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Physics Class 12th System of Particles and Rotational Motion Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A disc of mass 1kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $\sqrt[]{\frac{x}{3 R} r a d s^{- 1}}$ where x = __________.

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Vishal Baghel | Contributor-Level 10

2 months ago

Loss in P.E. = Gain in k.E

2 mg R = $\frac{1}{2} (\frac{1}{2} m R^{2} + m R^{2}) ω^{2}$

$ω = \sqrt[]{\frac{8 g}{3 R}} = 4 \sqrt[]{\frac{g}{2 \times 3 R}}$

$x = \frac{g}{2} = 5$

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