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Standard XII

Physics

Center of Mass as an Average Point

A point size ...

Question

A point size mass $100 g$ is rotated in a vertical circle using a cord of length $20 c m$ . When the string makes an angle $60^{o}$ with the vertical, the speed of the mass is $1.5 m / s$ . The tangential accelaration of the mass in that position is:

4.9 m s^{- 2}

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4.9 \sqrt{2} m s^{- 2}

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4.9 \sqrt{3} m s^{- 2}

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9.8 m s^{- 2}

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Solution

The correct option is B $4.9 \sqrt{3} m s^{- 2}$
Along tangent:
$m g s i n 60 = m a_{T}$
or, $g \frac{\sqrt{3}}{2} = a_{T}$
or, $g = 9.8$
or, $a_{T} = 4.9 \sqrt{3} m / s^{2}$

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A point size mass 100 g is rotated in a vertical circle using a cord of length 20 cm. When the string makes an angle 60o with the vertical, the speed of the mass is 1.5 m/s. The tangential accelaration of the mass in that position is:

The correct option is B 4.9√3 ms−2Along tangent:mgsin 60=maTor, g√32=aTor, g=9.8or, aT=4.9√3 m/s2

A point size mass $100 g$ is rotated in a vertical circle using a cord of length $20 c m$ . When the string makes an angle $60^{o}$ with the vertical, the speed of the mass is $1.5 m / s$ . The tangential accelaration of the mass in that position is:

The correct option is B $4.9 \sqrt{3} m s^{- 2}$
Along tangent:
$m g s i n 60 = m a_{T}$
or, $g \frac{\sqrt{3}}{2} = a_{T}$
or, $g = 9.8$
or, $a_{T} = 4.9 \sqrt{3} m / s^{2}$