Question

A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the angle made by the string when the mass comes to momentary rest
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Answer 1

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) $E=\frac{1}{2}mv^2+mgh$

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) $E=\frac{1}{2}mv^2$

The energy when the mass comes to a stop will be:

3) $E=mgh$

Setting equations 2 and 3 equal and solving for height h will give:

4) $h=\frac{v^2}{2g}$

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) $cos\phi=\frac{l-h}{l}$

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) $\phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})$