Question

Two strings of length/-0.5 m each are connected to a block of mass m = 2 kg at one
end and their ends are attached to the point A and B 0,5 m apart on a vertical pole which
rotates with a constant angular velocity (0) = 7 rad/sec. Find the ratio
of tension in the upper string (T,) and the lower string (T2). [Use g = 9.8 m/s?)​

Answer 1

Answer:

Ratio of the tension in upper string and lower string is 9 : 1

Explanation:

As we know that two strings are connected together with a block and made an equilateral triangle

so here net tension force on the block towards the center of the circle must be centripetal force

so here we can say

$(T_1 + T_2) cos30 = m\omega^2 Lcos30$

$T_1 + T_2 = m\omega^2 L$

$T_1 + T_2 = 2(7^2)(0.5)$

$T_1 + T_2 = 49$

now in vertical direction

$T_1 sin30 = mg + T_2 sin30$

$T_1 - T_2 = 2mg$

$T_1 - T_2 = 2(2\times 9.8)$

$T_1 - T_2 = 39.2$

now we can solve above two equations

$T_1 = 44.1 N$

$T_2 = 4.9 N$

So ratio of the tension in the upper string and lower string is

$T_1 : T_2 = 44.1 : 4.9 = 9 : 1$

#Learn

Topic : Uniform circular dynamics

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