Question

Q6: A ball of mass (m) 0.5 kg is attached to the end of a string having a length (L) 0.5 m. The ball is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of the angular velocity of ball (in radian/s) is :-​

Answer 1

Explanation:

REF.Image.

mω

2

Lsinθcosθ=mgsinθ

⇒cosθ=

ω

2

L

g

∴

sinθ=

ω

2

L

1

(ω

2

L)

2

−g

2

T=mgcosθ+mω

2

Lsin

2

θ

=mg

ω

2

L

g

+

(ω

2

L)

2

mω

2

L

((ω

2

L)

2

−g

2

)

=

ω

2

L

m

[g

2

+(ω

2

L)

2

−g

2

]

=mw

2

L=324 (given)

⇒ω=

0.5×0.5

324

ω=36rad/s

hope it helps you.