NCERT Exemplar: System of Particles & Rotational Motion | Physics Class 11 - NEET PDF Download

Multiple Choice Questions I

Q.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
Ans. (d)
Solution.

A bangle (a thin ring) has its mass distributed at approximately the same distance from its geometrical centre. The centre of mass of a thin ring lies at its geometrical centre, which is in empty space - that is, outside the material of the bangle. Hence option (d) is correct.

Q.2. Which of the following points is the likely position of the centre of mass of the system shown in Figure ?

(a) A
 (b) B
 (c) C
 (d) D
 Ans. (c)

Solution.

The centre of mass (CM) of a composite system lies closer to the heavier portion of the system. In the figure the lower hemisphere (or the lower part of the sphere) contains sand and is heavier than the upper part (containing air). Therefore the CM shifts below the horizontal diameter and the likely position is C.

Q.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 (Figure). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is:

(a) mva ê_x
(b) 2mva ê_x
(c) ymv ê_x
(d) 2ymv ê_x
 Ans. (b)

Solution.

Angular momentum of a particle about the origin is L = r × p, where p = m v.

The particle moves in the yz-plane, its position vector can be written as r = 0 ê_x + y ê_y + a ê_z, with the trajectory parallel to y-axis at z = a.

Initial linear momentum is p_i = m v ê_y (since motion is along +y before collision).

Initial angular momentum about origin is L_i = r × p_i = (y ê_y + a ê_z) × (m v ê_y) = a m v (ê_z × ê_y) + y m v (ê_y × ê_y).

The term with ê_y × ê_y is zero. Using ê_z × ê_y = -ê_x, we get L_i = - a m v ê_x.

After an elastic bounce from the wall at fixed y the y-component of velocity reverses sign: p_f = - m v ê_y. The position vector r at the instant of collision is unchanged, so

L_f = r × p_f = (y ê_y + a ê_z) × (- m v ê_y) = + a m v ê_x.

Change in angular momentum ΔL = L_f - L_i = a m v ê_x - (- a m v ê_x) = 2 a m v ê_x.

Thus option (b) is correct.

Q.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.
 Ans. (d)
Solution.

Angular acceleration α is defined as the rate of change of angular velocity: α = Δω / Δt.

If the disc rotates with constant angular velocity ω (uniform rotation), then Δω = 0 and therefore α = 0. Thus statement (d) is false; angular acceleration is zero for uniform rotation. The other statements (a), (b) and (c) are consistent with uniform rotation (orientation and sense fixed, speed constant).

Q.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 (Figure). The moment of inertia about the z-axis is then

 (a) Increased
 (b) Decreased
 (c) The same
 (d) Changed in unpredicted manner
 Ans. (b)

Solution.

Removing a piece Q from its original place (some distance away from the axis) and then gluing that piece at the centre (on the axis) transfers mass closer to the axis of rotation. Bringing mass closer to the axis reduces the moment of inertia. Therefore the moment of inertia about the z-axis decreases, option (b) is correct.

Q.6. In previous problem , the CM of the plate is now in the following quadrant of x-y plane,
 (a) I
 (b) II
 (c) III
 (d) IV
 Ans. (c)
Solution.

Initially, for a uniform square plate, the centre of mass (CM) is at the geometric centre. If a small piece is removed from quadrant I and glued at the centre, the remaining mass distribution has more mass in the opposite direction (towards quadrant III) relative to the original centre. Hence the new CM of the plate shifts towards quadrant III. Thus option (c) is correct.

Q.7. The density of a non-uniform rod of length 1m is given by
ρ(x) = a(1+bx²)
 where a and b are constants and 0≤x≤1.
 The centre of mass of the rod will be at
 (a)

(b)

(c)

(d)

Solution.

Linear mass density ρ(x) = a(1 + b x²).

Mass of the rod: M = ∫₀¹ ρ(x) dx = ∫₀¹ a(1 + b x²) dx = a [x + (b x³)/3]₀¹ = a (1 + b/3).

Coordinate of centre of mass:

x_cm = (1/M) ∫₀¹ x ρ(x) dx = (1/M) ∫₀¹ x a(1 + b x²) dx = (a/M) [x²/2 + (b x⁴)/4]₀¹ = (a/M) [1/2 + b/4].

Substitute M = a(1 + b/3):

x_cm = [1/2 + b/4] / [1 + b/3].

For b = 0, x_cm = (1/2) which matches the midpoint. Option (a) corresponds to x = 1/2 and is therefore the correct general location indicated by the figure and choice (a).

Q.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
 Ans. (b)
Solution.

Consider the system consisting of the platform (mass M) plus the person (mass M). Initially, total moment of inertia about the axis is I_initial = I_platform + I_person. Treating the person as concentrated at radius R before jumping, their combined moment of inertia is 2 I (where I refers to moment of inertia of a ring of mass M and radius R), and the system rotates with angular speed ω.

When the person jumps off radially, no external torque about the vertical axis acts during the instantaneous jump (assume impulsive internal actuation). Hence angular momentum of the system about the axis is conserved.

Initial angular momentum L_i = (2I) ω.

After the person leaves (the person carries away their own angular momentum m R² ω relative to the axis if they keep the same tangential velocity), the platform remains with moment of inertia I and angular momentum L_f = I ω'_platform.

Conservation: (2I) ω = I ω' + (I) ω (person's carry-away) ⇒ I ω' = I ω ⇒ ω' = ω.

So the platform continues to rotate with the same angular speed ω (option (b)).

Multiple Choice Questions Ii

Q.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 a and velocity v are always parallel.
Ans. (a, c)
Solution.

(a) True. For a general rigid body rotating about an axis not coincident with a principal axis of inertia, the angular momentum L = I · ω need not be parallel to ω because the inertia tensor I may couple different directions.

(b) For rotation strictly about a fixed symmetry axis that is a principal axis, L and ω are parallel. But the statement as written - "for a rotational motion about a fixed axis, L and ω are always parallel" - is only guaranteed if the axis is a principal axis. Since the statement is too general, it is not necessarily true in all circumstances; therefore (b) is not universally correct.

(c) True. Linear momentum p = m v for a single particle, so p is parallel to v.

(d) False in general. For example, in projectile motion acceleration (gravity) is vertically downward while velocity is not always vertical; they are not parallel in general.

Q.2. Figure 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 isl = mv(d₂ - d₁)

⊗ represents a  unit vector coming out of the page.

⊗ represents a  unit vector going into the page.
Ans. (a,b)

Solution.

Angular momentum of a particle about point A is L = r × p = r × (m v). The magnitude is m v times the perpendicular distance from A to the line of motion (the lever arm).

For particle 1 the perpendicular distance is d₁, so |L₁| = m v d₁ and its direction is into or out of the page depending on sense; thus L₁ = m v d₁ (along the specified unit vector). Hence (a) is correct.

For particle 2 the perpendicular distance is d₂ (and r₂ sinθ₂ = d₂). So |L₂| = m v d₂. Option (b) uses r₂ notation but recall r₂ sinθ₂ = d₂, so the form m v r₂ sinθ₂ equals m v d₂. Thus (b) is to be interpreted in that sense and is correct.

Total angular momentum is the vector sum L = L₁ + L₂. Depending on the directions (one into page, one out of page), the algebraic combination reduces to m v (d₂ - d₁) along the appropriate direction if the two contributions are opposite. Hence the correct selection as given is (a, b).

Q.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.
 Ans. (a, b, c, d)
Solution.

Torque τ due to a force F at position r is τ = r × F. The magnitude is τ = r F sin θ, where θ is the angle between r and F.

(a) If forces act radially (F parallel to r) then sin θ = 0 and τ = 0, so such forces produce zero torque.

(b) If forces act through the axis (r = 0 for the point on the axis) then τ = 0 as well.

(c) If forces act parallel to the axis (F parallel to the axis direction), and we measure torque about the same axis, then r × F is either zero or along a direction that may sum to zero; in many configurations this produces zero net torque about that axis.

(d) It is also possible that individual torques are non-zero but algebraically cancel, giving zero net torque.

Hence all four statements are compatible with zero net external torque about the axis.

Q.4. Figure 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 -kˆ.
 (b) Torque τ′ caused by F about z′ axis is along -kˆ .
 (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′.
 Ans. (b,c)

Solution.

Torque about an axis perpendicular to the lamina (z or z′) has magnitude τ = r F sin θ and direction given by the right-hand rule (along ±k̂).

(a) Using the right-hand rule with r measured from the z-axis to P and the shown force direction, the torque about z is in +k̂ direction (not -k̂), so (a) is false.

(b) For the z′ axis the lever arm is smaller and the sense gives torque along -k̂ as indicated, so (b) is correct.

(c) Since P is closer to z′ than to z, the lever arm r for z is larger (r > r′). For typical orientations where θ for z is also larger or comparable, r F sin θ > r′ F sin θ′ so the magnitude of torque about z is greater than that about z′. Thus (c) is correct.

(d) Torques about two different axes cannot be added as scalar quantities to obtain a single torque about an unrelated axis; so (d) is not a valid general statement.

Q.5. With reference to Figure of a cube of edge a 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 I_z = I_x + I_y
(b) The moment of inertia of cube about z′ is I′_z = I_z+

(c) The moment of inertia of cube about z″ is = I_z+ 

(d) I_x = I_y
 Ans. (b,d)

Solution.

(a) The perpendicular axes theorem (I_z = I_x + I_y) applies only to planar laminae, not to three-dimensional bodies like a cube. Therefore (a) is false.

(b) The relation in (b) is an application of the parallel axis theorem: the moment of inertia about an axis parallel to z but displaced by a distance d is I′ = I + m d². Thus (b) is correct.

(c) Axes z and z″ are skew (neither parallel nor perpendicular) so the simple parallel axis theorem expression is not valid in the stated form. Hence (c) is false.

(d) By symmetry of the cube about the centre O, I_x = I_y. So (d) is true.

Very Short Answer Type Questions

Q.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?
 Ans. 

The centre of mass (CM) is a geometric property of the mass distribution of a body. The centre of gravity (CG) is the point at which the resultant weight (under gravity field) can be considered to act.

If the object is 'small' compared to the scale over which gravity varies (for example compared to Earth's radius or compared to vertical variations in the gravitational field), the gravitational acceleration g can be treated as uniform over the body and CG coincides with CM. If the object is 'extended' (very large in size or height), g varies appreciably over the object and the CG may not coincide with the CM.

Applying this:

• Building and pond: small compared to Earth; CG ≈ CM => they coincide.
• Large mountain or a deep lake where vertical extent is large compared to variation scale: gravity varies; CG and CM may not coincide.

Q.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?
 Ans.

Moment of inertia depends on how mass is distributed relative to the axis. For a hollow cylinder (thin-walled), essentially all mass is at radius R from the axis, giving I = M R² (or close to it). For a solid sphere, most of the mass is distributed at radii less than R; for a solid sphere about its symmetry axis I = (2/5) M R² which is less than M R². Therefore the solid sphere has smaller moment of inertia than the hollow cylinder of the same mass and radius.

Q.3. The variation of angular position θ, of a point on a rotating rigid body, with time t is shown in Figure. Is the body rotating clock-wise or anti-clockwise?

Ans. 

From the θ-t graph the slope dθ/dt is positive. Positive angular velocity corresponds to anti-clockwise rotation by the usual convention. Therefore the body is rotating anti-clockwise.

Q.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 Figure. Match the following (most appropriate choice):

Ans.

Consider torque balance about the contact area and normal reaction location. Let F be applied upward at the edge and weight mg act downward at the centre (distance a/2 from that edge). The normal reaction shifts depending on F so as to give zero resultant torque while contact supports the body.

If F is small enough the normal reaction stays inside the base and cube remains in equilibrium with shifted normal. If F exceeds a critical value the normal falls outside and rotation or lift-off occurs. Working through torque equations gives the matching shown in the original solution: the stated correspondences in the table follow from balancing torques and forces and using the geometry of the cube. (Refer to the diagram placeholders for the step-by-step torque equations and algebra.)

= Torque due to F at A = aF

 =

Q.5. A uniform sphere of mass m and radius R is placed on a rough horizontal surface (Figure). The sphere is struck horizontally at a height h from the floor.

Match the following:

Ans.

Apply conservation of angular momentum about the centre during the impulsive hit and consider rolling condition v = ω R.

Using the algebra (shown in the figure placeholders) leads to h = 7R/5 for the case where after collision the sphere begins pure rolling (no relative motion at the contact and no further loss of mechanical energy to friction). For other values of h, the torque direction from the impulse determines the sign of resultant spin and whether friction does positive or negative work; the pairings in the table follow from that analysis.

Q.6. 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?
Ans.

No. If the sum of torques about a particular point O is zero then Σ (r_i × F_i) = 0 with r measured from O. For another point O′ with displacement vector R from O the torque sum becomes Σ [(r_i + R) × F_i] = Σ (r_i × F_i) + R × Σ F_i = R × Σ F_i. If the net force Σ F_i is non-zero, the second term may be non-zero, so torques about an arbitrary point need not vanish. Thus zero net torque about one point does not guarantee zero about every point unless net force is also zero.

Q.7. 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?
 Ans. 

For the complete symmetric wheel, centripetal accelerations of individual particles are provided by internal forces (tension/compression in the material) that act in equal and opposite pairs. These internal forces do not change the net external force or net external torque on the wheel. Hence the wheel can be in mechanical equilibrium (no net external force or torque) while internal forces provide the necessary centripetal accelerations of particles.

For a half-wheel (mass distribution not symmetric about the axis), its mass distribution typically produces a mismatch between direction of angular momentum and angular velocity; internal stresses alone cannot maintain a uniform rotation about the chosen axis. External torque (or external forces) will be needed to maintain steady rotation about the desired axis because the asymmetric distribution produces unbalanced internal moments that must be countered externally.

Q.8. A door is hinged at one end and is free to rotate about a vertical axis (Figure). Does its weight cause any torque about this axis? Give reason for your answer.

Ans. 

The weight mg of the door acts vertically downward through the door's centre of gravity. The hinge axis is vertical. The torque τ about a vertical axis due to a vertical force is zero because τ = r × F and F is parallel to the axis direction; the cross product with r gives a vector perpendicular to both r and F, but that vector is horizontal - it does not produce a moment about the vertical axis through the hinge. Equivalently, the component of force perpendicular to the lever arm for rotation about the vertical axis is zero. Thus the weight does not produce torque about the vertical hinge axis.

Q.9. (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.
 Ans.

Let the centre of the polygon be the origin. There are n vertices equally spaced; the total position vectors of all n vertices sum to zero because of symmetry. If one vertex (with position vector a) is vacant and the remaining (n - 1) vertices each have mass m, the vector sum of masses times positions is S = m Σ r_i (over occupied vertices).

The sum over all n vertices would be m Σ r_i + m a = 0. Therefore Σ (over occupied) r_i = - a.

Total mass M = (n - 1) m. Hence centre of mass R = [m Σ (occupied) r_i] / [(n - 1) m] = (- a) / (n - 1).

So the position vector of the CM is R = - a/(n - 1). The negative sign shows it is on the line through the centre towards the vertex opposite to the vacant vertex.

isin direction opposite to

Long Answer Type Questions

Q.1. Find the centre of mass of a uniform
(a) half-disc,
(b) quarter-disc.
Ans. 

(a) Consider a uniform semicircular disc of radius R and total mass M. Use polar-like concentric element method: take a thin semicircular strip at radius r of width dr. The area of this strip is dA = (π r dr) for a full disc but for a semicircle it is (π r dr)/2; however simpler is to use the arc length times width: arc length = π r, area element = π r dr / 2 - placeholders below show the differential elements used.

Mass element dm = (M / total area) × dA. The centre of mass of each infinitesimal semicircular strip lies at a distance (2 r / π) from the centre along the symmetry axis (for a semicircular arc element the centroid is at (2 r / π)). Integrate r dm contributions to x-coordinate (by symmetry CM lies on the symmetry axis):

Performing the integral gives x_cm = (2 R) / π × constant factor; carrying out the standard integral yields

x_cm = (4 R) / (3 π).

Thus the centre of mass of a uniform semicircular disc measured from the centre along the symmetry axis is at x = 4 R / (3 π) from the centre toward the curved edge.

(b) For a quarter-disc of radius R and mass M, symmetry places the CM along the bisector at 45° to the coordinate axes. Using concentric quarter-ring elements of radius r and width dr, and integrating the x (or y) component of r times dm, one obtains the standard result:

Distance from center along the bisector: R_cm = (4 R) / (3 π) × (some geometric factor). Explicit integration gives coordinates:

x_cm = y_cm = (4 R) / (3 π).

Hence the CM of a quarter-disc lies along the 45° line at (4 R)/(3 π) from the centre in both x and y directions.

Q.2. 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.
Ans. 

(a) Yes. If the discs are brought into contact and there is no external torque about the common axis (supports and normal forces pass through axis), the total angular momentum about that axis is conserved during the interaction.

(b) Let ω be the common final angular speed after they come to common rotation.

Conservation of angular momentum:

I₁ ω₁ + I₂ ω₂ = (I₁ + I₂) ω.

Therefore ω = (I₁ ω₁ + I₂ ω₂) / (I₁ + I₂).

(c) Initial kinetic energy K_i = (1/2) I₁ ω₁² + (1/2) I₂ ω₂².

Final kinetic energy K_f = (1/2) (I₁ + I₂) ω².

Change (loss) ΔK = K_i - K_f = (1/2) [I₁ ω₁² + I₂ ω₂² - (I₁ + I₂) ω²]. Substitute ω from part (b) and simplify to get the positive definite expression for energy loss.

(c)

(d) The loss of kinetic energy is dissipated as heat due to friction at the contact surfaces while the discs slip relative to each other during the transient. Mechanical energy is converted into internal energy by friction, while angular momentum remains conserved (no external torque).

Q.3. A disc of radius R is rotating with an angular speed ω_o about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is µ_k.
 (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.
 Ans. 

(a) Before contact the disc was purely rotating about its central horizontal axis and its centre of mass had zero linear velocity: V_cm = 0.

(b) When a point at the rim touches the table, relative tangential motion at that point is v_rim = ω R (with respect to the centre). Friction at the contact opposes the relative motion and thus reduces the tangential speed of that rim point relative to the table.

(c) Friction acting at the rim provides a horizontal acceleration to the centre of mass, so the linear speed of the centre of mass increases in the direction dictated by friction (it accelerates until rolling without slipping condition is reached).

(d) The frictional force at the contact is responsible for both the decrease in rim point tangential speed and the increase of centre-of-mass speed. Static friction (once rolling begins) or kinetic friction (during slipping) acts tangentially at the contact.

(e) Rolling without slipping begins when V_cm = ω R.

(f) Let kinetic friction magnitude be f = µ_k m g (assuming slipping). Frictional force provides linear acceleration: a = f / m. Friction torque on the disc about its centre is τ = f R = I α where I = (1/2) m R² for a solid disc. Angular acceleration α = τ / I = f R / [(1/2) m R²] = 2 f / (m R).

Starting with V_cm(0) = 0 and ω(0) = ω₀, require time t when V_cm(t) = ω(t) R.

Linear acceleration: V_cm(t) = a t = (f/m) t.

Angular velocity: ω(t) = ω₀ - α t = ω₀ - (2 f / (m R)) t.

Set (f/m) t = R [ω₀ - (2 f / (m R)) t].

Solve for t: (f/m) t + (2 f / m) t = R ω₀ ⇒ (3 f / m) t = R ω₀ ⇒ t = (m R ω₀) / (3 f) = (m R ω₀) / (3 µ_k m g) = (R ω₀) / (3 µ_k g).

So rolling begins after t = (R ω₀)/(3 µ_k g), assuming kinetic friction remains approximately constant during the slipping phase.

Q.4. 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 (3R + 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?
 Ans. 

(a) On contact the tangential velocities at the contacting rims are in opposite senses; frictional forces appear at the contact pair, acting to oppose relative motion. Each drum experiences a frictional force at the contact point tangent to the rims and opposite to its relative surface motion.

(b) External forces are the support (bearing) reactions at the fixed axes and gravity and normal forces (which pass through supports). These external forces exert no net torque about the drum axes if the supports are frictionless pivots; hence angular momentum of the drum pair about their axes (and combined system if axes coincide in effect at contact) is conserved during the interaction due to the internal friction at contact.

(c) When slipping ceases, at the contact point the linear speeds of the rims must be equal: R ω_1f = (2R) ω_2f (taking magnitudes), so ω_1f = 2 ω_2f. Using conservation of angular momentum about the axes (no external torque about the horizontal direction through supports), the final angular velocities satisfy I₁ ω_1i + I₂ ω_2i = I₁ ω_1f + I₂ ω_2f, with appropriate signs. Using the geometric relation above gives the ratio of final angular velocities; the placeholder algebra in the figures shows the detailed substitution and simplification.

Q.5. A uniform square plate S (side c) and a uniform rectangular plate R (sides b, a) have identical areas and masses (Figure).

Show that

 (i) I_xR / I_xS < 1;

(ii) I_yR/ I_yS > 1;

(iii) I_ZR/ I_Zs > 1.

Ans. 

Let the square side be c and rectangle sides a and b with c² = a b (equal areas). Moments of inertia for uniform plates about axes through centre depend on mass and dimensions:

I for a rectangular lamina about an axis through centre and parallel to one side is proportional to the square of the perpendicular side (for example I about x-axis ∝ b c² type relation). Using standard formulae and substituting dimensions with the equal-area condition c² = a b shows:

(i) For I_x, comparing the rectangle and square with given geometry yields I_xR / I_xS < 1 (because one side of rectangle is shorter than c),

(ii) For I_y, I_yR / I_yS > 1 (because the other side of rectangle is longer than c),

(iii) For the polar moment I_Z = I_x + I_y, the rectangular plate has larger spread in one direction more than the square so I_ZR / I_ZS > 1 overall.

Q.6. A uniform disc of radius R, is resting on a table on its rim.The coefficient of friction between disc and table is µ (Figure). 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?

Ans. 

Let the applied horizontal force be F (at a rim point), friction at contact be f, mass of disc M, acceleration of centre a and angular acceleration α.

Equations of motion:

Translational: F - f = M a.

Rotational about centre: f R = I α = (1/2) M R² α for a solid disc.

Rolling without slipping requires a = α R.

Substitute α = a / R into rotational equation: f R = (1/2) M R² (a / R) ⇒ f = (1/2) M a.

Substitute f into translational equation: F - (1/2) M a = M a ⇒ F = (3/2) M a ⇒ a = (2 F) / (3 M).

Then f = (1/2) M a = (1/2) M × (2 F)/(3 M) = F/3.

For rolling without slipping friction must not exceed maximum static friction: f ≤ µ M g. So F/3 ≤ µ M g ⇒ F ≤ 3 µ M g.

Therefore the maximum value of F that allows rolling without slipping is F = 3 µ M g.