A(1.25+a) kg bowling ball is hung on a (2.50+b) m long rope. it is then pulled back until the rope makes an angle of (12.0+c)o with the vertical and released. find the tension in the rope when the ball is at the lowest point. give your answer in n and with 3 significant figures.

F = 0.535 N

Explanation:

Let's use the concepts of energy, at the highest and lowest point of the trajectory

Higher

   Em₀ = U = mg y

Lower

    = K = ½ m v²

    Emo =

    mg y = ½ m v2

    v = √ 2gy

   y = L - L cos θ

  v = √ (2g L (1-cos θ))

Now let's use Newton's second law n at the lowest point where the acceleration is centripetal

     F = ma

     a = v² / r

In turning radius is the cable length r = L

    F = m 2g (1-cos θ)

Let's calculate

    F = 2  1.25  9.8 (1 - cos 12)

    F = 0.535 N

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