Question

A heavy stone hanging from a massless string of length 15 m is projected horizontally with speed 147 m/s. The speed of the particle at the point where the tension in the string equals the weight of the particle is :

Answer 1

Answer on Question#60215 -Physics- Mechanics -Relativity

A heavy stone hanging from a massless string of length $15\mathrm{m}$ is projected horizontally with speed $147\mathrm{m/s}$ . The speed of the particle at the point where the tension in the string equals the weight of the particle is:

Solution. The stone moves under the action of gravity $mg$ and the tension force of the thread $T$ .

According to the statement of the problem $v_{0} = 147\mathrm{m / s}$ , $L = 15\mathrm{m}$ . According to Newton's second law.

$$m \vec {a} = m \vec {g} + \vec {T}$$

We write it in the projection on the X-axis coinciding with the string at the moment when the string forms an angle $\alpha$ with the vertical (as shown in figure)

$$m a _ {x} = - m g c o s \alpha + T$$

As the stone moves in the radius of the circle $L$ , $a_{x}$ is the centripetal acceleration is equal to:

$$a _ {x} = \frac {v ^ {2}}{L}$$

where $v$ - velocity at this time. According to the statement of the problem $T = mg$ . Hence

$$m \frac {v ^ {2}}{L} = - m g c o s \alpha + m g \rightarrow \frac {v ^ {2}}{L} = g (1 - c o s \alpha) \rightarrow 1 - c o s \alpha = \frac {v ^ {2}}{g L}.$$

Using the law of conservation of energy:

In the initial moment of time the body will have only kinetic energy $\frac{mv_0^2}{2}$ , at some point in time the body has both potential and kinetic energy $\frac{mv^2}{2} + mgh$ , where $h = L - L\cos\alpha = L(1 - \cos\alpha)$ . (solve right triangle shown in figure). Therefore law of conservation of energy

$$\frac {m v _ {0} ^ {2}}{2} = \frac {m v ^ {2}}{2} + m g L (1 - \cos \alpha).$$ $$\frac {v _ {0} ^ {2}}{2} = \frac {v ^ {2}}{2} + g L (1 - \cos \alpha) \rightarrow v _ {0} ^ {2} = v ^ {2} + 2 g L (1 - \cos \alpha)$$

Using the formula $1 - \cos \alpha = \frac{v^2}{gL}$ get $v_0^2 = v^2 + 2gL\frac{v^2}{gL} \rightarrow v_0^2 = 3v^2$.

$$v = \frac {v _ {0}}{\sqrt {3}} \approx 84.87 \mathrm{m/s}.$$

Answer: $v = \frac{v_0}{\sqrt{3}} \approx 84.87 \, \mathrm{m/s}$.

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