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A block of mass m = 20 kg is kept is a d...

A block of mass m = 20 kg is kept is a distance R = 1m from central axis of rotation of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration, `alpha = 3 rad//sec^(2)`. The friction coefficient between block and table is `mu = 0.5`. At time `t = (x)/(30)` from starting of motion (i.e. t =0) the block is just about to slip. Find the value of x `(g = 10 m//s^(2))`

Text Solution

Verified by Experts

The correct Answer is:
20

(i) `because` Table starts rotation from rest
`:. omega_("initial") =0`
`omega` = Angular velocity at time t
`:. omega = omega_("initial") xx alpha t`
`omega =0 + 3 xx t = 3t`
(ii) Force diagram for block as seen from above is :

Block is just about to slip when friction force is limiting.
`:. f_(2)=sqrt((m alpha R)^(2)+(m omega^(2)R)^(2))`
or ` (mug)^(2)=m^(2)alpha^(2) R^(2)+m^(2) omega^(4)R^(2)`
`:. mu^(2) g = alpha^(2) R^(2)+omega^(4)R^(2)`
Putting valus
`(1)/(4)xx100 = 9xx1+(3t)^(4)(A)`
`:. (3t)^(4)=16`
`3t=2`
`t = (20)/(30)sec`
`:. x=2`
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