Pendulum question

  • #1
188
0

Homework Statement



A pendulum consists of mass M hanging at the bottom of a mass-less rod of length L with a frictionless pivot at its top. A bullet of mass m and velocity v collides with M and gets embedded. What is the smallest velocity v of the bullet sufficient to cause the pendulum with the bullet to clear the top of the arc.


Homework Equations



[tex]mv=(m+M)v'[/tex]
[tex]KE=\frac{1}{2}(m+M)v'^2[/tex]
[tex]PE=(m+M)gh[/tex]


The Attempt at a Solution



First I found out the velocity of the moving block.

[tex]mv=(m+M)v'[/tex]
[tex]\frac{mv}{(m+M)}=v'[/tex]

and [tex]KE_{bottom}=PE_{top}[/tex]

[tex]\frac{1}{2}(m+M)v'^2=(m+M)g(2L)[/tex]

plug in [tex]v'[/tex]

[tex]\frac{1}{2}(m+M)\frac{{m^2}{v^2}}{(m+M)^2}=(m+M)2gL[/tex]

cancel out [tex](m+M)[/tex]

[tex]\frac{1}{2}\frac{{m^2}{v^2}}{(m+M)}=(m+M)2gL[/tex]

cross multiply

[tex]{m^2}{v^2}={(m+M)^2}2gL[/tex]

divide by [tex]m^2[/tex]

[tex]v^2=\frac{{(m+M)^2}2gL}{m^2}}[/tex]

[tex]v=\frac{(m+M)\sqrt{4gL}}{m}[/tex]

[tex]v=\frac{(m+M)\sqrt{4}\sqrt{gL}}{m}[/tex]

[tex]v=\frac{(m+M)2\sqrt{gL}}{m}[/tex]
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
At the top of the arc, the velocity of M is not equal to zero because it clears the top.
 
  • #3
188
0
but is it more or less correct?
 
  • #4
rl.bhat
Homework Helper
4,433
8
The total energy at the top is
(m + M)*2gL + 1/2*(m + M)*v"^2
At the top acceleration v"^2/L = g. Or v"^2 = Lg. Sustitute this value in the above expression.
One more mistake.
Check the step from v^2 to v.
 

Related Threads on Pendulum question

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
575
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
922
Top