Answer:
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 / [tex]sec^{2}[/tex]
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× [tex]\frac{v^{2} }{r}[/tex]
and Force of friction Fs = 5.886 m
==> m× [tex]\frac{v^{2} }{r}[/tex] = 5.886 m
cancelling mass m on both sides;
==> [tex]\frac{v^{2} }{r}[/tex] = 5.886
==> [tex]v^{2}[/tex] = 5.886 × [tex]0.444^{2}[/tex] = 1.16 [tex](meter/sec)^{2}[/tex]
==> v =1.077 meter /sec
Now we can get angular velocity using
v= ω×r
From this formula we have;
ω = [tex]\frac{v}{r}[/tex] = 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
