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

#### Chemist116

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
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

#### Chemist116

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?. #### skeeter

Math Team
$\dfrac{2mv^2}{R} \ge 2mg \implies v \ge \sqrt{Rg}$, 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.

#### DarnItJimImAnEngineer

...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.

#### Fleurrose

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