Find the minimum speed
A small ball is suspended from a point by a light thread of length 1m. The ball is drawn aside so that the thread subtends an angle of 15 degrees with vertical. What minimum velocity must be imparted to the ball in the direction perpendicular to the plane of the thread and vertical so that the maximum deviation of the thread from vertical in the course of motion is 90 degrees, that is the thread becomes horizontal? Take acceleration due to gravity equal to 10m per second per second.
The answer is 4.55033225.
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1 solution
Let the magnitude of the velocity required be u, length of the thread be l, accn. due to gravity be g, initial deviation of the thread from vertical be α and the mass of the ball be m. Then from the angular momentum conservation we get mulsin(α)=mvl or v=usin(α), where v is the velocity of the ball when the thread is horizontal. Energy conservation principle gives (1/2)mu^2=(1/2)mv^2+mgl or u^2=(usin(α))^2+2gl or (ucos(α))^2=2gl or u=√(2gl/cos(α)). Substituting the values of g, l and α we get the value of u as 4.5503322525064