   Chapter 8, Problem 51P

Chapter
Section
Textbook Problem

A light rod of length ℓ = 1.00 m rotates about an axis perpendicular to its length and passing through its center as in Figure P8.51. Two particles of masses m1 = 4.00 kg and m2 = 3.00 kg are connected to the ends of the rod. (a) Neglecting the mass of the rod, what is the system’s kinetic energy when its angular speed is 2.50 rad/s? (b) Repeat the problem, assuming the mass of the rod is taken to be 2.00 kg. Figure P8.51 Problems 51 and 65

(a)

To determine
The kinetic energy of the system by neglecting mass of the rod.

Explanation

Given info: The masses at the ends of the rod are 4.00kg and 3.00kg , the length of the rod is 1.00m , and the angular speed of the masses is 2.50rad/s .

Explanation: Assume that the masses at the ends of the rod are point particle and the total moment of inertia of the system is I=Irod+I1+I2 . Since the mass of the rod is negligible, the moment of inertia of the rod is zero and only the moment of inertia of the point masses are such that I=0+I1+I2=m1(l/2)2+m2(l/2)2=(m1+m2)(l/2)2 and the rotational kinetic energy of the system is KEr=Iω2/2 .

The formula for the kinetic energy of the system is,

KEr=12[mrodl212+(m1+m2)(l/2)2]ω2

• ω is angular speed of the system

(b)

To determine
The kinetic energy of the system with mass of the rod is taken.

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