System Of Particles And Rotational Motion
Class 11th Physics NCERT Exemplar Solution
- For which of the following does the centre of mass lie outside the body?…
- Which of the following points is the likely position of the centre of mass of the system…
- A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory…
- When a disc rotates with uniform angular velocity, which of the following is not true?…
- A uniform square plate has a small piece Q of an irregular shape removed and glued to the…
- In problem 7.5, the CM of the plate is now in the following quadrant of x-y plane,…
- The density of a non-uniform rod of length 1m is given by ρ(x) = a(1+bx2) where an and b…
- A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with…
- Choose the correct alternatives:
- Figure 7.4 shows two identical particles 1 and 2, each of mass m, moving in opposite…
- The net external torque on a system of particles about an axis is zero. Which of the…
- Figure 7.5 shows a lamina in x-y plane. Two axes z and z′ pass perpendicular to its plane.…
- With reference to Fig. 7.6 of a cube of edge an and mass m, state whether the following…
- The centre of gravity of a body on the earth coincides with its centre of mass for a…
- Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass…
- The variation of angular position θ, of a point on a rotating rigid body, with time t is…
- A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A…
- A uniform sphere of mass m and radius R is placed on a rough horizontal surface (Fig.…
- The vector sum of a system of non-collinear forces acting on a rigid body is given to be…
- A wheel in uniform motion about an axis passing through its centre and perpendicular to…
- A door is hinged at one end and is free to rotate about a vertical axis (Fig. 7.10). Does…
- (n-1) equal point masses each of mass m are placed at the vertices of a regular n-polygon.…
- Find the centre of mass of a uniform A. half-disc, B. quarter-disc.…
- Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc…
- A disc of radius R is rotating with an angular speed ω0 about a horizontal axis. It is…
- Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating…
- A uniform square plate S (side c) and a uniform rectangular plate R (sides b, a) have…
- A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction…
Mcq I
Question 1.For which of the following does the centre of mass lie outside the body?
A. A pencil
B. A shotput
C. A dice
D. A bangle
Answer:
As centre of mass of bangle is in centre of its body which is outside the body of bangle. Hence correct option is d.
Question 2.
Which of the following points is the likely position of the centre of mass of the system shown in Fig. 7.1?
A. A
B. B
C. C
D. D
Answer:
According to the question hollow sphere and sand has equal volume i.e half the volume of sphere. But as
Mass of sand ≫ Mass of air
So, as the mass of sand is quite larger as compared to air. Hence the net centre of mass will be shifted downward (toward sand side).
Hence option c is correct.
Question 3.
A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis at z = a (Fig. 7.2). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is:
A. mva êxx
B. 2mva êx
C. ymv êx
D. 2ymv êx
Answer:
According to the question
Initial velocity
After reflection from the wall
Final velocity 
Trajectory equation for point of intersection is
Hence change in angular momentum is
{ As 
}
Hence option b is correct.
Question 4.
When a disc rotates with uniform angular velocity, which of the following is not true?
A. The sense of rotation remains same.
B. The orientation of the axis of rotation remains same.
C. The speed of rotation is non-zero and remains same.
D. The angular acceleration is non-zero and remains same.
Answer:
As disc is rotating with uniform angular velocity, ω is constant.
Therefore
Angular acceleration 
(Differentiation of constant is 0)
= 0
So, its angular acceleration is zero.
Hence, option d is correct answer.
Question 5.
A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind (Fig. 7.3). The moment of inertia about the z-axis is then
A. increased
B. decreased
C. the same
D. changed in unpredicted manner
Answer:
As the piece Q is removed and glued to centre, the moment of inertia decreases about z-axis as the
M.I
As the distance of that piece from z-axis is decreased to zero, hence its M.I will decrease. Hence, correct option is b.
Question 6.
In problem 7.5, the CM of the plate is now in the following quadrant of x-y plane,
A. I
B. II
C. III
D. IV
Answer:
According to problem as the mass at Q is displaced to initial centre of mass so new centre will shift toward other side of Q.
As more mass is now present in quadrant 3 than 1.
Hence correct option is C.
Question 7.
The density of a non-uniform rod of length 1m is given by ρ(x) = a(1+bx2) where an and b are constants and 0≤x≤1. The centre of mass of the rod will be at
A. 
B. 
C. 
D. 
Answer:
Given: 
As we know that
Centre of mass 
Question 8.
A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω. A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is
A. 2ω
B. ω
C. ω/2
D. 0
Answer:
As there is no external torque as jump is radially away.
Applying conservation of angular momentum
As the person jump with angular velocity w. Hence its angular momentum at that time is 
. So final angular momentum is for the person.
So
Hence option b is correct.
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Question 1.Choose the correct alternatives:
A. For a general rotational motion, angular momentum L and angular velocity ω need not be parallel.
B. For a rotational motion about a fixed axis, angular momentum L and angular velocity ω are always parallel.
C. For a general translational motion, momentum p and velocity v are always parallel.
D. For a general translational motion, acceleration an and velocity v are always parallel.
Answer:
For a general rotational motion angular momentum L and angular velocity ω are not parallel when the axis of rotation is not symmetric As 
and I is not a scalar quantity.
Whereas is translational motion, momentum 
, where m is scalar so direction of p and v are always same.
Hence correct option is A. and C.
Question 2.
Figure 7.4 shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant, r1 and r2 are their respective position vectors drawn from point A which is in the plane of the parallel lines. Choose the correct options:
A. Angular momentum l1 of particle 1 about A is l1 = mvd1
B. Angular momentum l2 of particle 2 about A is l2 = mvr2
C. Total angular momentum of the system about A is l = mv (r1 + r2)
D. Total angular momentum of the system about A is l = mv (d2 – d1) ⊗ represents a unit vector coming out of the page. ⊗ represents a unit vector going into the page.
Answer:
As we know that
And direction of L is perpendicular to both the vector 
According to question,
Angular momentum for particle1 about A is
(upward from right hand thumb rule)
Similarly,
(taking upward as positive)
Total angular momentum
(taking upward as positive)
(taking downward as positive)
Hence option an and d are correct.
Question 3.
The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it?
A. The forces may be acting radially from a point on the axis.
B. The forces may be acting on the axis of rotation.
C. The forces may be acting parallel to the axis of rotation.
D. The torque caused by some forces may be equal and opposite to that caused by other forces.
Answer:
As we know that
When forces are acting radially then θ=0. So, 
When forces are acting on axis of rotation then r=0. So,
When forces are acting parallel to axis of rotation then its component in plane r and F is 0. Hence F=0. So,
When torques are equal and opposite then net torque is zero.
Hence correct options are a, b, c, d.
Question 4.
Figure 7.5 shows a lamina in x-y plane. Two axes z and z′ pass perpendicular to its plane. A force F acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to z′-axis than the z-axis.)
A. Torque τ caused by F about z axis is along -
.
B. Torque τ′ caused by F about z′ axis is along -.
C. Torque τ caused by F about z axis is greater in magnitude than that about z axis.
D. Total torque is given be τ = τ + τ′.
Answer:
As we know that
And direction of z is perpendicular to plane of r and F by right hand thumb rule.
a) Torque caused by force F about z-axis is in +k direction by right hand thumb rule.
b) Similarly Torque caused by force F about z’-axis is in –k direction.
c) r>r’ and θ>θ’
So, rFsinθ > r’Fsinθ’
Hence 
d) There is no sense in adding torques about 2 different axes.
Question 5.
With reference to Fig. 7.6 of a cube of edge an and mass m, state whether the following are true or false. (O is the centre of the cube.)
A. The moment of inertia of cube about z-axis is Iz = Ix + Iy
B. The moment of inertia of cube about z′ is I’ z = I 2 + 
C. The moment of inertia of cube about z″ is = I 2 + 
D. I x = I y
Answer:
Option a is false since perpendicular axes theorem is applicable only for laminar/planar objects.
Distance between z and z’
Applying parallel axis theorem
So option b is correct.
Z’’ and z are skew lines. That is, they are neither parallel nor perpendicular. So option c is incorrect as it is using parallel axis theorem.
Since the cube is symmetric, moment of inertia about x and y axis are equal. So, correct options are b and d.
Vsa
Question 1.The centre of gravity of a body on the earth coincides with its centre of mass for a ‘small’ object whereas for an ‘extended’ object it may not. What is the qualitative meaning of ‘small’ and ‘extended’ in this regard? For which of the following the two coincides? A building, a pond, a lake, a mountain?
Answer:
Centre of gravity is the point through which the resultant of system of forces (due to weight) of all the particles constituting the body passes for all positions of body.
Centre of mass is a point where whole mass of a body can be concentrated.
When the vertical height of the object is very small as compared to earth’s radius, we call the object small otherwise it is extended.
a) Building and pond are small object so their centre of mass and centre of gravity coincide.
b) Deep lake and mountain are extended objects so their centre of mass doe of mass does not coincide with centre of gravity.
Question 2.
Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?
Answer:
As we know that
And all the mass in cylinder lies a distance R from axis of symmetry but most of the mass of solid sphere lies at smaller distance than R.
Therefore
Question 3.
The variation of angular position θ, of a point on a rotating rigid body, with time t is shown in Fig. 7.7. Is the body rotating clock-wise or anti-clockwise?
Answer:
According to question slope of θ-t graph is positive, i.e
And therefore slope indicate anti-clockwise which is traditionally taken as positive.
Question 4.
A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in Fig. 7.8. Match the following (most appropriate choice):
A. mg/4 < F mg < /2 (i) Cube will move up.
B. F > mg/2 (ii) Cube will not exhibit motion.
C. F > mg (iii) Cube will begin to rotate and slip at A.
D. F = mg/4 (iv) Normal reaction effectively at a/3 from A, no motion.
Answer:
From the problem
Cube will not move if 
So a-(ii)
Cube will rotate if 
So b-(iii)
If F>mg
Then cube will move up.
So, c-(i)
When the normal is a/3 from A.
So, for no rotation we balance the torque
Balancing force
Solving the above two equations
So, d-(iv)
Question 5.
A uniform sphere of mass m and radius R is placed on a rough horizontal surface (Fig. 7.9). The sphere is struck horizontally at a height h from the floor. Match the following:
A. h = R/2 (i) Sphere rolls without slipping with a constant velocity and no loss of energy.
B. h = R (ii) Sphere spins clockwise, loses energy by friction.
C. h = 3R/2 (iii) Sphere spins anti-clockwise, loses energy by friction.
D. h = 7R/5 (iv) Sphere has only a translational motion, loses energy by friction.
Answer:
Applying conservation of angular momentum just before and after collision.
And as there is rolling without slipping
So, sphere will roll with slipping (no loss of energy) when
So, d-(i)
Torque due to applied force about centre of mass
If h=R then 
Therefore, sphere will only have translational motion and loose energy by friction.
So, b-(iv)
If h>R, then torque is positive. Hence sphere will spin clockwise.
So, c-(ii)
If h<R, then torque is negative. Hence sphere will spin anticlockwise.
So, a-(iii)
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Question 1.The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?
Answer:
No,
The sum of all the torques about the given certain point o is zero.
So,
Now, for any other arbitrary point sum of torques can be written as
In the above expression the second term may or may not vanish.
Question 2.
A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium? How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?
Answer:
As the system is the symmetrical system, the centripetal acceleration in a wheel arise due to internal elastic which exists in pair and cancel each other.
In the second part of question, in half wheel the distribution of mass about centre of mass is not symmetrical.
Therefore, the direction of angular momentum doesn’t coincide with the direction of angular velocity and hence an external torque is required to maintain rotation.
Question 3.
A door is hinged at one end and is free to rotate about a vertical axis (Fig. 7.10). Does its weight cause any torque about this axis? Give reason for your answer.
Answer:
No,
As we know that
So force can be produce torque only along the direction normal to itself. So, when the door in x-y plane, the torque produced by gravity can only be along be positive or negative z direction, never about an axis passing through y direction.
Question 4.
(n-1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass.
Answer:
As we know that position of centre of mass is
Since the centre of mass of regular n-polygon lies in the geometrical centre.
And n-1 equal point masses each of mass m are placed at n-1 vertices of n polygon. Therefore, it’s centre of mass will be
where b = distance of n-1 masses from centre of mass
Negative sign indicates that 
is in direction opposite to 
Question 5.
Find the centre of mass of a uniform A. half-disc, B. quarter-disc.
Answer:
a) Let M be the mass of half disc and R be its radius.
Mass per unit length 
Let cut a semi-circular ring out of this semi disc at distance r and r+dr from centre.
So, Area of element
Mass of elementary ring
For this semi-circular ring centre of mass is 
So,
Hence centre of mass of semi-circular disc
b) Let M be the mass of quarter disc and R be its radius.
Mass per unit length 
Similarly, as above part
Area of element
Mass of elementary ring
For this semi-circular ring centre of mass is 
So,
Hence centre of mass of Quarter disc
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Question 1.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 speed ω1 and ω2 are brought into contact face to face with their axes of rotation coincident.
A. Does the law of conservation of angular momentum apply to the situation? why?
B. Find the angular speed of the two-disc system.
C. Calculate the loss in kinetic energy of the system in the process.
D. Account for this loss.
Answer:
a) Yes, since there is no external torque on system.
All external forces act through axis of rotation, hence produce no torque.
b) When two of them coincide let ω be their common angular velocity.
Applying conservation of angular momentum
Final angular momentum= Initial angular momentum
c) 
Change in kinetic energy= 
<0
d)The loss of kinetic energy is due to work against friction between two discs.
Question 2.
A disc of radius R is rotating with an angular speed ω0 about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is μk.
A. What was the velocity of its centre of mass before being brought in contact with the table?
B. What happens to the linear velocity of a point on its rim when placed in contact with the table?
C. What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
D. Which force is responsible for the effects in B. and C.
(e) What condition should be satisfied for rolling to begin?
(f) Calculate the time taken for the rolling to begin.
Answer:
a) As from the question the disc is rotating about its horizontal axis before coming in contact.
Hence it’s 
b) Linear velocity of point at rim decreases due to force of friction.
c) Linear speed of centre of mas of disc increases due to acceleration gained by it due to friction.
d) Friction is responsible for effects in b and C.
e) Rolling starts when 
where ω is angular speed of disc.
f) Acceleration produced in centre of mass due to friction
Angular acceleration produced by torque due to friction
As we know that
For rolling without slipping
So, 
Question 3.
Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities ω (anti-clockwise) and ω (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by (3 R + δ). They are now brought in contact (δ → 0).
A. Show the frictional forces just after contact.
B. Identify forces and torques external to the system just after contact.
C. What would be the ratio of final angular velocities when friction ceases?
Answer:
Here F and F’ are external forces through support.
Let ω1 and ω2 be final angular velocities (anticlockwise and clockwise respectively)
Since there will be no friction both drum has same linear velocity.
Hence, 
So, 
Question 4.
A uniform square plate S (side c) and a uniform rectangular plate R (sides b, a) have identical areas and masses (Fig. 7.11).
Show that
(i) I xr / Ixs < 1; (ii) Iyr/ Iys > 1; (iii) I2R/ I2s > 1.
Answer:
According to question
Area of square=Area of rectangular plate
a) As 
(as from diagram b<c)
b) 
(as from diagram a>c)
c) 
So, 
Question 5.
A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is μ (Fig 7.12). Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?
Answer:
According to question, let
f = force of friction
F = applied force
a = acceleration produced
So,
Solving above equations
We get 
And 
Therefore 
