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A simple pendilum with bob of mass m and...

A simple pendilum with bob of mass m and length l is held in position at an angle `theta` with t he verticle. Find its speed when it passes the lowest position.

Text Solution

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Hint : The height bob comes down is QR.
`h=QR=SQ-SR=l-lcos theta`
By law of conservation of mechanical energy
`mgh=(1)/(2)mv^(2)`

During motion of the bob from P to Q, its PE gets converted to KE. The height h, it comes in downward direction in QR
`h=QR = SQ - SR=l-lcos theta`
By the law of conservation of mechanical energy
`mgh=(1)/(2)mv^(2)`
`mg(l-lcos theta)=(1)/(2)mv^(2)`
`v=sqrt(2glsqrt((1-cos theta)))`
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