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A homogeneous sphere of radius R = 10 cm rotating about its vertical diameter is gently released on a hole of radius $r_{1} = 8.0 c m$ made in a fixed horizontal slab. Centre of the hole is vertically below the centre of the sphere. The sphere stops in a time interval $Δ t_{1} = 9.0 s .$ Calculate time interval required to stop the same sphere, if it is placed rotating with the same angular velocity on a hole of radius $r_{2} = 6.0 c m$ made in another identical horizontal slab.

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Detailed Solution

$N \sin 37^{0} = m g$

$N = \frac{5 m g}{3}$

$μ N 8 = \frac{2}{5} m {(10)}^{2} α$

$μ \frac{5 m g}{3} 8 = \frac{2}{5} m {(10)}^{2} α$

$μ \frac{μ 100 g}{3 (100)} = \frac{μ g}{3}$

$t = \frac{ω 3}{μ g} = 9 \sec$

i and $r_{2}$ = 6 cm

$N = \frac{5 m g}{4}$

$μ N 6 = \frac{2}{5} m {(10)}^{2} α$

$μ \frac{5 m g}{4} 6 = \frac{2}{5} m {(10)}^{2} α$

$α = \frac{3 μ g}{16}$

$t = \frac{ω 16}{3 μ g} = 16 x C$

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