Question b15fb

Dec 1, 2017

$T = 27.11 \setminus \quad N$

Explanation:

Use the fact that the horizontal component of the tension must provide the necessary centripetal force while its vertical component equilibrates with the weight of the ball.

We are not given the radius of the circle. We have to figure that out from the ball's speed and the rope length.

The forces acting on the ball are its weight ($\vec{w} = m \vec{g} = - m g$) and the rope tension ($\vec{T}$). Resolve the tension into its vertical and horizontal components.

As the ball goes in a horizontal circle, the rope traces a cone with an apex angle $\setminus \theta$.
vec T = vec T_{h} + vec T_{v}; \qquad T_v = T\cos\theta; \qquad T_h = T\sin\theta

Vertical Components: In the vertical direction, weight of the ball equilibrates with the vertical component of the tension.
vec T_v + m vec g = vec 0;
T_v - mg = 0; \qquad T\cos\theta = mg ...... (1)

Horizontal Components: The only force on the horizontal direction is the horizontal component of tension and that is the centripetal force making the object move in a circle.
We can write the orbital radius of the ball in terms of the rope length and apex angle as $r = L \setminus \sin \setminus \theta$.

vec T_h = mveca_c; \qquad T\sin\theta = m(v^2/r) = m(v^2)/(L\sin\theta) ...... (2)

Comparing (1) and (2),
g/(\cos\theta) = v^2/(L\sin^2\theta); \qquad (gL)/v^2 = (\cos\theta)/(\sin^2\theta) = \sqrt{1-\sin^2\theta}/\sin^2\theta

Let $x = \setminus {\sin}^{2} \setminus \theta$,
\sqrt{1-x}/x = (gL)/v^2; \qquad (1-x)/x^2 = ((gL)/v^2)^2;
${x}^{2} + {\left({v}^{2} / \left(g L\right)\right)}^{2} x - {\left({v}^{2} / \left(g L\right)\right)}^{2} = 0$

The solutions to this quadratic equation are -

${x}_{1 , 2} = \frac{1}{2} \left\{- {\left({v}^{2} / \left(g L\right)\right)}^{2} \setminus \pm \setminus \sqrt{{\left({v}^{2} / \left(g L\right)\right)}^{4} - 4 {\left({v}^{2} / \left(g L\right)\right)}^{2}}\right\}$

Since $x = \setminus {\sin}^{2} \setminus \theta$, negative solution is non physical,
${x}_{1} = \frac{1}{2} {\left({v}^{2} / \left(g L\right)\right)}^{2} \left\{\setminus \sqrt{1 + 4 {\left(\frac{g L}{v} ^ 2\right)}^{2}} - 1\right\} = 0.9957$

v= 15 m/s; \qquad L = 1.5m; g = 9.8 ms^{-2}

\sin^2\theta = 0.9957; \qquad \rightarrow \theta = 86.27^0#

Now use (1) to solve for $T$,
$T = \frac{m g}{\setminus} \cos \setminus \theta = 27.11 \setminus \quad N$