Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
LeoMessi
Group Title
Section 22: Concept Questions: how does the period of the uniform rod compare to the period of the simple pendulum for small oscillations?
 3 years ago
 3 years ago
LeoMessi Group Title
Section 22: Concept Questions: how does the period of the uniform rod compare to the period of the simple pendulum for small oscillations?
 3 years ago
 3 years ago

This Question is Closed

stan Group TitleBest ResponseYou've already chosen the best response.0
How'd this question get overlooked??? for a simple pendulum of length d and mass m acted on by gravity it is easy to show (by breaking the force mg into radial and tangential components) that \[F _{t}=mgsin(\theta)\]negative because it is a restoring force. Then\[a _{t} = \frac {F _{t}} {m}=gsin(\theta)=d \alpha=d \frac {d^{2} \theta} {dt^{2}}=\]or\[\frac {d^{2} \theta} {dt^{2}}+\frac {g} {d}\sin(\theta)=0\]In the small angle approximation\[\sin(\theta)=\theta\]so\[\frac {d^{2} \theta} {dt^{2}}+\omega^{2} \theta=0\]which has the solution \[\theta=\theta _{\max}\cos(\omega t + \phi)\]and since \[\omega T=2 \pi\]\[T=\frac {2 \pi} {\omega}=2 \pi \sqrt {\frac {d} {g}}\] for a physical pendulum like a rod with CM a distance d from the pivot\[\tau _{P} = I _{P} \alpha=mgdsin(\theta)\]or\[\frac {d^{2} \theta} {dt^{2}}= \frac {mgd} {I _{P}}\sin(\theta)=\frac {mgd} {I _{P}}\theta=\omega^{2} \theta\]in the small angle approximation. Once again the solution is \[\theta=\theta _{\max}\cos(\omega t + \phi)\]and since \[\omega T=2 \pi\]\[T=\frac {2 \pi} {\omega}=2 \pi \sqrt {\frac {I _{P}} {mgd}}\]\[I _{p} = \frac {1} {3} mL^{2}= \frac {1} {3} m(2d) ^{2}=\frac {4} {3}md^{2}\]Substituting you can see that \[T _{rod}=\frac {2 \pi} {\omega}={2 \pi} \sqrt{\frac {4d} {3g}}=\frac {2} {\sqrt{3}}T _{pendulum}=1.15 \times T _{pendulum}\]
 3 years ago

stan Group TitleBest ResponseYou've already chosen the best response.0
Of course in the small angle approximation\[T=\frac {2 \pi} {\omega}=2 \pi \sqrt {\frac {I _{P}} {mgd}}\]is a general solution, so you can just substitute in the I of any object including a simple pendulum \[md^{2}\]
 3 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.