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How can I solve this question: A parabolic bowl with its bottom at origin has the shape y =(x^2) /20. What is the maximum height where a bowl of mass m can be placed without slipping (x and y in metres, μ=0.5)?
Friction force would act tangentially obvious.Hence, f = umgNow, tanthetha = 2x/20 => x/10sinthetha = x/root(x^2 + 100)Hence,mgsin(thetha) = f = umg0.5 = x/root(x^2 + 100)x^2/(x^2 + 100) = ¼4x^2 = x^2 + 1003x^2 = 100x = 10/root(3) meter.
dy/dx=2x/10 i.e. x/5 which is equal to tan thetaAlso umg cos theta = mg sin thetaI.e. u = tan theta Hence 1/2= X/5Therefore value of X=2.5 mPutting value of X in equation y=x^2/10We get. Y= 2.5*2.5/10Y= 0.625 m or 62.5 cm
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