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Momentum/Conservation of Energy problem.

  • Thread starter Sefrez
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  • #1
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I am given the problem:

As shown below, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length script i and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use L for script i, g for gravity, and M and m as appropriate.)

And I derive minimum velocity initial as v = (2M/m)*sqrt(5gL) where as every solution I find is 4M/m*sqrt(gL). I am only able to get this when saying that K_i = 2m*g*L. That is, the initial kinetic energy of the suspended mass is the potential at the top of the swing.

Though, wouldn't this imply that the kinetic energy at the top of the swing is zero, and thus zero velocity? And if that was the case, wouldn't the mass free fall (no radial acceleration)? What I did was say the velocity at the top should be such that the downward radial acceleration is g, or v_top = sqrt(g*L).

EDIT:
Wait, I just caught something. It says stiff rod. They should bold that... So is my other solution correct given a "negligible" massed string then? :D
 

Answers and Replies

  • #2
993
13
Can you show how you got v = (2M/m)*sqrt(5gL)?
 
  • #3
NascentOxygen
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Science Advisor
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As shown below, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length script i and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? (Use L for script i, g for gravity, and M and m as appropriate.)
What is meant by "script i"?

And I derive minimum velocity initial as v = (2M/m)*sqrt(5gL) where as every solution I find is 4M/m*sqrt(gL). I am only able to get this when saying that K_i = 2m*g*L. That is, the initial kinetic energy of the suspended mass is the potential at the top of the swing.
K.E. = P.E. sounds good. 4M/m*sqrt(gL) seems right.

Though, wouldn't this imply that the kinetic energy at the top of the swing is zero, and thus zero velocity?
Yes.
 
  • #4
993
13
I got the same as NascentOxygen got. That is the reason for suggesting to the OP to show the method.
 

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