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(A) When a vehicle moves in a circular path,the bob experiences a pseudo force $F_p = \frac{mv^2}{r}$ acting outwards and the gravitational force $mg$

Vedclass · Accepted Answer

$	an^{-1}(0.5)$

Vedclass · Answer

(A) When a vehicle moves in a circular path,the bob experiences a pseudo force $F_p = \frac{mv^2}{r}$ acting outwards and the gravitational force $mg$ acting downwards.Let $	heta$ be the angle the string makes with the vertical.In the frame of the vehicle,the bob is in equilibrium under the tension $T$,the pseudo force,and the weight.Taking components,we have $T \sin 	heta = \frac{mv^2}{r}$ and $T \cos 	heta = mg$.Dividing the two equations: $	an 	heta = \frac{mv^2/r}{mg} = \frac{v^2}{rg}$.Given $v = 10 \ m/s$,$r = 20 \ m$,and $g = 10 \ m/s^2$.$	an 	heta = \frac{10^2}{20 	imes 10} = \frac{100}{200} = 0.5$.Therefore,$	heta = 	an^{-1}(0.5)$.

$A$ vehicle is moving with a constant speed of $10 \ m/s$ on a circular horizontal track of radius $20 \ m$. $A$ bob is suspended from the roof of the vehicle by a massless string. The angle made by the string with the vertical will be (acceleration due to gravity,$g = 10 \ m/s^2$).

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