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

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Question Number 24175 by NECx last updated on 13/Nov/17

$$Acyclistgoesroundacircular$$$$trackofradius70m.Thetotalmass$$$$ofbicycleandrideris70kg.Calculate$$$$thefrictionalforcewhichwill$$$$ensurethattheridersuccessfully$$$$negotiatesthetrackwithaspeed$$$$of25m/s.Whathappenstothe$$$$riderif\mu =0.3?$$

Commented by NECx last updated on 13/Nov/17

$$pleasehelp$$

Commented by ajfour last updated on 14/Nov/17

Commented by ajfour last updated on 14/Nov/17

$$f_{max}⩾{mv}^2/r$$$$⩾\frac{70\times 25\times 25}{70}$$$$\Rightarrow f_{max}⩾625N$$$$or\mu mg⩾625N$$$$\Rightarrow \mu ⩾\frac{625}{70\times 10}=\frac{6.25}{7}\approx 0.893$$$$if\mu =0.3theridershouldgo$$$$slowerthan25m/sorthe$$$$bicycletyreslipsoutward.$$$$$$

Commented by NECx last updated on 14/Nov/17

$$thankyousir.Youclearedmy$$$$doubt.$$$$$$

Commented by NECx last updated on 14/Nov/17

$$thankyousir.Youclearedmy$$$$doubt.$$$$$$

Commented by NECx last updated on 14/Nov/17

$$thankyousir.Youclearedmy$$$$doubt.$$$$$$

Answered by mrW1 last updated on 14/Nov/17

$$\frac{{mv}^2}{r}⩽\mu mg=f$$$$\Rightarrow \mu ⩾\frac{v^2}{rg}=\frac{{25}^2}{70\times 10}=0.89$$$$\Rightarrow f=0.89\times 70\times 10=625N$$

Commented by NECx last updated on 15/Nov/17

$$Thankyousir.$$$$$$

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