"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 1600π

^{2 }J. If the wheel is rotating at 20 rev/s, the moment of inertia of the wheel is
(a) 1 kgm

^{2}
(b) 2 kgm

^{2}
(c) 3 kgm

^{2}
(d) π

^{2}kgm^{2}
(e) 4π

^{2}kgm^{2}
The rotational kinetic energy

*E*is given by*E*= (½)

*I*

*ω*

^{2}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

1600π

^{2}= (½)*I*×(40π)^{2}from which*I*= 2 kgm^{2}
(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) 2

*I*/3
(d) 3

*I*/2
(e) 5

*I*/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*I*_{1}about a diameter is*I/*2 in accordance with the theorem of perpendicular axes. The moment of inertia*I*_{2}of the disc about an axis tangential to its edge and parallel to its plane is*I*_{1}+*MR*^{2}in accordance with the theorem of parallel axes:*I*

_{2}=

*I*

_{1}+

*MR*

^{2}= (

*MR*

^{2}/4) +

*MR*

^{2}

Or,

*I*_{2 }= 5*MR*^{2}/4
We know that

*I = MR*^{2}/2
Therefore,

*I*_{2 }= 5*I*/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

*R*_{g}. The diameter of the ring is
(a) 2

*R*_{g}
(b) 4

*R*_{g}
(c) √(2

*R*_{g})
(d) √(8

*R*_{g})
(e)

*R*_{g}/2
The moment of inertia of a circular ring about a central axis
perpendicular to the plane of the ring is

*MR*^{2 }where*M*is its mass and*R*is its radius. The moment of inertia of the ring about a diameter is*MR*^{2}/2 in accordance with the theorem of perpendicular axes. Since the radius of gyration of the ring about a diameter is*R*_{g}we have*MR*

^{2}/2 =

*MR*

_{g}

Therefore

*R*= √(2*R*_{g})
The diameter of the ring = 2

*R*= 2√(2*R*_{g}) = √(8*R*_{g})
(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) 3

*I*/2
(b) 2

*I*
(c) 5

*I*/2
(d) 3

*I*
(e) 4

*I*
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*=

*ML*

^{2}/12

Moment of inertia

*I*_{1}of the rod about a parallel axis through its end is given by*I*

_{1}= (

*ML*

^{2}/12) +

*M*(

*L*/2)

^{2}, in accordance with the theorem of parallel axes.

## No comments:

## Post a Comment