Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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.

$${Thank}\:{you}\:{sir}. \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com