A physical pendulum is constructed from a uniform thin of rodof length L suspend

Question

A physical pendulum is constructed from a uniform thin of rodof length L suspended from one end. The center of gravity for sucha rod lies at the mid point if its length.(Moment of Inertia(I)=1/3 ML2).
A) Derive the equation for this physical pendulum and ,from that,determine its period and angular frequency, assuming only smallangle oscillations.
B) What would be the length of a simple pendulum that has this sameperiod?
( Hint: Gravity is the only force acting on the singlependulum, by definition, can be considered to be acting as itscenter of gravity).

Explanation / Answer

A)Physical pendulum If a hanging object oscillates about a fixed axis that doesnot pass through its center of mass and the object cannot beapproximated as a point mass,we cannot treat the system as a simplependulum. Consider a rigid object pivoted at a point O that is adistance d from the center of mass.We model the object under a nettorque and use the rotational form of Newton's secondlaw, = I * ,where I is the moment of inertia ofthe object about the axis through O.The result is -mgL * sin = I *(d2/dt2) The negative sign indicates that the torque about O tends todecrease .If we again assume is small,theapproximation sin ˜ is valid and the equation ofmotion reduces to (d2/dt2) = -(mgL/I) * =-w2 * ---------------(1) The solution of equation (1) is given by =maxcos(wt + ),where max isthe maximum angular position and w = (mgd/I)1/2 ---------------(2) The period is T = (2/w) = 2 * (I/mgL)1/2----------------(3) This result can be used to measure the moment of inertia of aflat rigid object. Here,I = (1/3)mL2 Substituting the above value in equation (3),we get T = 2 * ((1/3)mL2/mgL)1/2 or T = 2 * (L/3g)1/2-----------------(4) The angular frequency of the physical pendulum is w = (mgL/I)1/2 Substituting the value of I in the above equation,we get w = (mgL/(1/3)mL2)1/2 or w = (3g/L)1/2 B)Let the length of a simple pendulum that has this sameperiod be L1. The time period of the simple pendulum is T = 2 * (L1/g)1/2-----------------(5) From equations (4) and (5),we get 2 * (L1/g)1/2 = 2 *(L/3g)1/2 or (L1/g)1/2 =(L/3g)1/2 or (L1/g) = (L/3g) or L1= (L/3) Therefore,the length of the simple pendulum that has this sameperiod is L1= (L/3). or (L1/g)1/2 =(L/3g)1/2 or (L1/g) = (L/3g) or L1= (L/3) Therefore,the length of the simple pendulum that has this sameperiod is L1= (L/3).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.