NCERT Solutions for Class 11th: Ch 7 System Of Particles And Rotational Motion Physics Science

Page No: 178

Excercises

7.1. Give the location of the centre of mass of a (i) sphere, (ii)
cylinder, (iii) ring, and (iv) cube, each of uniform mass density.
Does the centre of mass of a body necessarily lie inside the body?

In all the four cases, as the mass density is uniform, centreof mass is located at their respective geometrical centres.
No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

7.2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10–10
m). Find the approximate location of the CM of the molecule, given that
a chlorine atom is about 35.5 times as massive as a hydrogen atom and
nearly all the mass of an atom is concentrated in its nucleus.

Distance between H and Cl atoms = 1.27 Å

Mass of H atom = m
Mass of Cl atom = 35.5m
Let the centre of mass of the system lie at a distance x
from the Cl atom.
Distance of the centre of mass from the H atom = (1.27 – x)

Let us assume that the centre of mass of the given molecule lies
at the origin. Therefore, we can have:
[ m(1.27 – x) + 35.5mx ] / (m + 35.5m)  =  0
m(1.27 – x) + 35.5mx =  0
1.27 – x = -35.5x
x = -1.27 / (35.5 – 1)  =  -0.037 Å
Here, the negative sign indicates that the centre of mass lies at
the left of the molecule. Hence, the centre of mass of the HCl
molecule lies 0.037Å from the Cl atom.

7.3. A child sits stationary at one end of a long trolley moving
uniformly with a speed V on a smooth horizontal floor. If the
child gets up and runs about on the trolley in any manner, what is
the speed of the CM of the (trolley + child) system?

The child is running arbitrarily on a trolley moving with velocity
v. However, the running of the child will produce no effect on
the velocity of the centre of mass of the trolley. This is because
the force due to the boy’s motion is purely internal. Internal
forces produce no effect on the motion of the bodies on which they
act. Since no external force is involved in the boy–trolley
system, the boy’s motion will produce no change in the velocity
of the centre of mass of the trolley.

7.4. Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

Consider two vectors OK = vector |a| and OM = vector |b|, inclined at an angle θ, as shown in the following figure.

7.5. Show that a. (b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.

A parallelepiped with origin O and sides a, b, and c is shown in the following figure.

7.6. Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-yplane the angular momentum has only a z-component.

7.7. Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.

Let at a certain instant two particles be at points P and Q, as shown in the following figure.

Angular
momentum of the system about point P:

Lp = mv × 0 + mv × d  =  mvd   …(i)
Angular
momentum of the system about point Q:
LQ = mv × d + mv × 0   =  mvd   ….(ii)

Consider
a point R, which is at a distance y
from point Q, i.e.,
QR =
y
∴ PR
= d – y

Angular
momentum of the system about point R:
LR = mv × (dy) + mv × y
mvdmvy + mvy
= mvd  ….(iii)
Comparing equations
(i), (ii), and (iii), we get:
LP = LQ = LR    …(iv)

We
infer from equation (iv)
that the angular momentum of a system does not depend on the point
about which it is taken.

7.8. A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

The free body diagram of the bar is shown in the following figure.

Length
of the bar, l
= 2 m
T1
and T2
are the tensions produced in the left and right strings respectively.

At
translational equilibrium, we have:
T1Sin 36.90 = T2Sin 53.10
T1 / T2 = 4 / 3
T1 = (4/3) T2
For
rotational equilibrium, on taking the torque about the centre of
gravity, we have:
T1 (Cos 36.9) × d = T2Cos 53.1 (2 – d)
T1 × 0.800 d = T2 × 0.600 (2 – d)
(4/3) × T2 × 0.800d = T2 (0.600 × 2 – 0.600d)
1.067d + 0.6d = 1.2
d = 1.2 / 1.67
= 0.72 m
Hence,
the C.G. (centre
of gravity) of the given bar lies
0.72 m from its left end.

7.9. A
car weighs 1800 kg. The distance between its front and back axles is
1.8 m. Its centre of gravity is 1.05 m behind the front axle.
Determine the force exerted by the level ground on each front wheel
and each back wheel.

Mass
of the car, m
= 1800 kg

Distance
between the front and back axles, d
= 1.8 m
Distance
between the C.G. (centre
of gravity) and the back axle = 1.05
m
The
various forces acting on the car are shown
in the following figure.

Rf
and Rbare the forces exerted by the level
ground on the front and back wheels respectively.

At
translational equilibrium:
Rf + Rb = mg
= 1800 × 9.8
= 17640 N   ….(i)
For
rotational equilibrium, on taking the
torque about the C.G., we have:
Rf(1.05) = Rb(1.8 – 1.05)
Rb / Rf  =  7 / 5
Rb = 1.4 Rf    ……(ii)
Solving
equations (i)
and (ii),
we get:
1.4Rf + Rf = 17640
Rf = 7350 N
∴ Rb
= 17640 – 7350 = 10290 N
Therefore,
the force exerted on each front wheel = 7350 / 2  =  3675 N,
and

The
force exerted on each back wheel = 10290 / 2  =  5145 N

7.10. (a) Find the moment of
inertia of a sphere about a tangent to the sphere, given the moment
of inertia of the sphere about any of its diameters to be 2MR2/5,
where M is
the mass of the sphere and R
is the
radius of the sphere.

(b) Given the moment of
inertia of a disc of mass M
any of its diameters to be MR2/4,
find its moment of inertia about an axis normal to the disc and
passing through a point on its edge.

(a) The
moment of inertia (M.I.) of a sphere about its diameter = 2MR2/5

According
to the theorem of parallel axes, the moment
of inertia of a body about any axis is equal to the sum of the moment
of inertia of the body about a parallel axis passing through its
centre of mass and the product of its mass and the square of the
distance between the two parallel axes.

The
M.I. about a tangent of the sphere = 2MR2/5 + MR =  7MR2 / 5

(b) The
moment of inertia of a disc about its diameter = MR2 / 4
According
to the theorem of perpendicular axis, the
moment of inertia of a planar body (lamina) about an axis
perpendicular to its plane is equal to the sum of its moments of
inertia about two perpendicular axes concurrent with perpendicular
axis and lying in the plane of the body.
The
M.I. of the disc about its centre = MR2 / 4 + MR2 / 4  =  MR2 / 2
The
situation is shown in the given figure.

Applying
the theorem of parallel axes:

The
moment of inertia about an axis normal to the disc and passing
through a point on its edge
= MR2 / 2 + MR2 = 3MR2 / 2

Page No: 179

7.11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis, I1 = mr2
The moment of inertia of the solid sphere about an axis passing through its centre, I2 = (2/5)mr2
We have the relation:
τ = Iα
Where,
α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder, τ1 = I1α1

For the solid sphere, τn = Inαn

As an equal torque is applied to both the bodies, τ1 = τ2
α2 / α1  =  I1  /  I2  =  mr2  /  (2/5)mr2
α2 > α1     …(i)
Now, using the relation:
ω = ω0 + αt
Where,
ω0 = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω0 and t, we have:
ω α(ii)
From equations (i) and (ii), we can write:
ω2> ω1
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

7.12. A
solid cylinder of mass 20 kg rotates about its axis with angular
s–1.
The radius of the cylinder is 0.25 m. What is the kinetic energy
associated with the rotation of the cylinder? What is the magnitude
of angular momentum of the cylinder about its axis?

Mass
of the cylinder, m
= 20 kg
Angular
speed, ω
= 100 rad s–1
of the cylinder, r
= 0.25 m
The
moment of inertia of the solid cylinder:
I = mr2 / 2
= (1/2) × 20 × (0.25)2
= 0.625 kg m2
∴ Kinetic
energy = (1/2) I ω2
= (1/2) × 6.25 × (100)2 = 3125 J
∴Angular
momentum, L
=
= 6.25 × 100

= 62.5 Js

7.13. (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby blueuces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

(a) 100 rev/min

Initial angular velocity, ω1=
40 rev/min
Final angular velocity = ω2
The moment of inertia of the boy with stretched hands = I1
The moment of inertia of the boy with folded hands = I2
The two moments of inertia are related as:
I2 = (2/5) I1
Since no external force acts on the boy, the angular momentum L
is a constant.
Hence, for the two situations, we can write:
I2ω2  =  I1 ω1
ω2 = (I1/I2) ω1
= [ I1 / (2/5)I1] × 40  =  (5/2) × 40  =  100 rev/min

(b) Final K.E. = 2.5 Initial K.E.
Final kinetic rotation, EF = (1/2) I2ω22
Initial kinetic rotation, EI =  (1/2) I1ω12
EF / EI = (1/2) I2ω22 / (1/2) I1ω12
= (2/5) I1 (100)2 / I1 (40)2
= 2.5
EF = 2.5 E1
The increase in the rotational kinetic energy is attributed to the
internal energy of the boy.

7.14. A
rope of negligible mass is wound round a hollow cylinder of mass 3 kg
and radius 40 cm. What is the angular acceleration of the cylinder if
the rope is pulled with a force of 30 N? What is the linear
acceleration of the rope? Assume that there is no slipping.

Mass
of the hollow cylinder, m
= 3 kg
of the hollow cylinder, r
= 40 cm = 0.4 m
Applied
force, F =
30 N
The
moment of inertia of the hollow cylinder
about its geometric axis:

I
= mr2
=
3 × (0.4)2
= 0.48 kg m2
Torque, τ = F × r  =  30 × 0.4  = 12 Nm
For
angular acceleration α,
torque is also given by the relation:
τ = Iα
α = τ / I  =  12 / 0.48  = 25 rad s-2
Linear
acceleration = τα
= 0.4 × 25 = 10 m s–2

7.15. To maintain a rotor at a uniform angular speed of 200 rad s–1,
an engine needs to transmit a torque of 180 Nm. What is the power
requiblue by the engine?

(Note: uniform angular velocity in the absence of friction implies
zero torque. In practice, applied torque is needed to counter
frictional torque). Assume that the engine is 100 % efficient.

Angular speed of the rotor, ω =
Torque requiblue, τ = 180
Nm
The power of the rotor (P) is related to torque and angular
speed by the relation:
P = τω
= 180 × 200 = 36 × 103
= 36 kW
Hence, the power requiblue by the engine is 36 kW.

7.16. From a uniform disk of radius R, a circular hole of radius
R/2 is cut out. The centre of the hole is at R/2 from
the centre of the original disc. Locate the centre of gravity of the
resulting flat body.

AnswerR/6; from the original centre of the body and opposite to the centre of the cut portion.

Mass per unit area of the original disc = σ
Radius of the original disc = R
Mass of the original disc, M = πR2σ
The disc with the cut portion is shown in the following figure:

Radius of the smaller disc = R/2

Mass of the smaller disc, M = π (R/2)2σ = π R2σ / 4  =  M / 4
Let O and O′ be the respective centres of the original disc and the
disc cut off from the original. As per the definition of the centre of
mass, the centre of mass of the original disc is supposed to be
concentrated at O, while that of the smaller disc is supposed to be
concentrated at O′.
It is given that:
OO′= R/2
After the smaller disc has been cut from the original, the remaining
portion is consideblue to be a system of two masses. The two masses are:
M (concentrated at O), and –M′ (=M/4) concentrated at O′

(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centres of masses of two masses is given as:
x = (m1r1 + m2r2/  (m1 + m2)
For the given system, we can write:
x = [ M × 0 – M‘ × (R/2) ] /  ( M + (-M‘) )  =  –R / 6
(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)

7.17. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Let W and W′ be the respective weights of the metre stick and the coin.

The mass
of the metre stick is concentrated at its mid-point, i.e., at the 50
cm mark.

Mass
of the meter stick = m
Mass
of each coin, m
= 5 g
When the
coins are placed 12 cm away from the end P, the centre of mass gets
shifted by 5 cm from point R toward the end P. The centre of mass is
located at a distance of 45 cm from point P.
The
net torque will be conserved for rotational
equilibrium about point R.
10 × g(45 – 12) – m’g(50 – 45) = 0
m’ = 66 g
Hence, the
mass of the metre stick is 66 g.

7.18. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?

(a) Mass
of the sphere = m
Height
of the plane = h
Velocity
of the sphere at the bottom of the plane =
v

At
the top of the plane, the total energy of the sphere = Potential
energy = mgh
At
the bottom of the plane, the sphere has
both translational and rotational kinetic energies.

Hence,
total energy = (1/2)mv2 + (1/2) I ω2
Using
the law of conservation of energy, we can write:
(1/2)mv2 + (1/2) I ω2 = mgh
For
a solid sphere, the moment of inertia about its centre, I = (2/5)mr2

Hence,
equation (i)
becomes:
(1/2)mr2 + (1/2) [(2/5)mr22  =  mgh
(1/2)v2 + (1/5)r2ω2  =  gh
But we have the relation, v = rω
(1/2)v2 + (1/5)v2  =  gh
v2(7/10) = gh
v =   √(10/7)gh
Hence,
the velocity of the sphere at the bottom depends only on height (h)
and acceleration due to gravity (g). Both these values are constants.
Therefore, the velocity at the bottom remains the same from whichever
inclined plane the sphere is rolled.

(b) Consider
two inclined planes with inclinations θ1
and θ2,
related as:

θ1
< θ2
The
acceleration produced in the sphere when it rolls down the plane
inclined at θ1
is:
g
sin θ1
The various forces acting on the sphere are shown in the following
figure.

R1
is the normal reaction to the sphere.

Similarly,
the acceleration produced in the sphere when it rolls down the plane
inclined at θ2
is:

g
sin θ2

The
various forces acting on the sphere are
shown in the following figure.

R2
is the normal reaction to the sphere.
θ2
> θ1;
sin θ2
> sin θ1
… (i)

a2
> a1
(ii)
Initial
velocity, u =
0
Final
velocity, v
= Constant
Using
the first equation of motion, we can obtain
the time of roll as:
v
= u + at
t ∝ (1/α)
For inclination θ1 : t1∝ (1/α1)
For inclination θ2 : t2∝ (1/α2)

From
above equations,
we get:
t2
< t1
Hence, the sphere will take a longer time to reach the bottom of the
inclined plane having the smaller inclination.

7.19. A
hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor
so that its centre of mass has a speed of 20 cm/s. How much work has
to be done to stop it?

of the hoop, r
= 2 m
Mass
of the hoop, m
= 100 kg
Velocity
of the hoop, v
= 20 cm/s = 0.2 m/s
Total
energy of the hoop = Translational K.E. + Rotational K.E.
ET = (1/2)mv2 + (1/2) I ω2
Moment
of inertia of the hoop about its centre, I
= mr2
ET = (1/2)mv2 + (1/2) (mr2)ω2
But we have the relation, v = rω
ET = (1/2)mv2 + (1/2)mr2ω2
= (1/2)mv2 + (1/2)mv2 = mv2
The
work requiblue to be done for stopping the
hoop is equal to the total energy of the hoop.
∴ Requiblue
work to be done, W
= mv2
= 100 × (0.2)2
= 4 J.

7.20. The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94×10–46 kg m2 about
an axis through its centre perpendicular to the lines joining the two
atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and
that its kinetic energy of rotation is two thirds of its kinetic energy
of translation. Find the average angular velocity of the molecule.

Mass
of an oxygen molecule, m
= 5.30 × 10–26
kg

Moment
of inertia, I
= 1.94 × 10–46
kg m2
Velocity
of the oxygen molecule, v
= 500 m/s
The
separation between the two atoms of the oxygen molecule = 2r

Mass
of each oxygen atom = m/2
Hence,
moment of inertia I,
is calculated as:
(m/2)r2 + (m/2)r2 = mr2
r = ( I / m)1/2
(1.94 × 10-46 / 5.36 × 10-26 )1/2  =  0.60 × 10-10 m
It
is given that:
KErot = (2/3)KEtrans
(1/2) I ω2 = (2/3) × (1/2) × mv2
mr2ω2 = (2/3)mv2
ω = (2/3)1/2 (v/r)
= (2/3)1/2 (500 / 0.6 × 10-10) = 6.80 × 1012 rad/s.

7.21. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?

Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, θ = 30°
Let the cylinder go up the plane upto a height h.

From, 1/2 mv2 + 1/2 I ω2 = mgh
1/2mv2 + 1/2 (1/2 mr2) ω2 = mgh
3/4 mv2 = mgh
h = 3v2/4g = 3 × 52/4 × 9.8 = 1.913 m

If s is the distance up the inclined plane, then as
sin θ = h/s
s = h/sin θ  = 1.913/sin 30° = 3.826 m

Time taken to return to the bottom

7.22. As
shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m
long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40
kg is suspended from a point F, 1.2 m from B along the ladder BA.
Assuming the floor to be frictionless and neglecting the weight of the
ladder, find the tension in the rope and forces exerted by the floor on
the ladder. (Take g = 9.8 m/s2)

The given situation can be shown as:

NB = Force exerted on the ladder by the floor point B
NC = Force exerted on the ladder by the floor point C
T = Tension in the rope
BA = CA = 1.6 m
DE = 0. 5 m
BF = 1.2 m
Mass of the weight, m = 40 kg
Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H.
ΔABI and ΔAIC are similar
∴BI = IC
Hence, I is the mid-point of BC.
DE || BC
BC = 2 × DE = 1 m
AF = BA – BF = 0.4 m … (i)
D is the mid-point of AB.
Hence, we can write:
AD = (1/2) × BA  =  0.8 m   …(ii)
Using equations (i) and (ii), we get:
FE = 0.4 m
Hence, F is the mid-point of AD.
FG||DH and F is the mid-point of AD. Hence, G will also be the mid-point of AH.
ΔAFG and ΔADH are similar
∴ FG / DH  = AF / AD
FG / DH  =  0.4 / 0.8  =  1 / 2
FG = (1/2) DH
= (1/2) × 0.25  =  0.125 m

AH = (AD2 – DH2)1/2
= (0.82 – 0.252)1/2  =  0.76 m

For translational equilibrium of the ladder, the upward force should be equal to the downward force.
Nc + NB = mg = 392 … (iii)
For rotational equilibrium of the ladder, the net moment about A is:
-NB × BI + mg × FG + NC × CI + T × AG – T × AG  =  0
-NB × 0.5 + 40 × 9.8 × 0.125 + NC × 0.5  =  0
(NC – NB) × 0.5 = 49
NC – NB = 98   …..(iv)
Adding equations (iii) and (iv), we get:
NC = 245 N
NB = 147 N
For rotational equilibrium of the side AB, consider the moment about A.
-NB × BI + mg × FG + T × AG  =  0
-245 × 0.5 + 40 X 9.8 × 0.125 + T × 0.76  =  0
∴ T = 96.7 N.

Page No: 180

7.23. A
man stands on a rotating platform, with his arms stretched
horizontally holding a 5 kg weight in each hand. The angular speed of
the platform is 30 revolutions per minute. The man then brings his
arms close to his body with the distance of each weight from the axis
changing from 90cm to 20cm. The moment of inertia of the man together
with the platform may be taken to be constant and equal to 7.6 kg m2.

(a) What
is his new angular speed? (Neglect friction.)

(b) Is
kinetic energy conserved in the process? If not, from where does the

(a) Moment
of inertia of the man-platform system = 7.6 kg m2
Moment of inertia when the man stretches his hands to a distance of
90 cm:
2
×
m r2
=
2 ×
5 ×
(0.9)2
=
8.1 kg m2
Initial
moment of inertia of the system, Ii = 7.6 + 8.1 = 15.7 kg m2
Angular
speed, ωi = 300 rev/min
Angular
momentum, Li = Iiωi  =  15.7 × 30   ….(i)
Moment of inertia when the man folds his hands to a distance of 20
cm:
2
×
mr2
=
2 ×
5 (0.2)2
= 0.4 kg m2
Final
moment of inertia, If = 7.6 + 0.4 = 8 kg m2
Final
angular speed = ωf
Final
angular momentum, Lf = Ifωf = 0.79 ωf …… (ii)
From the conservation of angular momentum, we have:
Iiωi  =  Ifωf
ωf = 15.7 × 30  / 8  =  58.88 rev/min

(b) Kinetic
energy is not conserved in the given process. In fact, with the
decrease in the moment of inertia, kinetic energy increases. The
additional kinetic energy comes from the work done by the man to fold
his hands toward himself.

7.24. A
bullet of mass 10 g and speed 500 m/s is fiblue into a door and gets
embedded exactly at the centre of the door. The door is 1.0 m wide
and weighs 12 kg. It is hinged at one end and rotates about a
vertical axis practically without friction. Find the angular speed of
the door just after the bullet embeds into it.

(Hint:
The moment of inertia of the door about the vertical axis at one end
is ML2/3.)

Mass
of the bullet, m
= 10 g = 10 ×
10–3
kg
Velocity
of the bullet, v
= 500 m/s
Thickness
of the door, L
= 1 m
of the door,
r

= m / 2
Mass
of the door, M
= 12 kg
Angular
momentum imparted by the bullet on the door:
α =
mvr

= (10 × 10-3 ) × (500) × (1/2)  =  2.5 kg m2 s-1    …(i)
Moment
of inertia of the door:
I = ML2 / 3
= (1/3) × 12 × 12 = 4 kgm2
But α = Iω
ω = α / I
= 2.5 / 4
= 0.625 rad s-1

7.25. Two discs of moments of inertia I1 and I2
about their respective axes (normal to the disc and passing through
the centre), and rotating with angular speeds ω1
and ω2 are
brought into contact face to face with their axes of rotation
coincident. (a) What is the angular speed of the two-disc system? (b)
Show that the kinetic energy of the combined system is less than the
sum of the initial kinetic energies of the two discs. How do you
account for this loss in energy? Take ω1
≠ ω2.

(a) Moment of inertia of disc I = I1

Angular speed of disc I = ω1

Moment of inertia of disc I = I2
Angular speed of disc I = ω2
Angular momentum of disc I, L1 = I1ω1
Angular momentum of disc II, L2 = I2ω2
Total initial angular momentum Li = I1ω1 + I2ω2
When the two discs are joined together, their moments of inertia
Moment of inertia of the system of two discs, I = I1 + I2
Let ω be the angular speed of the system.
Total final angular momentum, LT = (I1 + I2) ω
Using the law of conservation of angular momentum, we have:
Li = LT
I1ω1 + I2ω2 = (I1 + I2)ω
ω = (I1ω1 + I2ω2) / (I1I2)

(b) Kinetic energy of disc I, E1 = (1/2) I1ω12
Kinetic energy of disc II, E1 = (1/2) I2ω22
Total initial kinetic energy, Ei = (1/2) ( I1ω12 + I2ω22)
When the discs are joined, their moments of inertia get added up.
Moment of inertia of the system, I = I1 + I2
Angular speed of the system = ω
Final kinetic energy Ef:  =  (1/2) ( I1 + I2) ω2
=  (1/2) ( I1 + I2) [ (I1ω1 + I2ω2) / (I1I2) ]2
= (1/2) (I1ω1 + I2ω2)2 / (I1 + I2)
EiEf
= (1/2) ( I1ω12 + I2ω22) – (1/2) (I1ω1 + I2ω2)2 / (I1 + I2)
Solving the equation, we get
= I1I2 (ω1ω2)2 / 2(I1 + I2)
All the quantities on RHS are positive
EiEf > 0
Ei > Ef
The loss of KE can be attributed to the frictional force that
comes into play when the two discs come in contact with each other.

7.26. (a) Prove
the theorem of perpendicular axes.

(Hint:
Square of the distance of a point (x,
y
)
in the x–y
plane
from an axis through the origin perpendicular to the plane is x2
+
y
2).

(b) Prove
the theorem of parallel axes.

(Hint:
If the centre of mass is chosen to be the origin ∑ miri = 0).

(a) The
theorem of perpendicular axes states that the moment of inertia of a
planar body (lamina) about an axis perpendicular to its plane is
equal to the sum of its moments of inertia about two perpendicular
axes concurrent with perpendicular axis and lying in the plane of the
body.
A
physical body with centre O and a point mass m,in the xy
plane at (x,
y) is
shown in the following figure.

Moment
of inertia about x-axis,
Ix
= mx2

Moment
of inertia about y-axis,
Iy
= my2
Moment
of inertia about z-axis,
Iz
= m(x2 + y2)1/2

Ix
+ Iy
= mx2
+ my2
= m(x2
+ y2)
= m [(x2
+ y2)1/2]1/2
Ix + Iy = Iz
Hence, the theorem is proved.

(b) The
theorem of parallel axes states that the moment of inertia of a body
about any axis is equal to the sum of the moment of inertia of the
body about a parallel axis passing through its centre of mass and the
product of its mass and the square of the distance between the two
parallel axes.

Suppose a rigid body is made up of n particles, having masses m1, m2, m3, … , mn, at perpendicular distances r1, r2, r3, … , rn respectively from the centre of mass O of the rigid body.
The moment of inertia about axis RS passing through the point O:

7.27. Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by v2 = 2gh / [1 + (k2/R2) ]
Using dynamical consideration (i.e. by consideration of forces and torques). Note k is
the radius of gyration of the body about its symmetry axis, and R is
the radius of the body. The body starts from rest at the top of the
plane.

A
body rolling on an inclined plane of height
h,is shown in the following figure:

m
= Mass of the body

R
= Radius of the body
K
= Radius of gyration of the body
v
= Translational velocity of the body
h
=Height
of the inclined plane
g =
Acceleration due to gravity
Total
energy at the top of the plane, E­1=
m
gh
Total
energy at the bottom of the plane, Eb = KErot + KEtrans
= (1/2) I ω2 + (1/2) mv2
But I = mk2 and ω = v / R
Eb = (1/2)(mk2)(v2/R2) + (1/2)mv2
= (1/2)mv2 (1 + k2 / R2)
From the
law of conservation of energy, we have:
ET = Eb
mgh = (1/2)mv2 (1 + k2 / R2)
v = 2gh / (1 + k2 / R2)
Hence,
the given result is proved.

9.28. A disc rotating about its axis with angular speed ωois
placed lightly (without any translational push) on a perfectly
frictionless table. The radius of the disc is R. What are the
linear velocities of the points A, B and C on the disc shown in Fig.
7.41? Will the disc roll in the direction indicated?

vA = Rωo;
vB = Rω0, vc = (R/2)ω0

The disc will not roll
Angular speed of the disc = ωo

Radius of the disc = R
Using the relation for linear velocity, v = ωoR
For point A:
vA = Rωo;
in the direction tangential to the right
For point B:
vB = Rωo;
in the direction tangential to the left
For point C:
vc = (R/2)ωo in
the direction same as that of vA
The directions of motion of points A, B, and C on the disc are
shown in the following figure

Since the disc is placed on a frictionless table, it will not roll. This is because the presence of friction is essential for the rolling of a body.

Page no: 181

7.29. Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated. (a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?

A torque is requiblue to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is requiblue for making the disc roll.

(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.
7.30.A solid disc and a ring, both of radius 10 cm are placed on a
horizontal table simultaneously, with initial angular speed equal to
10 π rad s-1. Which of the two will start to roll
earlier? The co-efficient of kinetic friction is μk
= 0.2.

Radii of the ring and the disc, r = 10 cm = 0.1 m
Initial angular speed, ω0
=10 π rad s–1
Coefficient of kinetic friction, μk = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per
Newton’s second law of motion, we have frictional force, f
= ma

μkmg= ma
Where,
a = Acceleration produced in the objects
m = Mass

∴ a = μkg(i)
As per the first equation of motion, the final velocity of the
objects can be obtained as:
v = u + at
= 0 + μkgt
= μkgt(ii)
The torque applied by the frictional force will act in
perpendicularly outward direction and cause blueuction in the initial
angular speed.
Torque, τ= –
α = Angular acceleration
μkmgr = –

α = –μkmgr / I     …..(iii)
Using the first equation of rotational motion to obtain the final
angular speed:
ω = ω0 + αt
= ω0 + (μkmgr / I)t    ….(iv)
Rolling starts when linear velocity, v = rω
v = r (ω0μkmgrt / I)    …(v)
Equating equations (ii) and (v), we get:
μkgt = r (ω0μkmgrt / I)
= rω0μkmgr2t / I    ….(vi)
For the ring:
I = mr2
μkgt = rω0μkmgr2t / mr2
= rω0μkgt
2μkgt = rω0
t = rω0 / 2μkg
= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8  =  0.80 s    ….(vii)
For the disc: I = (1/2)mr2
μkgt = rω0μkmgr2t / (1/2)mr2
= rω0 – 2μkgt
3μkgt = rω0
t = rω0 / 3μkg
= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8  =  0.53 s   …..(viii)
Since td > tr, the disc
will start rolling before the ring.

7.31. A
cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a
plane of inclination 30°.
The coefficient of static friction µs
= 0.25.

(a) How
much is the force of friction acting on the cylinder?

(b) What
is the work done against friction during rolling?

(c) If
the inclination θ
of
the plane is increased, at what value of θ does
the cylinder begin to skid, and not roll perfectly?

Mass
of the cylinder, m
= 10 kg
of the cylinder, r
= 15 cm = 0.15 m
Co-efficient
of kinetic friction, µk
= 0.25
Angle
of inclination, θ
= 30°
Moment
of inertia of a solid cylinder about its geometric axis, I = (1/2)mr2
The
various forces acting on the cylinder are
shown in the following figure:

The
acceleration of the cylinder is given as:

a = mg Sinθ / [m + (I/r2) ]
= mg Sinθ / [m + {(1/2)mr2/ r2} ]
= (2/3) g Sin 30°
= (2/3) × 9.8 × 0.5  =  3.27 ms-2

(a) Using
Newton’s second law of motion, we can write net force as:
fnet
= ma

mg Sin 30° – f = ma
f = mg Sin 30° – ma
= 10 × 9.8 × 0.5 – 10 × 3.27
49 – 32.7 = 16.3 N

(b) During
rolling, the instantaneous point of contact with the plane comes to
rest. Hence, the work done against frictional force is zero.

(c) For
rolling without skid, we have the relation:
μ = (1/3) tan θ
tan θ = 3μ = 3 × 0.25
θ = tan-1 (0.75) = 36.87°.

7.32. Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.
(b) The instantaneous speed of the point of contact during rolling is zero.
(c) The instantaneous acceleration of the point of contact during rolling is zero.
(d) For perfect rolling motion, work done against friction is zero.
(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

(a) False
Frictional force acts opposite to the direction of motion of
the centre of mass of a body. In the case of rolling, the direction of
motion of the centre of mass is backward. Hence, frictional force acts
in the forward direction.

(b) True
Rolling can be consideblue as the rotation of a body about an
axis passing through the point of contact of the body with the ground.
Hence, its instantaneous speed is zero.

(c) False
This is becausehen a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d) True
This is because once the perfect rolling begins, force of friction becomes zero. Hence work done against friction is zero.

(e) True
This is because rolling occurs only on account of friction which is a tangential force capable of providing torque. When the inclined plane is perfectly smooth, the wheel will simply slip under the effect of its own weight.