To understand the parallel-axis theorem and its applications To solve many probl

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To understand the parallel-axis theorem and its applications

To solve many problems about rotational motion, it is important to know the moment of inertia of each object involved. Calculating the moments of inertia of various objects, even highly symmetrical ones, may be a lengthy and tedious process. While it is important to be able to calculate moments of inertia from the definition (I=mir2i), in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of frequently occurring shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily available from any mechanics text, including your textbook. However, one must take into account that an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of tranlsational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moments of inertia that you can find in the textbooks are usually calculated with respect to an axis passing through the center of mass of the object. However, in many problems the axis of rotation does not pass through the center of mass. Does that mean that one has to go through the lengthy process of finding the moment of inertia from scratch? It turns out that in many cases, calculating the moment of inertia can be done rather easily if one uses the parallel-axis theorem. Mathematically, it can be expressed as I=Icm+md2, where Icm is the moment of inertia about an axis passing through the center of mass, mis the total mass of the object, and I is the moment of inertia about another axis, parallel to the one for which Icm is calculated and located a distance d from the center of mass. In this problem you will show that the theorem does indeed work for at least one object: a dumbbell of length 2r made of two small spheres of mass m each connected by a light rod (see the figure). NOTE: Unless otherwise noted, all axes considered are perpendicular to the plane of the page.

How far must the stone fall so that the pulley has 5.20 J of kinetic energy?

Express your answer numerically in meters to three significant figures.

Now calculate Ip for this object using the parallel-axis theorem. Express your answer in terms ofcm, m, and r. 2 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

a)

Ib = Icm + mr^2

where Icm = moment of inertia about the center of mass

b)

Similalrly, , Ic = Icm + mr^2

c)

Using conservation of energy , PE lost by the stone = KE gained by the system

So, mgh = 0.5*I*W^2 + 0.5*mv^2

here, W = angular speed of the disk = v/R

I = moment of inertia of the disk = 0.5*M*R^2 = 0.5*6*0.15^2

= 0.0675 kg.m2

So, 2.9*9.8*h = 0.0675 + 0.5*2.9*v^2

Now, KE of the pulley = 0.5*2.9*v^2 = 5.2 J

So, 2.9*9.8*h = 0.0675 + 5.2

So, h = 0.185 m <----------- answer

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