How to find the maximum tension a wire will experience when a sphere is tied to it in a circular motion?

Jun 2017
337
6
Lima, Peru
The problem is as follows:

A sphere of $0.5\,kg$ of mass is rotating around an axis in a vertical plane as shown in the picture from below. The minimum speed and the maximum speed of the sphere are $2\,\frac{m}{s}$ and $4\frac{m}{s}$ respectively. Find the maximum tension (measured in Newtons) that the wire can hold. Assume $g=10\,\frac{m}{s^2}$.



The alternatives are as follows:

$\begin{array}{ll}
1.&10\,N\\
2.&25\,N\\
3.&20\,N\\
4.&30\,N\\
\end{array}$

I'm confused exactly how to assess this problem. In order to find the maximum tension I believe the equation to get this is the conservation of mechanical energy.

My instinct tells me that the maximum tension will be on the bottom as follows:

Therefore:

At that point the forces acting will be the tension of the wire and the weight.

$T-mg=\frac{mv^2}{r}$

$T= \frac{mv^2}{r} + mg$

What's the speed at this point?. Should i assume $4\,frac{m}{s}$ or $2\frac{m}{s}$ and why?. what's the physical reason for it?.

I felt that (based on experience) that the speed will be at that point maximum. Hence I'll use $4\frac{m}{s}$.

Then the tension will be:

$T= \frac{0.5(4)^2}{r} + 0.5\times 10$

But the problem lies on what r should I use?.

Then I think this might come from the conservation of mechanical energy?.

$E_k+E_u=E'_k+E'_u$

$\textrm{E=Energy at the top}$

$\textrm{E'=Energy at the bottom}$

Now for this I'm assuming that the reference for establishing the height of the sphere is passing through the center or axis of rotation.

$\frac{1}{2}mv^2+mgr=\frac{1}{2}mu^2+mg(-r)$

$\frac{1}{2}(0.5)(2)^2+0.5(10)r=\frac{1}{2}(0.5)(4^2)+(0.5)(10)(-r)$

$r=0.3$

Therefore the radius is $0.3$

And the maximum tension of the wire will be: By replacing the earlier value in the above equation,

$T= \frac{0.5(4)^2}{0.3} + 0.5\times 10$

$T=31.66\,N$

However this answer does not appear within the alternatives.

But If I were to use this formula for the speed at the lowest point=

$v_2=\sqrt{5rg}$ (How was this formula derived?)

Then $r= \frac{v^2_2}{5g}=\frac{4^2}{5*10}=\frac{8}{25}$

Introducing this in the earlier equation becomes into:

$T= \frac{0.5(4)^2}{\frac{8}{25}} + 0.5\times 10 = 30\,N\\$

Which does appear in one of the alternatives.

But If I were to use this formula instead:

$v_1=\sqrt{rg}$ (How was this formula derived?)

$r=\frac{v^2_1}{g}=\frac{4}{10}=\frac{2}{5}$

Then the radius is different? Why is it so?.

Replacing this value in the earlier equation:

$T= \frac{0.5(4)^2}{\frac{2}{5}} + 0.5\times 10 = 25\,N\\$

And this results into $25\,N$ which is the correct answer. But can somebody help me here?. Why am I obtaining two different values for the force and for the radius?. My original method did not worked, why?.

Why does it exist a discrepancy between the radius and the maximum speed and the minimum speed?. What am I doing wrong?.
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
In my opinion, whoever "engineered" this problem just picked 4 as the max speed and 2 as the minimum speed without taking energy into consideration.

The given minimum speed at the top of the arc is that necessary to make it through the top of the arc where the sphere's weight is the only force providing the centripetal acceleration, i.e. tension is essentially zero at the top.

$\dfrac{mv_{min}^2}{r} = mg \implies v_{min} = \sqrt{rg} \implies r = \dfrac{v_{min}^2}{g} = 0.4 \text{ m}$

At the bottom, where max tension occurs, the calculated radius value, $r = 0.4 \text{ m}$, makes the tension ...

$T = \dfrac{mv_{max}^2}{r} + mg \implies T = 25 \text{ N}$

However, the rub is that energy conservation doesn't work out.

Using $U_g = 0$ at the bottom, the energy at the top works out as $mg(2r) + \dfrac{1}{2}mv_{min}^2 = 5 \text{ J}$

Energy at the bottom works out to be $\dfrac{1}{2}mv_{max}^2 = 4 \text{ J}$, hence the discrepancy in energy conservation.
 
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