Question #7ea53

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Question #7ea53

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Answer 1 · 2016-12-30T14:48:42

Drawn

Here radius of the circular turn $r = 10 m$ $r=10m$

Speed of the car during turn $v = 36 km/hr = \frac{36 \times 10^{3} m}{3600 s} = 10 \frac{m}{s}$ $v=36"km/hr"=(36xx10^3m)/(3600s)=10m/s$

Let $m$ $m$ represents mass of the bob of the pendulum suspended from the ceiling of the car then the magnitude of pseudo force acting on the bob will be same as the centripetal force on it during its motion in a circular path of radius $r$ $r$ with speed $v$ $v$

During dynamic equilibrium let $T$ $T$ be the tension on string and $θ$ $theta$ be the angle made by the string with the vertical.

$T sin θ$ $Tsintheta$ will balance the weight of the bob $m g$ $mg$ and $T cos θ$ $Tcostheta$ will be equal to centripetal force $\frac{m v^{2}}{r}$ $(mv^2)/r$

So $Tsintheta=mg.......[1]$
where $g -> "acceleration due to gravity"=9.8m/s^2$

$Tcostheta=(mv^2)/r.........[2]$

Dividing [1] by [2] we get

$tantheta=(rg)/v^2=(10xx9.8)/10^2=0.98$

$=>theta=tan^-1(0.98)=44.42^@$

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Question #7ea53

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