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InterviewSolution
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A particle of mass 0.2 kg attached to a massless string in a verticle circle of radius 1.2 m. It is imparted a speed of `8 ms^(-1)` at the lowest point of its circular path. Does the particle complete the verticle circle ? What is the change in tension in the string when the particle moves from the position where the string is vertical to the position where the string is horizontal ? |
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Answer» Here, ` m = 0.2kg , r =1.2 m , upsilon_(L) = 8 ms^(-1)` Minimum speed at lowest point for completing the verticle circle ` =sqrt(5 g r) = sqrt(5 xx 10 xx 1.2) = 7.75 m^(-1) ` As ` upsilon = 8m//s ` is greater than` 7.75ms^(-1)`, therefore , the particle completes the vertical circle . From ` T = (m upsilon^(2))/(r) + mg cos theta` At the lowest point , string is vertical ` T_(L) = (m upsilon_(L)^(2))/(r) + mg cos 0^(@) = (m upsilon_(L)^(2))/(r) + mg ` When the string is horizontal say at A ` T_(A) = (m upsilon_(A)^(2))/(r) + mg cos90^(@) = (m upsilon_A^(2))/(r) ` ` T_(L) - T_(A) = (m)/(r) (upsilon_(L)^(2) - upsilon_(A)^(2)) + mg = (m)/(r) xx (2 g r ) + mg = 3 mg = 3 xx 0.2 xx 10 = 6 N ` . |
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