- #1
Satvik Pandey
- 591
- 12
Homework Statement
A rod of length 2 m(##l##. and mass 16 kg (##M##) is projected vertically upwards from the ground, such that the top of the rod reaches a height ##h##. At the moment that it reaches its apex, a ball of mass 3 kg(##m## hits the top of the rod with a velocity of 10 m/s(##v##) and binds to the very end of the rod. The rod, now starts to spin. It perfectly completes one full rotation, and then hits the ground. Find ##h##.
Homework Equations
The Attempt at a Solution
As no external torque was acting on the system just before the collision so I think I can conserve angular momentum of the system about point ##O##.
Initial momentum will be ##0.5mvl=30##
After collision the ball sticks to the rod so position of CoM will change. Let the CoM be located at ##x## distance from point O.
So ##x=\frac{ml}{2(M+m)}=\frac{3}{19}##
Now final can angular momentum can be find by adding the angular momentum of the CoM about the point O and angular momentum of the system about it's CoM.
So Final angular momentum = ##(M+m)v_{0}x+I_{CoM} \omega##
As no external force acts on the system in horizontal direction so we can conserve linear momentum.
##mv=(M+m)v_{0}##
Now ##I_{CoM}##= I{rod about CoM}+I {m about CoM}
=##\frac { M{ l }^{ 2 } }{ 12 } +M{ x }^{ 2 }+m(1-x)^{ 2 }##
On putting values of ##x## I got it's value =##\frac { 16\times 28 }{ 3\times 19 } ##
Putting this in the equation framed by conserving angular momentum I got
##\omega =\frac { 15(3) }{ 14 } ##
Time taken to complete one rotation is ##\frac{2 \pi}{\omega}##
So ##t=\frac{88}{45}##
Now the rod will fall by distance ##h-2## before hitting the ground. So time required will be ##\sqrt { \frac { h-2 }{ 4.9 } } ##
Equating these I got ##h =20.738##. But this is not correct? I am unable to find my mistakes. Please help me!