How do I find the minimum speed of a bullet so that a sphere hanging from a ceiling performs a circular orbit?

Jun 2017
300
6
Lima, Peru
The problem is as follows:

The diagram from below shows a sphere is hanging from a ceiling by a cable whose length is R as is shown in the diagram from below. A bullet whose mass is $m$ is fired and has a speed of $v$ and travels in a straight line and gets embedded in the sphere whose mass is also $m$. As a result of the collision the sphere swings and makes a circular orbit. What should be the minimum speed of the bullet such the sphere does a circular orbit?.



The alternatives given are as follows:

$\begin{array}{ll}
1.&2\sqrt{2Rg}\\
2.&2\sqrt{Rg}\\
3.&4\sqrt{Rg}\\
4.&3\sqrt{Rg}\\
5.&\sqrt{0.5Rg}\\
\end{array}$

What I thought to solve this problem was that I can relate the momentum before and after the collision and with that speed I can relate it with the conservation of the mechanical energy when the bullet is embedded in the sphere.

This would become into:

$mv= \left(m+m\right) v_f$

$v_f=\frac{1}{2}v$

This speed can be used to relate it with the conservation of mechanical energy as follows:

$E_{k1}=E_u+E_{k2}$

$\frac{2m}{2}\left(\frac{1}{2}v\right)^2=(2m)gR+\frac{2m}{2}v_{t}^2$

$v_{t}=\textrm{Tangential speed when the ball swings}$

Then:

$\frac{v^2}{4} = 2gR+v_{t}^2$

However I end up with not knowing how to get the tangential speed or relating it to the speed which should had the bullet. How can I find it?.
 

skeeter

Math Team
Jul 2011
3,211
1,734
Texas
speed at the top of the circular arc needs to satisfy $F_c = \dfrac{2mv^2}{R} \ge 2mg$

Energy at the top of the circular arc = energy at the bottom of the circular arc (post collision)

initially, note that linear momentum is conserved in the inelastic collision
 
Jun 2017
300
6
Lima, Peru
speed at the top of the circular arc needs to satisfy Fc=2mv2R≥2mgFc=2mv2R≥2mg

Energy at the top of the circular arc = energy at the bottom of the circular arc (post collision)

initially, note that linear momentum is conserved in the inelastic collision
I'm having a confusion on identifying what are the energies that you are referring to post collision. I mean the energy at the top of the circular arc. Can you state the equation for that?.

Would it be this?

$\frac{1}{2}2mv^2=2mg(2R)$

Then:

$v=2\sqrt{Rg}$

But then, do we need the equations for momentum and the other equation which you mentioned. I mean the centripetal force with that of the weight.

Can you clear this for me please?. :rolleyes:
 

skeeter

Math Team
Jul 2011
3,211
1,734
Texas
$\dfrac{2mv^2}{R} \ge 2mg \implies v \ge \sqrt{Rg}$, so the minimum speed of the 2m mass at the top of the circular arc is $v_{top} = \sqrt{Rg}$

minimum total energy at the top of the circular arc is $KE + GPE = \dfrac{1}{2}(2m)(v_{top})^2 + 2mg(2R) = 5mgR$

KE immediately after the collision at the bottom of the circular arc has to be ar least $5mgR$ for the combined masses to complete the circle.
 
Jun 2019
493
261
USA
...Not to mention, the bullet, sphere, and cable need to be able to pass through the ceiling without resistance while the pivot remains attached. That's an interesting engineering challenge.
 
Dec 2019
5
1
Algerie
speed at the top of the circular arc needs to satisfy $F_c = \dfrac{2mv^2}{R} \ge 2mg$

Energy at the top of the circular arc = energy at the bottom of the circular arc (post collision)

initially, note that linear momentum is conserved in the inelastic collisionusa today protonmail
it's not the complete answer
 

skeeter

Math Team
Jul 2011
3,211
1,734
Texas
it's not the complete answer
It wasn't meant to be ... with the post you quoted and post #4, the OP should have enough information to solve it on his/her own.