A copper block rests 44.4 cm from the center of a steel turntable. The coefficient of static friction between the block and the surface is 0.60. The turntable starts from rest and rotates with a constant angular acceleration of 0.60 rad/s2 . After what time interval will the block start to slip on the turntable? The acceleration of gravity is 9.81 m/s 2 . Answer in units of s.

4.041 sec

Explanation:

Let

m = mass of the block

r = Radius of circle in which the block moves = 44.4 cm = 0.444 m

μ = coefficient of friction =0.60

We know

Foprce of friction

Fs= coefficient of friction × Normal reaction force = μ×weight of block =μ mg=0.6×9.81 m =5.886 m

We are given angular acceleration = 0.6 rad /

The block will slip when centripetal force on block will become equal or greater than force of friction

Centripetal Force = Force of friction

But centripetal force F= m×

and Force of friction Fs = 5.886 m

==>  m× = 5.886 m

cancelling mass m on both sides;

==>    = 5.886

==> = 5.886 × = 1.16

==> v =1.077 meter /sec

Now we can get angular velocity using

v= ω×r

From this formula we have;

ω = = 1.077 / 0.444=2.43 rad / sec

But we know that

angular acceleration = ω / t

==> 0.6 = 2.43 / t

==> t= 4.04 sec

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