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## Thursday, November 8, 2012

### AP Physics C - Multiple Choice Practice Questions on Moment of Inertia

"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."
Albert Einstein

As you know, moment of inertia in rotational motion takes the role of mass in translational motion. Even though mass of an object is a constant in Newtonian mechanics, its moment of inertia is not a constant since it depends on the axis of rotation and the position of its constituent particles relative to the axis. Today we shall discuss a few multiple choice practice questions on moment of inertia. Many questions in this section were discussed on this site earlier. You can access them by clicking on the label ‘moment of inertia’ below this post or by trying a search for ‘moment of inertia’ using the search box provide on this page.

Here are the questions:
(1) Moment of inertia in mechanical systems is comparable to one of the following in electrical systems. Pick it out:
(a) Resistance
(b) Magnetic moment
(c) Time constant
(d) Capacitance
(e) Inductance
Because of moment of inertia a mechanical system opposes any change in its state of rotation. Because of inductance an electrical system opposes any change in the current flowing in it. Therefore moment of inertia in mechanical systems is comparable to inductance in electrical systems.
(2) The rotational kinetic energy of a wheel is 1600π2 J. If the wheel is rotating at 20 rev/s, the moment of inertia of the wheel is
(a) 1 kgm2
(b) 2 kgm2
(c) 3 kgm2
(d) π2 kgm2
(e) 4π2 kgm2
The rotational kinetic energy E is given by
E = (½) Iω2 where I is the moment of inertia and ω is the angular velocity (in radian per second). Since the wheel is rotating at 20 rev/s, its angular velocity ω in radian/second is 2π×20 = 40π.
Therefore we have
1600π2 = (½) I×(40π)2 from which I = 2 kgm2
(3) A thin circular disc has moment of inertia I about an axis passing through its centre and perpendicular to its plane. What is its moment of inertia about an axis tangential to its edge and parallel to its plane?
(a) I/4
(b) I/2
(c) 2I/3
(d) 3I/2
(e) 5I/2

Since the moment of inertia of the circular disc about an axis passing through its centre and perpendicular to its plane is I, the moment of inertia I1 about a diameter is I/2 in accordance with the theorem of perpendicular axes. The moment of inertia I2 of the disc about an axis tangential to its edge and parallel to its plane is I1 + MR2 in accordance with the theorem of parallel axes:
I2 = I1 + MR2 = (MR2/4) + MR2
Or, I2 = 5MR2/4
We know that I = MR2/2
Therefore, I2 = 5I/2
The adjoing figure will make things very clear for you.
(4) The radius of gyration of a thin circular ring about a diameter is Rg. The diameter of the ring is
(a) 2 Rg
(b) 4 Rg
(c) √(2Rg)
(d) √(8Rg)
(e) Rg/2
The moment of inertia of a circular ring about a central axis perpendicular to the plane of the ring is MR2 where M  is its mass and R is its radius. The moment of inertia of the ring about a diameter is MR2/2 in accordance with the theorem of perpendicular axes. Since the radius of gyration of the ring about a diameter is Rg we have
MR2/2 = MRg
Therefore R = √(2Rg)
The diameter of the ring = 2R = 2√(2Rg) = √(8Rg)
(5) The moment of inertia of a thin uniform rod about an axis perpendicular to the rod and passing through its centre is I. Moment of inertia of the rod about a parallel axis through its end is
(a) 3I/2
(b) 2I
(c) 5I/2
(d) 3I
(e) 4I
If the mass of the rod is M an the length is L, the moment of inertia I of the rod about an axis perpendicular to the rod and passing through its centre is given by
I = ML2/12
Moment of inertia I1 of the rod about a parallel axis through its end is given by
I1 = (ML2/12) + M(L/2)2, in accordance with the theorem of parallel axes.
Or, I1 = ML2/3 = 4I

Now, see these questions (with solution) involving moment of inertia.