A-lorry-goes-round-an-unbanked-curve-If-the-radius-of-the-curve-is-30m-and-the-coefficient-of-friction-between-the-ground-and-the-tyre-is-0-6-Calculate-the-maximum-speed-of-the-lorry-g-10m-s-2-

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Question Number 24177 by NECx last updated on 13/Nov/17

$$Alorrygoesroundanunbanked$$$$curve.Iftheradiusofthecurve$$$$is30mandthecoefficientof$$$$frictionbetweenthegroundand$$$$thetyreis0.6.Calculatethe$$$$maximumspeedofthelorry.$$$$(g=10m/s^2)$$$$$$$$$$$$pleasebuddieshelp$$

Commented by NECx last updated on 14/Nov/17

$$pleasehelp$$

Answered by mrW1 last updated on 14/Nov/17

$$\frac{{mv}^2}{r}⩽\mu mg$$$$\Rightarrow v⩽\sqrt{\mu gr}=\sqrt{0.6\times 10\times 30}=13.4m/m=48km/h$$

Commented by NECx last updated on 14/Nov/17

$$butIthoughtthatthecentripetal$$$$forcewasgoingtobegreaterthan$$$$thefrictionalorequalnotless$$$$becauseitisalreadyinmotion$$$$thusovercomingthefrictional$$$$force.$$$$$$$$Pleaseexplainwhy\frac{{mv}^2}{r}⩽\mu mg$$

Commented by mrW1 last updated on 14/Nov/17

$$Thecentripetalforce\frac{{mv}^2}{2}is$$$$providedbythefrictionforcef,$$$$andthefrictionforcefmaynot$$$$bemorethan\mu mg,thus$$$$\frac{{mv}^2}{2}⩽\mu mg$$

Commented by NECx last updated on 15/Nov/17

$$Inowunderstand\dots .Thankssir.$$

Answered by ajfour last updated on 14/Nov/17

$$\mu mg={mv}^2/r$$$$\Rightarrow v=\sqrt{\mu gr}$$$$=\sqrt{0.6\times 10\times 30}=6\sqrt{5}m/s.$$

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