0
$\begingroup$

The problem is as follows:

A bob is hanging from the ceiling of a room as indicated in the figure from below. The bob is spinning around the axis indicated. The wires are tied to two brass spheres whose have a mass $M$ and they make $37^{\circ}$ and $53^{\circ}$ with the vertical respectively. The lengths of the wires are $5\,m$ each. Given this information find the angular speed of the pendulum. Assume that the acceleration due gravity is $g=10\frac{m}{s^2}$.

Sketch of the problem

My book list the answer to be $\omega=1.38\frac{rad}{s}$

My attempt to solve this problem is summarized using the FBD seen in the sketch from below. As indicated the bob in the middle is affected by the bob hanging beneath it.

Sketch of the solution

Is the FBD correct?

What I tried to do is to decompose each force acting on the spheres. Given this I could spot that:

In the second sphere:

$T_2\cos 53^{\circ}=Mg$

$T_2\sin 53^{\circ}+T_{1}\sin 37^{\circ}=\frac{Mv^2}{R}=M\omega ^2 R$

$Mg \tan 53^{\circ}+T_{1}\sin 37^{\circ}=M\omega ^2 R$

For the sphere in the middle:

$T_1\cos 37^{\circ}=Mg+T_2 \cos 53^{\circ}= 2 Mg$

Then this makes the equation from above to be:

$Mg \tan 53^{\circ}+2Mg \tan 37^{\circ}=M\omega ^2 R$

Masses cancel and by plugin the given values this reduces to:

Assuming from the graphic that the radius is $R=5\sin 53^{\circ} + 5 \sin 37^{\circ}= 7\,m$

$g \tan 53^{\circ}+2g \tan 37^{\circ}=\omega ^2 (7)$

Since it mentions that the acceleration due gravity is $g=10\frac{m}{s^2}$

$10\cdot \frac{4}{3} + 20 \cdot \frac{3}{4}=\omega^2 (7)$

Solving this results into:

$\omega=\sqrt{\frac{85}{21}}\approx 2.01 \frac{rad}{s}$

Thus is way off from the given answer. What could be happening here?.

The part where I'm stuck on is if the method which I used is correct?

Assuming that the commanding angular speed for this system is made possible by the sphere in the middle the equation becomes into:

$Mg \tan 53^{\circ}+2Mg \tan 37^{\circ}=M\omega ^2 R$

$2Mg \tan 37^{\circ}-Mg \tan 53^{\circ}=M\omega ^2 R$

Which by following the same procedure indicated above becomes into:

$20 \cdot \frac{3}{4} - 10\cdot \frac{4}{3} =\omega^2 (7)$

which in the end produces $\omega \approx 0.48 \frac{rad}{s}$ which is conflicting with what I obtained above. Since all the results seem to don't get close. Can someone help me on exactly what sort of procedure should be used here?. Help please?.

$\endgroup$
0
$\begingroup$

Hint.

Consider only the lower ball so

$$ T_2(\sin(a_2),\cos(a_2))+\omega^2 l(\sin(a_1)+\sin(a_2))M(-1,0)+M g(0,-1) = 0 $$

Solving for $\omega$ we obtain

$$ \omega = 1.37665\approx 1.38 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.