   Chapter 8, Problem 86AP

Chapter
Section
Textbook Problem

A uniform thin rod of length L and mass M is free to rotate on a frictionless pin passing through our end (Fig. P8.47). The rod is released from rest in the horizontal position. (a) What is the speed of its center of gravity when the rod reaches its lowest position? (b) What is the tangential .speed of the lowest point on the rod when it is in the vertical position?

(a)

To determine
of the centre of gravity when the rod reaches its lowest position.

Explanation

Given Info:

The mass of the rod is M , length of the rod is L

Refer figure P8.47 and take the reference point as O.

The expression for the total mechanical energy of the centre of mass of the rod before releasing is,

E=Ep+ER

• E total mechanical energy of the rod before released
• Ep is the potential energy of the rod
• ER is the rotational energy of the rod

At the reference point, total mechanical energy is zero.

E=0 (I)

The expression for the total mechanical energy of the center of mass of the rod after it is released is,

E=Ep+ER

• E is the total mechanical energy of the center of mass of the rod after it is released
• Ep is the potential energy of the rod after released
• ER is the rotational energy of the rod after released

Formula to calculate the potential energy of rod after release is,

Ep=mg(L2)

Formula to calculate the kinetic energy of rod after release is,

ER=12Iω2

• ω is the angular speed of the center of mass of the rod after it is released

Use mg(L/2) for Ep and (1/2)Iω2 for ER in E=Ep+ER to rewrite E .

E'=12Iω2MgL2 (II)

According to conservation of energy theorem the total energy of the center of mass is constant that is total energy of the rod before releasing is equal to total energy after releasing

(b)

To determine
The tangential speed of the rod when it is in the vertical position.

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