Understanding Pendulum Swing: Physics 11 Help for Equal Height on Right Side

  • Thread starter Thread starter roxxyroxx
  • Start date Start date
  • Tags Tags
    Physics Physics 11
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
5 replies · 2K views
roxxyroxx
Messages
48
Reaction score
0

Homework Statement



A pendulum bob is pulled sideways to the left. When released, the pendulum string will contact a horizontal rod causing the pendulum to swing in an arc of smaller radius.

the height to which the bob swings on to the right is equal to the original height but i don't understand why-- please help ?


Homework Equations



N/A

The Attempt at a Solution



?
 
Physics news on Phys.org
The relevant equations have to do with kinetic energy KE, potential energy PE, and total energy TE.

Show us the equations, and tell us how they could be interpreted to give you your answer.
 
berkeman said:
The relevant equations have to do with kinetic energy KE, potential energy PE, and total energy TE.

Show us the equations, and tell us how they could be interpreted to give you your answer.

KE = 0.5mv^2
PE = mgh
TE = 0.5mv^2 + mgh
= mv^2 + gh
 
roxxyroxx said:
KE = 0.5mv^2
PE = mgh
TE = 0.5mv^2 + mgh
= mv^2 + gh

Good start. (Not sure what you mean by the last line, but the other lines are correct.)

Now, what common characteristic does the ball have when it is first released, and when it stops at the top of its arc after the string contacts the peg? What is the KE in each of those two cases? Answer that, and then answer the whole problem!
 
berkeman said:
Good start. (Not sure what you mean by the last line, but the other lines are correct.)

Now, what common characteristic does the ball have when it is first released, and when it stops at the top of its arc after the string contacts the peg? What is the KE in each of those two cases? Answer that, and then answer the whole problem!

umm I am not 2 sure but:
mgh = 0.5mv^2 + mgh

??
 
roxxyroxx said:
umm I am not 2 sure but:
mgh = 0.5mv^2 + mgh

??

No, how fast is the object moving when FIRST released? How fast when it reaches the top of its arc? What do those velocities tell you about the KE at those two points? Can you then write the equation that answers the original question?