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Question

A small block is placed on a turn table. The coefficient of static friction between the block and the table is μs=0.5. If the turn table is rotating with constant angular acceleration α=3 rad/s2, what will be the angular speed at which the block begins to slip on the table ?
Given r=1 m, g=10 m/s2.


A
1 rad/s
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B
2 rad/s
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C
3 rad/s
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D
4 rad/s
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Solution

The correct option is B 2 rad/s
Since the turn table is rotating with constant acceleration α=3 rad/s2, so at any instant, the block (rotating with table) will have two components of acceleration:

Tangential acceleration: at=rα

Centripetal acceleration: ac=ω2r

So, total acceleration a of the body will be:
a=a2t+a2c

F.B.D of block:

Vertical equilibrium gives N=mg
Applying the condition of just slipping, friction will act at limiting value
fmax=μsN=μsmg
and friction will act in a direction opposite to that of total acceleration a

ma= Pseudoforce on block, corresponding to total acceleration a
[ from frame attached with the body ]

From FBD, applying the condition of equilibrium when the block is just about to slip.

fmax=ma
μsmg=ma2t+a2c
μsg=(rα)2+(ω2r)2

Putting α=3 rad/s2, r=1 m, μs=0.5
0.5×10=9+ω4
ω4=259=16
ω=2 rad/s

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