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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15 kg and moves at v = 4.83 m/s. The circular path has a radius of R = 0.93 m

1) What is the magnitude of the tension in the string when the ball is at the bottom of the circle?

2) What is the magnitude of the tension in the string when the ball is at the side of the circle?

3) What is the magnitude of the tension in the string when the ball is at the top of the circle?

4) What is the minimum velocity so the string will not go slack as the ball moves around the circle?  Edit
Added Mon, 09 May '16 Comment  Solutions
0 Consider above figure, Suppose that the object moves from point $A$, where its tangential velocity is $v$, to point $B$, where its tangential velocity is $v'$. Let us, first of all, obtain the relationship between $v$ and $v'$. This is most easily achieved by considering energy conservation. At point $A$, the object is situated a vertical distance $r$ below the pivot, whereas at point $B$ the vertical distance below the pivot has been reduced to $r \cos\theta$. Hence, in moving from $A$ to $B$ the object gains potential energy

$m g r (1-\cos\theta)$.

This gain in potential energy must be offset by a corresponding loss in kinetic energy. Thus,

$\frac{1}{2} m v^2 - \frac{1}{2} m {v'}^2 = m g r (1-\cos\theta)$

which reduces to

${v'}^2 = v^2 - 2 r g (1-\cos\theta).$

Let us now examine the radial acceleration of the object at point $B$. The radial forces acting on the object are the tension $T$ in the rod, or string, which acts towards the centre of the circle, and the component

$m g \cos\theta$

of the object's weight, which acts away from the centre of the circle. Since the object is executing circular motion with instantaneous tangential velocity $v'$, it must experience an instantaneous acceleration

${v'}^2/r$

towards the centre of the circle. Hence, Newton's second law of motion yields

$\frac{m {v'}^2}{r} = T - m g \cos\theta.$

Equations (291) and (292) can be combined to give

$T = \frac{m v^2}{r} + m g (3 \cos\theta -2)$

From the above derivations:

$v=4.83m/s$

$m=0.15kg$

$r=0.93m$

1)magnitude of the tension in the string when the ball is at the bottom of the circle

$\theta=0^o$

$\therefore T= \frac{mv^2}{r}+mg(3cos\theta -2)$

$\therefore T= \frac{0.15*4.83^2}{0.93}+0.15*9.8(3cos0 -2)$

$\therefore T= 3.76+1.47=5.23 N$

$\therefore T= 5.23 N$

2) magnitude of the tension in the string when the ball is at the side of the circle

$\theta=90^o$

$\therefore T= \frac{mv^2}{r}+mg(3cos\theta -2)$

$\therefore T= \frac{0.15*4.83^2}{0.93}+0.15*9.8(3cos90 -2)$

$\therefore T= 3.76-2.94=0.82 N$

$\therefore T= 0.82 N$

3) magnitude of the tension in the string when the ball is at the top of the circle

$\theta=180^o$

$\therefore T= \frac{mv^2}{r}+mg(3cos\theta -2)$

$\therefore T= \frac{0.15*4.83^2}{0.93}+0.15*9.8(3cos180 -2)$

$\therefore T= 3.76-7.35=-3.59N$

Hence the string will become slack before becoming vertical.

4) minimum velocity so the string will not go slack as the ball moves around the circle?

$\theta =180^o$

For string to remain taught at the vertical position, the tension in the string must be greater than zero.

$\Rightarrow T> 0$

$\therefore T= \frac{mv^2}{r}+mg(3cos\theta -2)=0$

$\Rightarrow\frac{mv^2}{r}+mg(3cos180-2)> 0$

$\Rightarrow\frac{mv^2}{r}-5mg> 0$

$\Rightarrow\frac{mv^2}{r}> 5mg$

$\Rightarrow\frac{v^2}{r}> 5g$

$\Rightarrow v^2> 5rg$

$\Rightarrow v> \sqrt{5rg}$

$\Rightarrow v> \sqrt{5*0.93*9.8}$

$\Rightarrow v> 6.751 m/s$

Hence the minimum velocity is:

$\Rightarrow v_{min} =6.751 m/s$  Edit
Added Mon, 09 May '16 Comment  Close Choose An Image
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