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Parallel Axis Thereom to find angular velocity |
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| Jul21-11, 10:58 PM | #1 |
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Parallel Axis Thereom to find angular velocity
1. The problem statement, all variables and given/known data
A meter stick is freely pivoted about a horizontal axis at the 94.7 cm mark. Find the (angular) frequency of small oscillations, in rad/s 2. Relevant equations I=Icm+md^2 [itex]\Sigma[/itex] [itex]\tau[/itex]=I [itex]\alpha[/itex] mg*sin([itex]\Theta[/itex])=-I(d^2[itex]\Theta[/itex]/dt^2) 3. The attempt at a solution 5.37 rad/s |
| Jul24-11, 07:41 PM | #2 |
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Hi,
The torque equation you wrote should be: -mgd*sin[itex]\theta[/itex]=Id^2[itex]\theta[/itex]/dt^2 (you forgot the 'd' on the left-side.) For small angles, sin[itex]\theta[/itex][itex]\approx[/itex][itex]\theta[/itex] The torque equation can then be rewritten as: -(mgd*[itex]\theta[/itex])/I=d^2[itex]\theta[/itex]/dt^2 This is a common differential equation that arises in physics, and it describes a type of oscillatory motion known as "simple harmonic motion". The period is: T=2[itex]\pi[/itex]*[itex]\sqrt{I/mgd}[/itex] The angular frequency can be easily found from here. |
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