A civil engineer wishes to redesign the curved roadway in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a ramp is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the ramp is to be 10.0 m/s and the radius of the curve is 20.0 m. At what angle should the curve be banked?

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Solution

The correct option is D

$\sum F_{r} = N s i n θ = \frac{m v^{2}}{r}$
$\sum F_{g} = N c o s θ - m g = 0$
$\Rightarrow t a n θ = \frac{v^{2}}{r g}$
$\Rightarrow θ = t a n^{- 1} (\frac{10^{2}}{(20) (10)}) = t a n^{- 1} (\frac{1}{2})$

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Similar questions

Two cars of masses M and 2M travel at the same speed around a banked curve, of radius r, which is covered with ice. If the first car negotiates the curve without slipping, the second car will :

Q. The maximum speed with which a car can be driven round a curve of radius

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m without skidding (when

g = 10 m s^{- 2}

and the coefficient of friction between rubber tyres and the roadway is 0.2) is:

Q. An automobile negotiates a turn of radius

\frac{40}{\sqrt{3}} m

on a road which is banked at the curve at angle

30^{\circ}

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then maximum speed at which the car can negotiate the turn without skidding will be

(g = 10 {m/s}^{2})

Q. The maximum speed with which a vehicle can negotiate a curved road, which is banked at the angle,

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. If another road is flat and the vehicle has to negotiate a curve with the same maximum speed, coefficient of friction between road and tyre should be
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The correct option is D ∑Fr=Nsinθ=mv2r ∑Fg=Ncosθ−mg=0 ⇒tanθ=v2rg ⇒θ=tan−1(102(20)(10))=tan−1(12)

The correct option is D

$\sum F_{r} = N s i n θ = \frac{m v^{2}}{r}$
$\sum F_{g} = N c o s θ - m g = 0$
$\Rightarrow t a n θ = \frac{v^{2}}{r g}$
$\Rightarrow θ = t a n^{- 1} (\frac{10^{2}}{(20) (10)}) = t a n^{- 1} (\frac{1}{2})$