At time t = 0 , a horizontal disc starts rotating with angular acceleration 1 rad s - 2 ⁡ about an axis perpendicular to its plane and passing through its Centre. A small block is lying on this disc at a distance 0 . 5 m from the centre, coefficient of friction between the surface of block and disc is 0 . 255 . The block will start slipping on the disc at time t , which is approximately equal to

Question

At time t = 0 , a horizontal disc starts rotating with angular acceleration 1 rad s - 2 ⁡ about an axis perpendicular to its plane and passing through its Centre. A small block is lying on this disc at a distance 0 . 5 m from the centre, coefficient of friction between the surface of block and disc is 0 . 255 . The block will start slipping on the disc at time t , which is approximately equal to, 2 3 s, 2 2 s, 6 s, 5 s

MARKS App · Accepted Answer

The block will start slipping on the table when friction will reach its limiting value. f = m α r 2 + α t 4 r 2 = μ m g m α r 2 + α t 4 r 2 = μ m g ⇒ α 2 r 2 + α 4 r 2 t 4 = μ 2 g 2 ⇒ t = μ 2 g 2 - α 2 r 2 α 4 r 2 1 / 4 ⇒ t = 5 s

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