Physics Notes - System of Particles and Rotational Motion
Centre of Mass
The centre of mass of a system is the point where the entire mass is concentrated and all external forces act.
For rigid bodies, the centre of mass is independent of the body's state (rest or accelerated motion).
For a system of n particles with masses m1, m2, m3, … mn and position vectors r1, r2, r3, … rn, the position vector of the centre of mass is given by:
r{cm} = (m1r1 + m2r2 + m3r3 + … + mnrn) / (m1 + m2 + m3 + … + m_n)For a two-particle system:
Position of centre of mass from m1 = (m2d) / (m1 + m2)
Position of centre of mass from m2 = (m1d) / (m1 + m2)
If position vectors of particles of masses m1 and m2 are r1 and r2 respectively, then
If, in a two-particle system, particles of masses m1 and m2 are moving with velocities v1 and v2 respectively, then the velocity of the centre of mass is:
If the accelerations of the particles are a1 and a2 respectively, then the acceleration of the centre of mass is:
The centre of mass of an isolated system has a constant velocity.
An isolated system will remain at rest if initially at rest, or move with the same velocity if initially in motion.
The position of the centre of mass depends on the shape, size, and mass distribution of the body.
The centre of mass of an object need not lie within the object.
In symmetrical bodies with homogeneous mass distribution, the centre of mass coincides with the geometrical centre.
The position of the centre of mass of an object changes in translatory motion but remains unchanged in rotatory motion.
Translational Motion
A rigid body performs pure translational motion if each particle of the body undergoes the same displacement in the same direction in a given interval of time.
Rotational Motion
A rigid body performs pure rotational motion if each particle of the body moves in a circle, and the centre of all the circles lies on a straight line called the axis of rotation.
Rigid Body
If the relative distance between the particles of a system does not change upon applying force, then it is called a rigid body.
The general motion of a rigid body consists of both translational and rotational motion.
Moment of Inertia
The inertia of rotational motion is called the moment of inertia, denoted by I.
Moment of inertia is the property of an object by virtue of which it opposes any change in its state of rotation about an axis.
The moment of inertia of a body about a given axis is equal to the sum of the products of the masses of its constituent particles and the square of their respective distances from the axis of rotation.
Its unit is kg \\cdot m^2 and its dimensional formula is [ML^2].
The moment of inertia of a body depends upon:
Position of the axis of rotation
Orientation of the axis of rotation
Shape and size of the body
Distribution of mass of the body about the axis of rotation
The physical significance of the moment of inertia is the same in rotational motion as mass in linear motion.
Radius of Gyration
The root mean square distance of its constituent particles from the axis of rotation is called the radius of gyration of a body, denoted by K.
The product of the mass of the body (M) and square of its radius of gyration (K) gives the same moment of inertia of the body about the rotational axis.
Therefore, moment of inertia I = MK^2, which implies K = \sqrt{I/M}.
Parallel Axes Theorem
The moment of inertia of any object about any arbitrary axis is equal to the sum of the moment of inertia about a parallel axis passing through the centre of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.
Mathematically, I = I{CM} + Mr^2, where I is the moment of inertia about the arbitrary axis, I{CM} is the moment of inertia about the parallel axis through the centre of mass, M is the total mass of the object, and r is the perpendicular distance between the axes.
Perpendicular Axes Theorem
The moment of inertia of any two-dimensional body about an axis perpendicular to its plane (I_z) is equal to the sum of the moments of inertia of the body about two mutually perpendicular axes lying in its own plane and intersecting each other at a point where the perpendicular axis passes through it.
Mathematically, Iz = Ix + Iy, where Ix and I_y are the moments of inertia of the plane lamina about perpendicular axes X and Y respectively, which lie in the plane lamina and intersect each other.
The theorem of parallel axes is applicable for any type of rigid body whether it is two-dimensional or three-dimensional, while the theorem of perpendicular axes is applicable for laminar-type or two-dimensional bodies only.
Moment of Inertia of Homogeneous Rigid Bodies
Thin Circular Ring
About an axis passing through its center and perpendicular to its plane: I = MR^2
About a tangent perpendicular to its plane: I = 3/2MR^2
About a tangent in its plane: I = 5/4MR^2
About a diameter: I = 1/2MR^2
Thin Rod
About an axis passing through its center and perpendicular to its length: I = ML^2/12
About an axis passing through one end and perpendicular to its length: I = ML^2/3
Solid Cylinder
About its geometrical axis: I = MR^2/2
About an axis passing through its outer face along its length: I = MR^2
About an axis passing through its center and perpendicular to its length: I = M(L^2/12 + R^2/4)
About an axis passing through its diameter of circular surface: I= M L^2/3 + MR^2/4
Rectangular Plate
About an axis passing through its centre and perpendicular to its plane: I = M(a^2 + b^2)/12
Thin Spherical Shell
About its any diameter: I = 2/3MR^2
About its any tangent: I = 5/3MR^2
Solid Sphere
About its diameter: I = 2/5MR^2
Equations of Rotational Motion
\omega = \omega_0 + \alpha t
\theta = \omega_0 t + 1/2 \alpha t^2
\omega^2 = \omega0^2 + 2\alpha \theta where \theta is the displacement in rotational motion, \omega0 is the initial velocity, \\omega is the final velocity, and \alpha is the acceleration.
Torque
Torque or moment of a force about the axis of rotation is given by: \tau = r \times F = rFsin\theta \hat{n}
It is a vector quantity.
If the nature of the force is to rotate the object clockwise, then torque is called negative, and if it rotates the object anticlockwise, then it is called positive.
Its SI unit is ‘newton-metre’ and its dimension is [ML^2T^{-2}].
In rotational motion, torque, \tau = I\alpha, where \alpha is angular acceleration and I is the moment of inertia.
Angular Momentum
The moment of linear momentum is called angular momentum, denoted by L.
Angular momentum, L = I\omega = mvr
In vector form, L = I\omega = r \times mv
Its unit is ‘joule-second’ and its dimensional formula is [ML^2T^{-1}].
Torque, \tau = dL/dt
Conservation of Angular Momentum
If the external torque acting on a system is zero, then its angular momentum remains conserved.
If \tau{ext} = 0, then L = I\omega = constant \Rightarrow I1\omega1 = I2\omega_2
Angular Impulse
The total effect of a torque applied on a rotating body in a given time is called angular impulse.
Angular impulse is equal to the total change in the angular momentum of the system in a given time.
Thus, angular impulse = change in angular momentum.
MCQ I
7.1 For which of the following does the centre of mass lie outside the body ? (d) A bangle
7.2 Which of the following points is the likely position of the centre of mass of the system shown in Fig. 7.1? (b) B
7.3 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 atz = a (Fig. 7.2). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is: (b) 2mva êx
7.4 When a disc rotates with uniform angular velocity, which of the following is not true? (d) The angular acceleration is non-zero and remains same.
7.5 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 (b) decreased
7.6 In problem 7.5, the CM of the plate is now in the following quadrant of x-y plane, (a) I
7.7 The density of a non-uniform rod of length 1m is given by (x) = a(1+bx2) where a and b are constants and o x 1. The centre of mass of the rod will be at (a) 4(3 + b) / 3(2 + b)
7.8 A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed . 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 (asseen from the round). The speed of the round afterwards is (c) 2
MCQ II
7.9 Choose the correct alternatives: ((a) (b) (c))
For a general rotational motion, angular momentum L and angular velocity need not be parallel.
For a rotational motion about a fixed axis, angular momentum L and angular velocity are always parallel.
For a general translational motion , momentum p and velocity v are always parallel.
7.10 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) (c))
Angular momentum l of particle 1 about A is l = mvd 1v
Total angular momentum of the system about A is l = mv(r + r )
7.11 The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it ? ((a) (b) (c) (d))
The forces may be acting radially from a point on the axis.
The forces may be acting on the axis of rotation.
The forces may be acting parallel to the axis of rotation.
The torque caused by some forces may be equal and opposite to that caused by other forces.
7.12 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? ((a) (b))
Torque caused by F about z axisis along -k ˆ .
Torque caused by F about z axis is along -k ˆ .
7.13 With reference to Fig. 7.6 of a cube of edge a and mass m,state whether the following are true or false. ((a) (c) (d))
The moment of inertia of cube about z-axis is I z = I x + I y
The moment of inertia of cube about z is m a 2 /2 = I z + m a 2 / 2
I x = I y
VSA
7.14 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?
Small means that the object's dimensions are small enough that the gravitational field can be considered uniform across the object. Extended means that the object is large enough that the gravitational field varies noticeably across it.
The centre of gravity and centre of mass coincide for: a pond, a lake
7.15 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?
Because in a solid sphere, more of the mass is concentrated closer to the axis of rotation compared to a hollow cylinder where all the mass is at the radius.
7.16 The variation of angular position , 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?
The body is rotating clockwise.
Short Answer
7.17 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 - (iv) Normal reaction effectively at a/3 from A, no motion.
(b) F > mg/2 - (iii) Cube will begin to rotate and slip at A.
(c) F > mg - (i) Cube will move up.
(d) F = mg/4 - (ii) Cube will not exhibit motion.
7.18 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 - (iii) Sphere spins anti-clockwise, loses energy by friction.
(b) h = R - (iv) Sphere has only a translational motion, looses energy by friction.
(c) h = 3R/2 - (ii) Sphere spins clockwise, loses energy by friction.
(d) h = 7R/5 - (i) Sphere rolls without slipping with a constant velocity and no loss of energy.
7.19 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?
No, it doesn't mean that it is necessary zero about any arbitrary point. The torques will only be zero if all the forces are acting along the same line.
7.20 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?
The wheel is in equilibrium because the net external force and net external torque are zero. The equilibrium of the wheel as a whole does no imply that each particle is in equilibrium. To set a half-wheel into uniform motion you would require external forces to sustain the wheel.
7.21 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.
No, the weight of the door acts through the centre of mass, which lies on the axis of rotation. Therefore, the torque due to the weight about the hinge is zero. The force of gravity is balanced by the reaction force at the hinge.
Long Answer
7.22 (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.
r_{cm} = -a/(n-1)
7.23 Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.
(a) half-disc: r_{cm} = 4R/(3\pi)
(b) quarter-disc: r_{cm} = (4R/(3\pi), 4R/(3\pi))
7.24 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 1 and 2 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?
Yes, since not net external torque acts in the situation the total angular momentum is conserved.
(b) Find the angularspeed of the two-disc system.
\omega = (I1 \omega1 + I2 \omega2) / (I1 + I2)
(c) Calculate the loss in kinetic energy of the system in the process.
KE{loss} = (I1 I2) / (2(I1 + I2) (\omega1 - \omega_2)^2
(d) Account for thisloss.
The loss in kinetic energy is due to the work done to the heat generated by the friction between the two disks.
7.25 A disc of radius R is rotating with an angular speed o about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is k.
(a) What wasthe velocity of its centre of mass before being brought in contact with the table?
Zero, as it’s only rotating.
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?
The linear velocity decreases due to friction.
(c) What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
The linear speed increases from zero until rolling begins.
(d) Which force is responsible for the effects in (b) and (c).
Kinetic friction is responsible.
(e) What condition should be satisfied for rolling to begin?
v = R\omega
(f) Calculate the time taken for the rolling to begin.
t = R\omega0 / (3\muk g)
7.26 Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities (anti-clockwise) and (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by (3R + ) . They are now brought in contact ( → 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?
The ratio of final angular velocities is \omega1 / \omega2 = 2
7.27 A uniform square plate S (side c) and a uniform rectangular plate R (sides b, a) have identical areas and masses (Fig. 7.11).
7.28 A uniform disc of radius R, is resting on a table on its rim.The coefficient of friction between disc and table is (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 ?
F_{max} = 1/3 \mu mg
Centre of Mass
The centre of mass of a system is the point where the entire mass is concentrated and all external forces act. It's a crucial concept in physics for analyzing the motion of complex objects.
For rigid bodies, the centre of mass is independent of the body's state (rest or accelerated motion). This means whether the object is stationary or moving, its center of mass remains the same relative to the object.
For a system of n particles with masses m1, m2, m3, … mn and position vectors r1, r2, r3, … rn, the position vector of the centre of mass is given by:
r{cm} = (m1r1 + m2r2 + m3r3 + … + mnrn) / (m1 + m2 + m3 + … + m_n)
This formula calculates the weighted average of the positions of all particles, where the weights are their masses.
For a two-particle system:
Position of centre of mass from m1 = (m2d) / (m1 + m2)
Position of centre of mass from m2 = (m1d) / (m1 + m2)
These equations give the location of the center of mass relative to each particle in a two-particle system, where d is the distance between the particles.
If position vectors of particles of masses m1 and m2 are r1 and r2 respectively, then
If, in a two-particle system, particles of masses m1 and m2 are moving with velocities v1 and v2 respectively, then the velocity of the centre of mass is:
If the accelerations of the particles are a1 and a2 respectively, then the acceleration of the centre of mass is:
The centre of mass of an isolated system has a constant velocity.
An isolated system will remain at rest if initially at rest, or move with the same velocity if initially in motion. This is a direct consequence of Newton's first law.
The position of the centre of mass depends on the shape, size, and mass distribution of the body.
The centre of mass of an object need not lie within the object. A classic example is a ring, where the center of mass is at the center of the ring, which is empty space.
In symmetrical bodies with homogeneous mass distribution, the centre of mass coincides with the geometrical centre. Examples include a perfect sphere or a uniform cube.
The position of the centre of mass of an object changes in translatory motion but remains unchanged in rotatory motion.
Translational Motion
A rigid body performs pure translational motion if each particle of the body undergoes the same displacement in the same direction in a given interval of time. Imagine a box sliding down a ramp without rotating.
Rotational Motion
A rigid body performs pure rotational motion if each particle of the body moves in a circle, and the centre of all the circles lies on a straight line called the axis of rotation. Think of a spinning top.
Rigid Body
If the relative distance between the particles of a system does not change upon applying force, then it is called a rigid body. This is an idealization, as all real bodies deform to some extent.
The general motion of a rigid body consists of both translational and rotational motion. A ball rolling down a hill is an example of combined translational and rotational motion.
Moment of Inertia
The inertia of rotational motion is called the moment of inertia, denoted by I.
Moment of inertia is the property of an object by virtue of which it opposes any change in its state of rotation about an axis.
The moment of inertia of a body about a given axis is equal to the sum of the products of the masses of its constituent particles and the square of their respective distances from the axis of rotation.
Its unit is kg cdot m^2 and its dimensional formula is [ML^2].
The moment of inertia of a body depends upon:
Position of the axis of rotation
Orientation of the axis of rotation
Shape and size of the body
Distribution of mass of the body about the axis of rotation
The physical significance of the moment of inertia is the same in rotational motion as mass in linear motion. It measures the resistance of an object to changes in its rotational speed.
Radius of Gyration
The root mean square distance of its constituent particles from the axis of rotation is called the radius of gyration of a body, denoted by K.
The product of the mass of the body (M) and square of its radius of gyration (K) gives the same moment of inertia of the body about the rotational axis.
Therefore, moment of inertia I = MK^2, which implies K = \sqrt{I/M}.
Parallel Axes Theorem
The moment of inertia of any object about any arbitrary axis is equal to the sum of the moment of inertia about a parallel axis passing through the centre of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.
Mathematically, I = I{CM} + Mr^2, where I is the moment of inertia about the arbitrary axis, I{CM} is the moment of inertia about the parallel axis through the centre of mass, M is the total mass of the object, and r is the perpendicular distance between the axes.
Perpendicular Axes Theorem
The moment of inertia of any two-dimensional body about an axis perpendicular to its plane (I_z) is equal to the sum of the moments of inertia of the body about two mutually perpendicular axes lying in its own plane and intersecting each other at a point where the perpendicular axis passes through it.
Mathematically, Iz = Ix + Iy, where Ix and I_y are the moments of inertia of the plane lamina about perpendicular axes X and Y respectively, which lie in the plane lamina and intersect each other.
The theorem of parallel axes is applicable for any type of rigid body whether it is two-dimensional or three-dimensional, while the theorem of perpendicular axes is applicable for laminar-type or two-dimensional bodies only.
Moment of Inertia of Homogeneous Rigid Bodies
Thin Circular Ring
About an axis passing through its center and perpendicular to its plane: I = MR^2
About a tangent perpendicular to its plane: I = 3/2MR^2
About a tangent in its plane: I = 5/4MR^2
About a diameter: I = 1/2MR^2
Thin Rod
About an axis passing through its center and perpendicular to its length: I = ML^2/12
About an axis passing through one end and perpendicular to its length: I = ML^2/3
Solid Cylinder
About its geometrical axis: I = MR^2/2
About an axis passing through its outer face along its length: I = MR^2
About an axis passing through its center and perpendicular to its length: I = M(L^2/12 + R^2/4)
About an axis passing through its diameter of circular surface: I= M L^2/3 + MR^2/4
Rectangular Plate
About an axis passing through its centre and perpendicular to its plane: I = M(a^2 + b^2)/12
Thin Spherical Shell
About its any diameter: I = 2/3MR^2
About its any tangent: I = 5/3MR^2
Solid Sphere
About its diameter: I = 2/5MR^2
Equations of Rotational Motion
\omega = \omega_0 + \alpha t
\theta = \omega_0 t + 1/2 \alpha t^2
\omega^2 = \omega0^2 + 2\alpha \theta
where \theta is the displacement in rotational motion, \omega0 is the initial velocity, \omega is the final velocity, and \alpha is the acceleration. These equations are analogous to the equations of linear motion.
Torque
Torque or moment of a force about the axis of rotation is given by: \tau = r \times F = rFsin\theta \hat{n}
It is a vector quantity. Torque has both magnitude and direction.
If the nature of the force is to rotate the object clockwise, then torque is called negative, and if it rotates the object anticlockwise, then it is called positive. The sign convention depends on the direction of rotation.
Its SI unit is ‘newton-metre’ and its dimension is [ML^2T^{-2}].
In rotational motion, torque, \tau = I\alpha, where \alpha is angular acceleration and I is the moment of inertia. This equation is analogous to Newton's second law in linear motion, F = ma.
Angular Momentum
The moment of linear momentum is called angular momentum, denoted by L.
Angular momentum, L = I\omega = mvr
In vector form, L = I\omega = r \times mv
Its unit is ‘joule-second’ and its dimensional formula is [ML^2T^{-1}].
Torque, \tau = dL/dt This equation relates the torque to the rate of change of angular momentum.
Conservation of Angular Momentum
If the external torque acting on a system is zero, then its angular momentum remains conserved.
If \tau{ext} = 0, then L = I\omega = constant \Rightarrow I1\omega1 = I2\omega_2
Angular Impulse
The total effect of a torque applied on a rotating body in a given time is called angular impulse.
Angular impulse is equal to the total change in the angular momentum of the system in a given time.
Thus, angular impulse = change in angular momentum.
MCQ I
7.1 For which of the following does the centre of mass lie outside the body ? (d) A bangle
7.2 Which of the following points is the likely position of the centre of mass of the system shown in Fig. 7.1? (b) B
7.3 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 atz = a (Fig. 7.2). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is: (b) 2mva êx
7.4 When a disc rotates with uniform angular velocity, which of the following is not true? (d) The angular acceleration is non-zero and remains same.
7.5 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 (b) decreased
7.6 In problem 7.5, the CM of the plate is now in the following quadrant of x-y plane, (a) I
7.7 The density of a non-uniform rod of length 1m is given by (x) = a(1+bx2) where a and b are constants and o x 1. The centre of mass of the rod will be at (a) 4(3 + b) / 3(2 + b)
7.8 A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed . 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 (asseen from the round). The speed of the round afterwards is (c) 2
MCQ II
7.9 Choose the correct alternatives: ((a) (b) (c))
For a general rotational motion, angular momentum L and angular velocity need not be parallel.
For a rotational motion about a fixed axis, angular momentum L and angular velocity are always parallel.
For a general translational motion , momentum p and velocity v are always parallel.
7.10 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) (c))
Angular momentum l of particle 1 about A is l = mvd 1v
Total angular momentum of the system about A is l = mv(r + r )
7.11 The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it ? ((a) (b) (c) (d))
The forces may be acting radially from a point on the axis.
The forces may be acting on the axis of rotation.
The forces may be acting parallel to the axis of rotation.
The torque caused by some forces may be equal and opposite to that caused by other forces.
7.12 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? ((a) (b))
Torque caused by F about z axisis along -k ˆ .
Torque caused by F about z axis is along -k ˆ .
7.13 With reference to Fig. 7.6 of a cube of edge a and mass m,state whether the following are true or false. ((a) (c) (d))
The moment of inertia of cube about z-axis is I z = I x + I y
The moment of inertia of cube about z is m a 2 /2 = I z + m a 2 / 2
I x = I y
VSA
7.14 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?
Small means that the object's dimensions are small enough that the gravitational field can be considered uniform across the object. Extended means that the object is large enough that the gravitational field varies noticeably across it.
The centre of gravity and centre of mass coincide for: a pond, a lake
7.15 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?
Because in a solid sphere, more of the mass is concentrated closer to the axis of rotation compared to a hollow cylinder where all the mass is at the radius.
7.16 The variation of angular position , 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?
The body is rotating clockwise.
Short Answer
7.17 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 - (iv) Normal reaction effectively at a/3 from A, no motion.
(b) F > mg/2 - (iii) Cube will begin to rotate and slip at A.
(c) F > mg - (i) Cube will move up.
(d) F = mg/4 - (ii) Cube will not exhibit motion.
7.18 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 - (iii) Sphere spins anti-clockwise, loses energy by friction.
(b) h = R - (iv) Sphere has only a translational motion, looses energy by friction.
(c) h = 3R/2 - (ii) Sphere spins clockwise, loses energy by friction.
(d) h = 7R/5 - (i) Sphere rolls without slipping with a constant velocity and no loss of energy.
7.19 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?
No, it doesn't mean that it is necessary zero about any arbitrary point. The torques will only be zero if all the forces are acting along the same line.
7.20 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?
The wheel is in equilibrium because the net external force and net external torque are zero. The equilibrium of the wheel as a whole does no imply that each particle is in equilibrium. To set a half-wheel into uniform motion you would require external forces to sustain the wheel.
7.21 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.
No, the weight of the door acts through the centre of mass, which lies on the axis of rotation. Therefore, the torque due to the weight about the hinge is zero. The force of gravity is balanced by the reaction force at the hinge.
Long Answer
7.22 (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.
r_{cm} = -a/(n-1)
7.23 Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.
(a) half-disc: r_{cm} = 4R/(3\pi)
(b) quarter-disc: r_{cm} = (4R/(3\pi), 4R/(3\pi))
7.24 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 1 and 2 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?
Yes, since not net external torque acts in the situation the total angular momentum is conserved.
(b) Find the angularspeed of the two-disc system.
\omega = (I1 \omega1 + I2 \omega2) / (I1 + I2)
(c) Calculate the loss in kinetic energy of the system in the process.
$$KE{loss} = (I1 I2) / (2(I1 + I2) (