Question

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.

Answer 1

Answer:

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

    $Em_{f}$ = K = ½ m v²

    Emo =$Em_{f}$

    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