A bob of mass m is suspended from point O by a massless string of length l as shown. At the bottommost point it is given a velocity u=√12gl for l=1m and m=1kg, match the following two columns when string becomes horizontal (g=10ms−2)
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Solution
Given : u=√12glm=1kgl=1 m g=10m/s2
Let the velocity of the bob be v when the string becomes horizontal.
Using work-energy theorem : W=ΔK.E
∴−mgl=12mv2−12mu2
OR −2gl=v2−(12gl)⟹v=√10gl
∴v=√10×10×1=10 m/s
Radial acceleration, ar=v2l=1021=100m/s2
Tension in the string, T=mar=1×100=100N
Tangential acceleration, at=mg=1×10=10N
∴ Total acceleration, a=√a2r+a2t=√(100)2+(10)2=10√101=100.5m/s2