Menu Close

A-cyclist-goes-round-a-circular-track-of-radius-70m-The-total-mass-of-bicycle-and-rider-is-70kg-Calculate-the-frictional-force-which-will-ensure-that-the-rider-successfully-negotiates-the-track-with-a

Tinku Tara 0 Comments
Pin




Question Number 24175 by NECx last updated on 13/Nov/17
A cyclist goes round a circular track of radius 70m.The total mass of bicycle and rider is 70kg.Calculate the frictional force which will ensure that the rider successfully negotiates the track with a speed of 25m/s.What happens to the rider if μ=0.3?
$${A}\:{cyclist}\:{goes}\:{round}\:{a}\:{circular} \\ $$$${track}\:{of}\:{radius}\:\mathrm{70}{m}.{The}\:{total}\:{mass} \\ $$$${of}\:{bicycle}\:{and}\:{rider}\:{is}\:\mathrm{70}{kg}.{Calculate} \\ $$$${the}\:{frictional}\:{force}\:{which}\:{will} \\ $$$${ensure}\:{that}\:{the}\:{rider}\:{successfully} \\ $$$${negotiates}\:{the}\:{track}\:{with}\:{a}\:{speed} \\ $$$${of}\:\mathrm{25}{m}/{s}.{What}\:{happens}\:{to}\:{the} \\ $$$${rider}\:{if}\:\mu=\mathrm{0}.\mathrm{3}? \\ $$
Commented by NECx last updated on 13/Nov/17
please help
$${please}\:{help} \\ $$
Commented by ajfour last updated on 14/Nov/17
Commented by ajfour last updated on 14/Nov/17
f_(max) ≥ mv^2 /r ≥ ((70×25×25)/(70)) ⇒ f_(max) ≥ 625 N or μmg ≥ 625 N ⇒ μ ≥ ((625)/(70×10)) =((6.25)/7) ≈ 0.893 if μ =0.3 the rider should go slower than 25m/s or the bicycle tyre slips outward.
$${f}_{{max}} \:\geqslant\:{mv}^{\mathrm{2}} /{r} \\ $$$$\:\:\:\geqslant\:\frac{\mathrm{70}×\mathrm{25}×\mathrm{25}}{\mathrm{70}}\: \\ $$$$\Rightarrow\:\:{f}_{{max}} \:\geqslant\:\mathrm{625}\:{N} \\ $$$${or}\:\:\mu{mg}\:\geqslant\:\mathrm{625}\:{N} \\ $$$$\:\Rightarrow\:\mu\:\geqslant\:\frac{\mathrm{625}}{\mathrm{70}×\mathrm{10}}\:=\frac{\mathrm{6}.\mathrm{25}}{\mathrm{7}}\:\approx\:\mathrm{0}.\mathrm{893} \\ $$$${if}\:\mu\:=\mathrm{0}.\mathrm{3}\:{the}\:{rider}\:{should}\:{go} \\ $$$${slower}\:{than}\:\mathrm{25}{m}/{s}\:\:{or}\:{the} \\ $$$${bicycle}\:{tyre}\:{slips}\:{outward}. \\ $$$$ \\ $$
Commented by NECx last updated on 14/Nov/17
thank you sir. You cleared my doubt.
$${thank}\:{you}\:{sir}.\:{You}\:{cleared}\:{my} \\ $$$${doubt}. \\ $$$$ \\ $$
Commented by NECx last updated on 14/Nov/17
thank you sir. You cleared my doubt.
$${thank}\:{you}\:{sir}.\:{You}\:{cleared}\:{my} \\ $$$${doubt}. \\ $$$$ \\ $$
Commented by NECx last updated on 14/Nov/17
thank you sir. You cleared my doubt.
$${thank}\:{you}\:{sir}.\:{You}\:{cleared}\:{my} \\ $$$${doubt}. \\ $$$$ \\ $$
Answered by mrW1 last updated on 14/Nov/17
((mv^2 )/r)≤μmg=f ⇒μ≥(v^2 /(rg))=((25^2 )/(70×10))=0.89 ⇒f=0.89×70×10=625 N
$$\frac{{mv}^{\mathrm{2}} }{{r}}\leqslant\mu{mg}={f} \\ $$$$\Rightarrow\mu\geqslant\frac{{v}^{\mathrm{2}} }{{rg}}=\frac{\mathrm{25}^{\mathrm{2}} }{\mathrm{70}×\mathrm{10}}=\mathrm{0}.\mathrm{89} \\ $$$$\Rightarrow{f}=\mathrm{0}.\mathrm{89}×\mathrm{70}×\mathrm{10}=\mathrm{625}\:{N} \\ $$
Commented by NECx last updated on 15/Nov/17
Thank you sir.
$${Thank}\:{you}\:{sir}. \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *