A person with his hand in his pocket is skati | vedclass

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(B) The centripetal force required for circular motion on a frictionless surface is provided by the horizontal component of the normal reaction when t

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$	an^{-1}(1/5)$

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(B) The centripetal force required for circular motion on a frictionless surface is provided by the horizontal component of the normal reaction when the person leans at an angle $	heta$ with the vertical.For a body moving in a circle of radius $r$ with speed $v$,the angle of inclination $	heta$ is given by the formula: $	an 	heta = \frac{v^2}{rg}$.Given values are $v = 10 \ m/s$,$r = 50 \ m$,and $g = 10 \ m/s^2$.Substituting these values into the formula:$	an 	heta = \frac{(10)^2}{50 	imes 10} = \frac{100}{500} = \frac{1}{5}$.Therefore,the angle of inclination is $	heta = 	an^{-1}(1/5)$.

$A$ person with his hand in his pocket is skating on ice at the rate of $10 \ m/s$ and describes a circle of radius $50 \ m$. What is his inclination to the vertical? (Take $g = 10 \ m/s^2$)

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