
#1
Jul2111, 10:58 PM

P: 4

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 



#2
Jul2411, 07:41 PM

P: 6

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 leftside.) 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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