How Is the Pivot Distance Calculated in a Physical Pendulum?

[sophzilla]

"A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 5.27 s. Find d."


I know that the period for a physical pendulum is T = 2pi * sqrt (I/mgL).

I'm really stuck on how to start out this one. Should I define the center of mass first? Any help would be appreciated. Thank you.

 

[Andrew Mason]

"A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 5.27 s. Find d."


I know that the period for a physical pendulum is T = 2pi * sqrt (I/mgL).

I'm really stuck on how to start out this one. Should I define the center of mass first? Any help would be appreciated. Thank you.

What is the mass in the pendulum? Can you treat the mass as a point mass located some distance from the fulcrum? If so, where would it be? (Hint: what is the moment of inertia of the metre stick?. I think you are to ignore the width of the metre stick).

AM

 

[kyle8921]

Greetings, fellow scientist. To answer your question, yes, it would be helpful to define the center of mass first. The center of mass for a meter stick is located at the 50 cm mark, since the mass is evenly distributed along the stick.

To find the distance d, we can use the equation for period and rearrange it to solve for d. The equation becomes d = (T/2pi)^2 * m * g * L / I.

To find the moment of inertia (I) for a meter stick, we can use the equation I = (1/12) * m * L^2, where m is the mass of the meter stick and L is its length.

Plugging in the given values, we get d = (5.27 s / 2pi)^2 * 1 kg * 9.8 m/s^2 * 1 m / ((1/12) * 1 kg * 1 m^2). This gives us a value of d = 0.458 m.

Therefore, the distance d from the 50 cm mark to the pivot point is approximately 0.458 m. I hope this helps in your calculations. Keep up the good work in your scientific pursuits!