Question 15 Marks
A turn of radius 20m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?
Answer
View full question & answer→In the question it is given radius of turn r = 20m and banked with angle $\theta$ for speed$\upsilon=\frac{36\text{km}}{\text{h}}=\frac{36{\times}5}{18}=10\text{m}/\text{ sec}$
And coefficient of friction $\mu=0.4$ From here we can find the value of $\tan\theta$ We know $\tan\theta=\frac{\upsilon^2}{\text{rg}}$$\Rightarrow\tan\theta=\frac{10^2}{20\times10}=\frac12\dots(1)$
We assume two conditions one when a vehicle moves with maximum speed in this condition the vehicle may skid up if speed will increase. In this situation friction between road and tyre is opposite to skidding direction as shown in figure We mark all forces on vehicle as shown in figure and dividing them in their component From figure clearly see that if vehicle is in equilibrium so the force acting on vehicle are equal along the road and perpendicular to the road Force along the road
$\Rightarrow{\text{mg}\sin\theta}+\mu\text{N}=\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta$
Rearranging this equation$\Rightarrow\mu\text{N}=\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta-\text{mg}\sin\theta\dots(2)$
Now take forces perpendicular to road$\Rightarrow\text{N}=\text{mg}\cos\theta+\frac{\text{m}\upsilon^2}{\text{r}}\sin\theta\dots(3)$
Divide (2) by (3)$\Rightarrow\mu=\frac{\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta-\text{mg}\sin\theta}{\text{mg}\cos\theta+\frac{\text{m}\upsilon^2}{\text{r}}\sin\theta}$
Solving this equation we get$\Rightarrow\mu=\frac{\upsilon^2\cos\theta-\text{r}\sin\theta}{\text{rg}\cos\theta+\upsilon^2\sin\theta}$
Right hand side divided by $\cos\theta$$\Rightarrow\mu(\text{rg}+\text{v}^2\tan\theta)=\text{v}^2-\text{rg}\tan\theta$
Solving this we can find value of vv maximum velocity of vehicle where it will not skid$\Rightarrow\upsilon=\sqrt{\frac{\text{rg}\tan\theta+\mu\text{rg}}{1-\mu\tan\theta}}$
Put the given values in this equation$\Rightarrow\upsilon=\sqrt{\frac{20\times10\times\frac12+0.4\times20\times10}{1-0.4\times\frac12}}$
$\Rightarrow\upsilon=\sqrt{225}$
$\upsilon=15\text{m}/\text{ sec}$
Convert into $\text{km/ h}$ Maximum speed $\text{v = 54km/ h}$ Now we take the minimum speed case in this case friction applied in upward direction along the road as shown in figure.
Equate Forces along road$\Rightarrow\mu\text{N}=\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta=\text{mg}\sin\theta $
$\Rightarrow\mu\text{N}=\text{mg}\sin\theta-\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta\dots(4)$
Forces perpendicular to road$\Rightarrow\text{N}=\text{mg}\cos\theta-\frac{\text{m}\upsilon^2}{\text{r}}\sin\theta\dots(5)$
Divide (4) by (5)$\Rightarrow\mu=\frac{\text{mg}\sin\theta-\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta}{\text{mg}\cos\theta+\frac{\text{m}\upsilon^2}{\text{r}}\sin\theta}$
Solving this we get$\Rightarrow\mu=\frac{\text{rg}\sin\theta-\upsilon^2\cos\theta}{\text{rg}\cos\theta-\upsilon^2\sin\theta}$
Divide right hand side by $\cos\theta$$\Rightarrow\mu=\frac{\text{rg}\tan\theta-\upsilon^2}{\text{rg}-\upsilon^2\tan\theta}$
Further solving it we get$\Rightarrow\mu=\sqrt{\frac{\text{rg}\tan\theta-\mu\text{rg}}{\mu\tan\theta+1}}$
Putting given values in above equation$\Rightarrow\upsilon=\sqrt{\frac{20\times10\times\frac12+0.4\times20\times10}{0.4\times\frac12+1}}$
By solving this$\Rightarrow\upsilon=\sqrt{16.66}$
Minimum speed$\therefore\text{v}=4.08\text{m/ sec}$
Convert into $\text{km/ h}$$\upsilon=4.08\times\frac{18}5=14.68\text{km/ h}$
Hence the minimum speed is $14.7\text{km/ h}$
And coefficient of friction $\mu=0.4$ From here we can find the value of $\tan\theta$ We know $\tan\theta=\frac{\upsilon^2}{\text{rg}}$$\Rightarrow\tan\theta=\frac{10^2}{20\times10}=\frac12\dots(1)$
We assume two conditions one when a vehicle moves with maximum speed in this condition the vehicle may skid up if speed will increase. In this situation friction between road and tyre is opposite to skidding direction as shown in figure We mark all forces on vehicle as shown in figure and dividing them in their component From figure clearly see that if vehicle is in equilibrium so the force acting on vehicle are equal along the road and perpendicular to the road Force along the road
$\Rightarrow{\text{mg}\sin\theta}+\mu\text{N}=\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta$Rearranging this equation$\Rightarrow\mu\text{N}=\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta-\text{mg}\sin\theta\dots(2)$
Now take forces perpendicular to road$\Rightarrow\text{N}=\text{mg}\cos\theta+\frac{\text{m}\upsilon^2}{\text{r}}\sin\theta\dots(3)$
Divide (2) by (3)$\Rightarrow\mu=\frac{\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta-\text{mg}\sin\theta}{\text{mg}\cos\theta+\frac{\text{m}\upsilon^2}{\text{r}}\sin\theta}$
Solving this equation we get$\Rightarrow\mu=\frac{\upsilon^2\cos\theta-\text{r}\sin\theta}{\text{rg}\cos\theta+\upsilon^2\sin\theta}$
Right hand side divided by $\cos\theta$$\Rightarrow\mu(\text{rg}+\text{v}^2\tan\theta)=\text{v}^2-\text{rg}\tan\theta$
Solving this we can find value of vv maximum velocity of vehicle where it will not skid$\Rightarrow\upsilon=\sqrt{\frac{\text{rg}\tan\theta+\mu\text{rg}}{1-\mu\tan\theta}}$
Put the given values in this equation$\Rightarrow\upsilon=\sqrt{\frac{20\times10\times\frac12+0.4\times20\times10}{1-0.4\times\frac12}}$
$\Rightarrow\upsilon=\sqrt{225}$
$\upsilon=15\text{m}/\text{ sec}$
Convert into $\text{km/ h}$ Maximum speed $\text{v = 54km/ h}$ Now we take the minimum speed case in this case friction applied in upward direction along the road as shown in figure.
Equate Forces along road$\Rightarrow\mu\text{N}=\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta=\text{mg}\sin\theta $$\Rightarrow\mu\text{N}=\text{mg}\sin\theta-\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta\dots(4)$
Forces perpendicular to road$\Rightarrow\text{N}=\text{mg}\cos\theta-\frac{\text{m}\upsilon^2}{\text{r}}\sin\theta\dots(5)$
Divide (4) by (5)$\Rightarrow\mu=\frac{\text{mg}\sin\theta-\frac{\text{m}\upsilon^2}{\text{r}}\cos\theta}{\text{mg}\cos\theta+\frac{\text{m}\upsilon^2}{\text{r}}\sin\theta}$
Solving this we get$\Rightarrow\mu=\frac{\text{rg}\sin\theta-\upsilon^2\cos\theta}{\text{rg}\cos\theta-\upsilon^2\sin\theta}$
Divide right hand side by $\cos\theta$$\Rightarrow\mu=\frac{\text{rg}\tan\theta-\upsilon^2}{\text{rg}-\upsilon^2\tan\theta}$
Further solving it we get$\Rightarrow\mu=\sqrt{\frac{\text{rg}\tan\theta-\mu\text{rg}}{\mu\tan\theta+1}}$
Putting given values in above equation$\Rightarrow\upsilon=\sqrt{\frac{20\times10\times\frac12+0.4\times20\times10}{0.4\times\frac12+1}}$
By solving this$\Rightarrow\upsilon=\sqrt{16.66}$
Minimum speed$\therefore\text{v}=4.08\text{m/ sec}$
Convert into $\text{km/ h}$$\upsilon=4.08\times\frac{18}5=14.68\text{km/ h}$
Hence the minimum speed is $14.7\text{km/ h}$
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