“If
I have seen a little further it is by standing on the shoulders of Giants.”

– Sir Isaac Newton

A significant number of students in any class
have the wrong notion that circular motion and rotation (or, generally angular
motion) is a somewhat difficult section. But you can rest assured that this
section is as interesting and simple as any other section of your choice,
provided you have a clear understanding of the basic principles. You will find
the useful formulae and many questions in this section, posted earlier on this site, which can be
accessed by clicking on the label, ‘circular motion and rotation’ below this
post. Since the number of posts
displayed in each screen is limited, you will have to make use of the ‘older
posts’ tabs to access all the posts.

Here are some more multiple choice practice
questions (with solution) on circular motion and rotation:

(1) A thin uniform iron rod of mass

*M*and length*L*is resting on a smooth horizontal surface. Two small objects each of mass*m*and traveling with uniform speed*v*in opposite directions at right angles to the length of the rod collide with the two ends of the rod simultaneously as shown in the adjoining figure. If the objects get stuck to the rod after the collision and continue to move along with the rod, the angular impulse received by the rod is
(a)

*mvL*
(b) 2

*mvL*
(c)

*mvL*/2
(d)

*mvL*/4
(e)

*mvL*/12
The angular impulse received by the rod is equal
to the change in the angular momentum of the rod. Since the rod is initially at
rest, the change in the angular momentum of the rod is equal to the angular
momentum supplied by the two objects.

Because of the collision, the rod will rotate
about a normal axis through its middle and the total angular momentum of the objects
about this axis of rotation is

*mvL*/2 +*mvL*/2, which is equal to*mvL*.
Thus the angular
impulse received by the
rod is

*mvL*.
(2) In the above
question, if

*m*=*M/*6 and*v = L*ms^{–1}, then the time taken by the rod to rotate through Ï€ radian is
(a) 1 sec

(b) 2 sec

(c) Ï€/2 sec

(d) Ï€ sec

(e) 2Ï€ sec

Because of the collision, the rod will rotate
about a normal axis through its middle with an angular velocity Ï‰ given by

IÏ‰ =

*m*vL where ‘I’ is the moment of inertia of the rod and the masses*m*and*m*at its ends.
[Note that we have equated the final angular
momentum of the system (containing the rod and the masses) to the initial
angular momentum].

Since v = L the above equation gets modified as

IÏ‰ =

*m*L^{2}
After the collision, the rod and the masses move
together and the total angular momentum is given by

IÏ‰ = [(ML

^{2}/12) + 2*m*(L/2)^{2}]Ï‰
[The first term
within the square bracket above is the moment of inertia of the rod about a
central axis perpendicular to its length and the second term is the moment of
inertia of the two masses].

From the above two
equations, we have

*m*L

^{2}= [(ML

^{2}/12) +

*m*L

^{2}/2

^{ }]Ï‰

Since

*m =*M/6 the above equation becomes.
M/6 = [(M/12) + (M/12)]Ï‰ = (M/6)Ï‰

Therefore Ï‰ = 1
radian/sec.

Therefore, the time
taken by the rod to rotate through Ï€
radian is Ï€ sec.

(3) The angular momentum of the electron due to its orbital
motion in the hydrogen atom is directed

(a) along the direction of motion of the electron.

(b) opposite to the
direction of motion of the electron.

(c) radially outwards.

(d) radially inwards.

(e) normal to the plane of the orbit.

The
correct option is (e). The angular momentum vector is the vector product (cross
product) of the radius vector

**r**and the linear momentum vector**p**(of the electron). The angular momentum vector**r****×p**is perpendicular to both**r**and**p**. Therefore, the angular momentum of the electron due to its orbital motion must be directed normal to the plane of the orbit.
(4) A solid cylinder of
mass

*M*and radius*R*is mounted on frictionless bearings so that it can rotate about its horizontal axis (Fig.). A light inextensible string, one end of which is fixed to the cylinder, is wound round the cylinder and carries a sphere of mass*M*. Initially the system is at rest. On releasing the sphere, it moves down under gravity and the cylinder starts rotating. What is the speed of the sphere after it has descended through a distance*h*?
(a) √(2

*gh*)
(b) √(4

*Rgh*/3)
(c) √(4

*gh*/3)
(d) √(2

*Rgh*)
(e) √(4

*gh*)
The gravitational potential
energy lost by the sphere (on falling through the height

*h*) is*mgh*. This can be equated to the kinetic energy gained by the cylinder and the sphere. Therefore we have*Mgh*= ½

*Mv*

^{2}+ ½

*IÏ‰*

^{2}where

*v*is the speed of the sphere on falling through the distance

*h*,

*I*is the moment of inertia of the cylinder about its axis and

*Ï‰*is the angular velocity of the cylinder.

[Note that the
sphere has

*translational*kinetic energy and the cylinder has*rotational*kinetic energy].
Since

*I =*½*MR*^{2}*and**Ï‰ = v/R*we can rewrite the above equation as*Mgh*= ½

*Mv*

^{2}+ ½ (½

*MR*

^{2})(

*v*

^{2}/

*R*

^{2})

Or,

*gh =*(*v*^{2}/2) + (*v*^{2}/4) = 3v^{2}/4
This gives

*v =*√(4*gh*/3)
Now, click here to see more questions in this
section.