# A 4.05 kg object is attached to a 3.00 m pole by two 2 m strings. The object rotates in a horizontal circle at a constant speed 4.60 m/s. What is the tension in the strings?

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### 1 Answer

The question is not verclearly stated, however, the prolem is being solved based on the following:

mass of object = m = 4.05 kg

distance between the ends of the strings attached to pole = 3.00 m

length of strings = 2.00 m

Speed of rotation = v = 4.6 m/s

Solution:

Radius of rotation = r = sqrt(2^2 - (3/2)^2)

= sqrt(4-9/4)=2*sqrt(7) = 5.29

Centripetal force exerted by 2 strings in horizontal direction

= m*v^2/r = 4.05*(4.6^2)/5.29 = 453.34 Newton

Therefore combined horizontal component of Tension in both strings = 453.34 Newton

Downward force due to gravity = 4.05* 9.8 = 44.69 Newton

Angle of strings with pole = cos-1(3/2 / 2) = 41.41 degree

Suppose the tnesion in upper & lower strings is Tu and Tl

Then:

Tu*sin(41.41)+Tl*sin(41.41) = 453.34

Tu*cos(41.41)-Tl*cos(41.41) = 44.69

Tu+Tl = 453.34/sin(41.41) = 685.41

Tu-Tl = 44.69/cos(41.41) = 59.58

Solving above 2 equations in Tu and Tl,

Tu = 372.50 Newton

Tl = 322.91 Newton

**Tension in upper & lower string is 372.50 & 322.91 Newton respectively.**