How to obtain the mass of a particle rotating around an axis when the difference of tensions in a wire are known?

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

A particle of mass $m$ is tied to a very thin wire. Assume the wire is inflexible and of negligible mass. The particle is spinning about a fixed axis as shown located in the center of the circle. Let $T_a$ and $T_b$ be the modulus of the tensions in the string when the particle is located in the points $a$ and $b$ respectively. Find the mass $m$ of the particle if the difference between the tensions $T_{b}-T_{a}=39.2\,N$. Assume $g=9.8\frac{m}{s^2}$



The alternatives are as follows:

$\begin{array}{ll}
1.&0.8\,kg\\
2.&2.0\,kg\\
3.&5.8\,kg\\
4.&1.0\,kg\\
5.&0.5\,kg\\
\end{array}$

For this problem what I've attempted to do was to use the force at the top to be as follows:

$T+mg=\frac{mv_a^2}{R}$

At this point the Tension must be zero (I don't know if this statement is correct.)

This reduces the top equation to:

$v_a=\sqrt{Rg}$

Then to obtain the speed in the lowest point would be by the conservation of mechanical energy:

$E_i=E_f$

$\frac{1}{2}mv^2_{a}+mgR=mg(-R)+\frac{1}{2}mv_{b}^2$

Then inserting in the above equation would give the speed for the bottom:

Cancelling masses and multiplying by $2$ to both terms:

$v^2_{a}+4gR=v^2_{b}$

Since it is known $v_{a}$ then:

$v_{b}^2=Rg+4Rg=5Rg$

Finally I'll use these in the given statements:

The tension in the top:

$T_a+mg=\frac{mv_{a}^2}{R}$

Tension in the bottom.

$T_b-mg=\frac{mv_{b}^2}{R}$

Doing a difference between these:

$T_b-T_a=2mg+\frac{mv_{b}^2}{R}-\frac{mv_{a}^2}{R}$

Replacing the known values:

$T_b-T_a=2mg+\frac{m(5Rg)}{R}-\frac{m(Rg)}{R}$

$T_b-T_a=2mg+4mg=6mg$

Then:

$39.2=6m(10)$

Which results into:

$m=0.67\,kg$

But the answers sheet indicates that the mass is $2\,kg$.

For doing that what it should happenned is that the "$2mg$" is negative in the right side of the equation but I can't find a way to do that. Can someone help me here?. Is it me?, or did I overlooked anything?. Help!. Please.
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
I agree with your solution ...

$T_b-T_a = \dfrac{m}{r}(v_b^2-v_a^2) + 2mg$

using energy conservation yields $v_b^2-v_a^2= 4rg$