx last updated on 25/Feb/18
$${can}\:{I}\:{know}\:{how}\:{you}\:{got}\:{the} \\ $$$${formula}\:{please}? \\ $$
Commented by mrW2 last updated on 25/Feb/18
$${the}\:{car}\:{can}\:{reach}\:{an}\:{acceleration}\:{from} \\ $$$${a}=\mathrm{12}\:{m}/{s}^{\mathrm{2}} ,\:{that}\:{means}\:{a}\:{friction} \\ $$$${coeefficient}\:{between}\:{tires}\:{and}\:{ground} \\ $$$${is}\:\mu=\frac{{a}}{{g}}=\mathrm{1}.\mathrm{2}.\:{when}\:{the}\:{car}\:{goes} \\ $$$${into}\:{the}\:{curve}\:{and}\:{its}\:{speed}\:{increases}, \\ $$$${a}\:{part}\:{of}\:{the}\:{friction}\:{should}\:{be}\:{left}\:{for} \\ $$$${the}\:{centrifugal}\:{force},\:{that}\:{means} \\ $$$${with}\:{increasing}\:{speed}\:{the}\:{acceleration} \\ $$$${should}\:{be}\:{decreased}.\:{when}\:{the}\:{car} \\ $$$${obtains}\:{the}\:{max}.\:{speed}\:{the}\:{acceleration} \\ $$$${should}\:{be}\:{zero}\:{and}\:{the}\:{complete} \\ $$$${frictionforce}\:{can}\:{be}\:{avaiable}\:{for}\:{the} \\ $$$${centrifugal}\:{force},\:{i}.{e}. \\ $$$${m}\frac{{v}_{{max}} ^{\mathrm{2}} }{{r}}=\mu{mg}={ma}_{{max}} \\ $$$$\Rightarrow{v}_{{max}} =\sqrt{{ra}_{{max}} }=\sqrt{\mathrm{20}×\mathrm{12}}=\mathrm{15}.\mathrm{5}\:{m}/{s} \\ $$
Commented by mrW2 last updated on 25/Feb/18
Commented by mrW2 last updated on 25/Feb/18
$${at}\:{point}\:\mathrm{1}: \\ $$$${f}={ma}_{{max}} \\ $$$${at}\:{point}\:\mathrm{2}: \\ $$$${f}=\sqrt{\left({ma}\right)^{\mathrm{2}} +\left({m}\frac{{v}^{\mathrm{2}} }{{r}}\right)^{\mathrm{2}} }={m}\sqrt{{a}^{\mathrm{2}} +\left(\frac{{v}^{\mathrm{2}} }{{r}}\right)^{\mathrm{2}} } \\ $$$${at}\:{point}\:\mathrm{3}: \\ $$$${f}={m}\frac{{v}_{{max}} ^{\mathrm{2}} }{{r}} \\ $$
Commented by NECx last updated on 25/Feb/18
$${wow}...\:{Thanks}\:{so}\:{much}. \\ $$$${I}\:{really}\:{love}\:{it}\:{when}\:{you}\:{explain} \\ $$$${with}\:{the}\:{aid}\:{of}\:{diagrams}. \\ $$$${Thanks}\:{once}\:{again}. \\ $$
Commented by mrW2 last updated on 27/Feb/18
$${Does}\:{my}\:{answer}\:{match}\:{with}\:{that}\:{in} \\ $$$${your}\:{book}? \\ $$
Commented by NECx last updated on 06/Mar/18
$${yes}\:{sir}. \\ $$