System of Particles & Rotational Motion — NCERT Exercises

(a) Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring and (iv) cube, each of uniform mass density.

(b) Does the centre of mass of a body necessarily lie inside the body?

Answer

(a) For rigid bodies of regular geometrical shape having a uniform distribution of mass, the centre of mass is at their geometric centre. Hence,

(i) Sphere — the centre of mass is at the centre of the sphere.

(ii) Cylinder — the centre of mass is at the mid-point of its axis of symmetry.

(iii) Ring — the centre of mass is at the centre of the ring.

(iv) Cube — the centre of mass is at the point of intersection of its diagonals, that is, at its centre.

(b) No, the centre of mass of a body does not necessarily lie inside the body.

The centre of mass is the weighted average location of all the mass in the system, and it depends on the distribution of mass and not on the physical boundaries of the body. Hence for a body having no material at that average position, the centre of mass lies outside the material of the body. For example, the centre of mass of a ring is at its centre, where there is actually no matter. The same is true for a hollow sphere and a hollow cylinder.

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å. Find the approximate location of the centre of mass of the molecule, given that a chlorine atom is about 35.5 times as massive as hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Answer

Given,

• Separation between the nuclei of the two atoms, d = 1.27 Å
• Mass of the chlorine atom, m₂ = 35.5 m₁, where m₁ is the mass of the hydrogen atom

Nearly all the mass of an atom is concentrated in its nucleus, so the two atoms may be treated as point masses placed at their nuclei. The centre of mass lies on the line joining the H and Cl atoms. Let this line be the X-axis with the H atom at the origin.

Then x₁ = 0 and x₂ = 1.27 Å.

The centre of mass relative to the H atom is given by

$\text x_{cm} = \dfrac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}$

Substituting the values,

$\text x_{cm} = \dfrac{(\text m_1 \times 0) + (35.5\text m_1 \times 1.27)}{\text m_1 + 35.5\text m_1} \\[1em] = \dfrac{35.5 \times 1.27}{36.5} = \dfrac{45.085}{36.5}$

$\text x_{cm} = 1.235 \text{\AA}$

Hence, the centre of mass of the HCl molecule is located at a distance of 1.235 Å from the hydrogen atom, on the line joining the two nuclei. Being very close to the chlorine nucleus, it shows that the centre of mass lies nearer to the heavier particle.

A child is sitting at one end of a long trolley moving with a uniform speed v on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what will be the speed of the centre of mass of the system (trolley + child)?

Answer

The speed of the centre of mass of the system remains unchanged and is equal to v.

The child and the trolley together form a single system. When the child gets up and runs about on the trolley, the forces involved between the child and the trolley, such as the action and reaction at his feet and the friction between his feet and the floor of the trolley, are all internal forces of the system.

Internal forces always occur in equal and opposite pairs, in compliance with Newton's third law of motion, so their vector sum is zero and they cancel out. Since the floor is smooth, no external force acts on the system in the horizontal direction, that is, F_ext = 0. Therefore

$\dfrac{\text{d}\vec{\text P}}{\text{dt}} = \vec{\text F}_{ext} = 0 \quad \Rightarrow \quad \vec{\text P} = \text{M}\vec{\text v}_{cm} = \text{a constant}$

Hence the velocity of the centre of mass remains constant. Without external forces, a system's internal dynamics alone cannot induce acceleration. The speed of the centre of mass therefore continues to be v, however the child may run about on the trolley.

Show that the area of the triangle contained between the vectors $\vec{\text A} \space$and $\vec{\text B} \space$is one-half the magnitude of $\vec{\text A} \times \vec{\text B}$.

Answer

Let the vectors $\vec{\text A} \space$and $\vec{\text B} \space$represent the sides OP and OR of a triangle OPR, and let θ be the angle between them. Let RS be the perpendicular dropped from R on OP.

The area of the triangle OPR is

$\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2}(\text{OP})(\text{RS})$

From the right-angled triangle ORS,

$\text{RS} = (\text{OR})\sin \theta$

Substituting this value,

$\text{Area} = \dfrac{1}{2}(\text{OP})(\text{OR})\sin \theta = \dfrac{1}{2}\text{AB}\sin \theta$

where A and B are the magnitudes of the two vectors.

But by the definition of the vector product,

$|\vec{\text A} \times \vec{\text B}| = \text{AB}\sin \theta$

Therefore,

$\text{Area of the triangle} = \dfrac{1}{2}|\vec{\text A} \times \vec{\text B}|$

Hence, the area of the triangle contained between the vectors $\vec{\text A} \space$and $\vec{\text B} \space$is one-half the magnitude of $\vec{\text A} \times \vec{\text B}$.

Show that $\vec{\text A} \cdot (\vec{\text B} \times \vec{\text C}) \space$is equal in magnitude to the volume of the parallelopiped formed on the vectors $\vec{\text A}$, $\vec{\text B} \space$and $\vec{\text C}$.

Answer

Let a parallelopiped be formed on the three vectors $\vec{\text A}$, $\vec{\text B} \space$and $\vec{\text C}$, with θ the angle between $\vec{\text B} \space$and $\vec{\text C}$, and $\phi$ the angle between $\vec{\text A} \space$and the normal to the base.

The vector product of $\vec{\text B} \space$and $\vec{\text C} \space$is

$\vec{\text B} \times \vec{\text C} = \text{BC}\sin \theta\ \hat{\text n}$

where $\hat{\text n}$ is a unit vector normal to the plane containing $\vec{\text B} \space$and $\vec{\text C}$. Its magnitude is

$|\vec{\text B} \times \vec{\text C}| = \text{BC}\sin \theta$

Now taking the scalar product of $\vec{\text A} \space$with $\vec{\text B} \times \vec{\text C}$,

$\vec{\text A} \cdot (\vec{\text B} \times \vec{\text C}) = \text A(\text{BC}\sin \theta)\cos \phi \\[1em] = (\text{BC}\sin \theta)(\text A \cos \phi)$

Here BC sin θ is the area of the parallelogram which forms the base of the parallelopiped, and A cos φ is the projection of $\vec{\text A} \space$along $\vec{\text B} \times \vec{\text C}$, that is, the height of the parallelopiped.

Therefore,

$\vec{\text A} \cdot (\vec{\text B} \times \vec{\text C}) = \text{base area} \times \text{height}$

Hence, $\vec{\text A} \cdot (\vec{\text B} \times \vec{\text C}) \space$is equal in magnitude to the volume of the parallelopiped formed on the vectors $\vec{\text A}$, $\vec{\text B} \space$and $\vec{\text C}$.

Find the components along the X-, Y-, Z-axes of the angular momentum $\vec{\text L} \space$of a particle, whose position vector is $\vec{\text r} \space$with components x, y, z and linear momentum is $\vec{\text p} \space$with components p_x, p_y and p_z. Show that if the particle moves only in X-Y plane the angular momentum has only a z-component.

Answer

The position vector and the linear momentum of the particle in terms of their x, y, z-components are

$\vec{\text r} = \text x\hat{\text i} + \text y\hat{\text j} + \text z\hat{\text k} \quad \text{and} \quad \vec{\text p} = \text p_x\hat{\text i} + \text p_y\hat{\text j} + \text p_z\hat{\text k}$

By definition, the angular momentum of the particle is the moment of linear momentum,

$\vec{\text L} = \vec{\text r} \times \vec{\text p} = \begin{vmatrix} \hat{\text i} & \hat{\text j} & \hat{\text k} \\ \text x & \text y & \text z \\ \text p_x & \text p_y & \text p_z \end{vmatrix}$

Expanding the determinant,

$\vec{\text L} = \hat{\text i}(\text y\text p_z - \text z\text p_y) + \hat{\text j}(\text z\text p_x - \text x\text p_z) + \hat{\text k}(\text x\text p_y - \text y\text p_x) \qquad \dots(\text i)$

The angular momentum may also be written in terms of its own components as

$\vec{\text L} = \text L_x\hat{\text i} + \text L_y\hat{\text j} + \text L_z\hat{\text k} \qquad \dots(\text{ii})$

Comparing equations (i) and (ii), the components of the angular momentum along the X-, Y- and Z-axes are

$\text L_x = \text y\text p_z - \text z\text p_y \\[1em] \text L_y = \text z\text p_x - \text x\text p_z \\[1em] \text L_z = \text x\text p_y - \text y\text p_x$

When the particle moves only in the X-Y plane : In this case the particle has no z-coordinate and no z-component of momentum, so

$\text z = 0 \quad \text{and} \quad \text p_z = 0$

Substituting these in the components obtained above,

$\text L_x = (\text y \times 0) - (0 \times \text p_y) = 0 \\[1em] \text L_y = (0 \times \text p_x) - (\text x \times 0) = 0 \\[1em] \text L_z = \text x\text p_y - \text y\text p_x$

Therefore,

$\vec{\text L} = (\text x\text p_y - \text y\text p_x)\hat{\text k}$

Hence, if the particle moves only in the X-Y plane, the angular momentum has only a z-component, that is, it is directed perpendicular to the plane of motion.

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.

Answer

Let AB and CD be the two parallel lines separated by a distance d, and let P and Q be the instantaneous positions of the two particles, each of mass m, travelling in opposite directions with speed v along these lines.

Let O be any point dividing the separation between the lines in the ratio d₁ : d₂, about which the angular momentum is to be taken, so that

$\text d_1 + \text d_2 = \text d$

Taking O as the origin, the instantaneous coordinates of P and Q are (x₁, d₁) and (x₂, − d₂) respectively. Their position vectors are

$\vec{\text r}_1 = \text x_1\hat{\text i} + \text d_1\hat{\text j} \quad \text{and} \quad \vec{\text r}_2 = \text x_2\hat{\text i} - \text d_2\hat{\text j}$

Since the particles travel in opposite directions, their momenta are $\vec{\text p}_1 = \text p_x\hat{\text i}$ and $\vec{\text p}_2 = -\text p_x\hat{\text i}$, where p_x = mv.

The angular momentum of particle P about O is

$\vec{\text L}_P = \vec{\text r}_1 \times \vec{\text p}_1 = (\text x_1\hat{\text i} + \text d_1\hat{\text j}) \times \text p_x\hat{\text i} \\[1em] = -\text p_x \text d_1\hat{\text k} = -\text{mv}\text d_1\hat{\text k}$

Similarly, the angular momentum of particle Q about O is

$\vec{\text L}_Q = \vec{\text r}_2 \times \vec{\text p}_2 = (\text x_2\hat{\text i} - \text d_2\hat{\text j}) \times (-\text p_x\hat{\text i}) \\[1em] = -\text p_x \text d_2\hat{\text k} = -\text{mv}\text d_2\hat{\text k}$

The total angular momentum of the two particles about O is

$\vec{\text L} = \vec{\text L}_P + \vec{\text L}_Q = -\text{mv}\text d_1\hat{\text k} - \text{mv}\text d_2\hat{\text k} \\[1em] = -\text{mv}(\text d_1 + \text d_2)\hat{\text k}$

But d₁ + d₂ = d, therefore

$\vec{\text L} = -\text{mvd}\hat{\text k}$

The magnitude of the total angular momentum is mvd, directed perpendicular to the two lines.

Since the result is independent of d₁ and d₂, it does not depend upon the position of the origin O. Hence the vector angular momentum of the two-particle system is the same whatever be the point about which it is taken.

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure. 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.

Given : sin 36.9° = cos 53.1° = 0.6 and sin 53.1° = cos 36.9° = 0.8.

Answer

Given,

• Length of the bar = 2 m
• Angle made by the first string with the vertical, θ₁ = 36.9°
• Angle made by the second string with the vertical, θ₂ = 53.1°
• sin 36.9° = cos 53.1° = 0.6 and sin 53.1° = cos 36.9° = 0.8

Let T₁ and T₂ be the tensions in the two strings and let d be the distance of the centre of gravity G of the bar from its left end.

Condition of translational equilibrium : Resolving the tensions horizontally, the net force must be zero,

$\text T_1 \sin \theta_1 = \text T_2 \sin \theta_2 \qquad \dots(\text i)$

Condition of rotational equilibrium : Taking moments of the forces about the centre of gravity G, the net torque must be zero,

$\text T_1 \cos \theta_1 \times \text d = \text T_2 \cos \theta_2 \times (2 - \text d) \qquad \dots(\text{ii})$

From equation (i),

$\text T_2 = \text T_1 \dfrac{\sin \theta_1}{\sin \theta_2}$

Substituting this value of T₂ in equation (ii),

$\text T_1 \cos \theta_1 \times \text d = \text T_1 \dfrac{\sin \theta_1}{\sin \theta_2}\cos \theta_2 \times (2 - \text d)$

$\text d = \dfrac{\sin \theta_1 \cos \theta_2}{\cos \theta_1 \sin \theta_2}(2 - \text d)$

Substituting the given values,

$\text d = \dfrac{0.6 \times 0.6}{0.8 \times 0.8}(2 - \text d) = \dfrac{9}{16}(2 - \text d)$

$16\text d = 18 - 9\text d \\[1em] 25\text d = 18$

$\text d = \dfrac{18}{25} = 0.72\ \text m$

Hence, the centre of gravity of the bar is at a distance of 0.72 m, that is, 72 cm, from its left end.

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity G is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. (g = 9.8 N/kg)

Answer

Given,

• Weight of the car, W = 1800 kg-wt
• Distance between the front and back axles = 1.8 m
• Distance of the centre of gravity G behind the front axle = 1.05 m
• g = 9.8 N/kg

The distance of the centre of gravity from the back axle is

$1.8 - 1.05 = 0.75\ \text m$

Let R₁ be the reactionary force exerted by the ground on each front wheel and R₂ that on each back wheel. Since there are two wheels on each axle, the total upward force on the front axle is 2R₁ and that on the back axle is 2R₂.

Condition of translational equilibrium : For the vertical equilibrium of the car,

$2\text R_1 + 2\text R_2 = \text W \qquad \dots(\text i)$

Condition of rotational equilibrium : Taking moments of the forces about the centre of gravity G,

$2\text R_1 \times 1.05 = 2\text R_2 \times 0.75$

$\text R_1 \times 1.05 = \text R_2 \times 0.75 \\[1em] \text R_1 = \dfrac{0.75}{1.05}\text R_2 = \dfrac{5}{7}\text R_2 \qquad \dots(\text{ii})$

Substituting the value of R₁ from equation (ii) in equation (i),

$2 \times \dfrac{5}{7}\text R_2 + 2\text R_2 = 1800\ \text{kg-wt} \\[1em] \dfrac{10}{7}\text R_2 + 2\text R_2 = \dfrac{24}{7}\text R_2 = 1800\ \text{kg-wt}$

$\text R_2 = 1800 \times \dfrac{7}{24} = 525\ \text{kg-wt}$

Converting to newton,

$\text R_2 = 525 \times 9.8 = 5145\ \text N$

Now from equation (ii),

$\text R_1 = \dfrac{5}{7}\text R_2 = \dfrac{5}{7} \times 5145 = 3675\ \text N$

Hence, the force exerted by the level ground on each front wheel is 3675 N and that on each back wheel is 5145 N.

The force on each back wheel is larger because the centre of gravity of the car lies nearer to the back axle.

Torques of equal magnitudes are applied to a hollow cylinder and a solid sphere, both having 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?

Answer

Let M be the mass and R the radius of both the hollow cylinder and the solid sphere.

The moment of inertia of the hollow cylinder about its standard axis of symmetry is

$\text I_1 = \text{MR}^2$

and the moment of inertia of the solid sphere about an axis passing through its centre is

$\text I_2 = \dfrac{2}{5}\text{MR}^2$

Let τ be the magnitude of the equal torque applied to each body, producing angular accelerations α₁ in the cylinder and α₂ in the sphere. From the relation between torque, moment of inertia and angular acceleration,

$\tau = \text I_1 \alpha_1 = \text I_2 \alpha_2$

Therefore,

$\dfrac{\alpha_1}{\alpha_2} = \dfrac{\text I_2}{\text I_1} = \dfrac{(2/5)\text{MR}^2}{\text{MR}^2} = \dfrac{2}{5}$

$\alpha_2 = \dfrac{5}{2}\alpha_1$

Thus the angular acceleration produced in the sphere is 2.5 times that produced in the cylinder. Both bodies start from rest and, from the equation of rotational kinematics ω = ω₀ + αt, the angular speed acquired in a given time t is directly proportional to the angular acceleration.

Hence, the solid sphere will acquire a greater angular speed after a given time. This is because the sphere has a smaller moment of inertia, its mass being distributed nearer to the axis of rotation, and so it offers less resistance to the change in its rotational motion.

A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad 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?

Answer

Given,

• Mass of the solid cylinder, M = 20 kg
• Angular speed, ω = 100 rad s^-1
• Radius of the cylinder, R = 0.25 m

The moment of inertia of a solid cylinder about its own axis is

$\text I = \dfrac{1}{2}\text{MR}^2$

Substituting the values,

$\text I = \dfrac{1}{2} \times 20 \times (0.25)^2 = \dfrac{1}{2} \times 20 \times 0.0625 \\[1em] = 0.625\ \text{kg m}^2$

Kinetic energy of rotation :

$\text K = \dfrac{1}{2}\text I \omega^2 = \dfrac{1}{2} \times 0.625 \times (100)^2 \\[1em] = \dfrac{1}{2} \times 0.625 \times 10000 = 3125\ \text J$

Angular momentum about the axis :

$\text L = \text I \omega = 0.625 \times 100 = 62.5\ \text{kg m}^2\text s^{-1}$

Hence, the kinetic energy associated with the rotation of the cylinder is 3125 J and the magnitude of its angular momentum about its axis is 62.5 kg m² s^-1.

(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 reduces his moment of inertia to $\dfrac{2}{5}$ 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?

Answer

Given,

• Initial angular speed of the turntable, ω₁ = 40 rev/min
• Final moment of inertia, $\text I_2 = \dfrac{2}{5}\text I_1$, where I₁ is the initial moment of inertia

(a) The turntable rotates without friction, so no external torque acts on the system. Hence the angular momentum of the system is conserved,

$\text I_1 \omega_1 = \text I_2 \omega_2$

$\omega_2 = \dfrac{\text I_1 \omega_1}{\text I_2}$

Substituting the values,

$\omega_2 = \dfrac{\text I_1 \times 40}{(2/5)\text I_1} = \dfrac{5}{2} \times 40 \\[1em] = 100\ \text{rev/min}$

Hence, the new angular speed of the child is 100 rev/min.

(b) The kinetic energy of rotation is $\text K = \dfrac{1}{2}\text I \omega^2$. Therefore

$\dfrac{\text{Final K.E. of rotation}}{\text{Initial K.E. of rotation}} = \dfrac{\dfrac{1}{2}\text I_2 \omega_2^2}{\dfrac{1}{2}\text I_1 \omega_1^2} = \dfrac{(2/5)\text I_1 (100)^2}{\text I_1 (40)^2}$

$= \dfrac{2}{5} \times \dfrac{10000}{1600} = \dfrac{2}{5} \times 6.25 = 2.5$

Hence, the new kinetic energy of rotation is 2.5 times the initial kinetic energy of rotation, that is, it has increased.

Accounting for the increase : The angular momentum is conserved, but the kinetic energy is not. To fold his hands back, the child has to do work against the centrifugal effect in pulling his arms inwards. This work is done by the child using his own internal (chemical) energy, and it appears as the additional kinetic energy of rotation. Hence there is no violation of the conservation of energy.

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.

Answer

Given,

• Mass of the hollow cylinder, M = 3 kg
• Radius of the cylinder, R = 40 cm = 0.40 m
• Force with which the rope is pulled, F = 30 N

Every particle of a hollow cylinder is at the same distance R from its own central axis, so its moment of inertia about that axis is

$\text I = \text{MR}^2 = 3 \times (0.40)^2 \\[1em] = 3 \times 0.16 = 0.48\ \text{kg m}^2$

The rope is wound round the rim, so the force acts tangentially at a perpendicular distance R from the axis. The torque exerted on the cylinder is

$\tau = \text F \times \text R = 30 \times 0.40 = 12\ \text{N m}$

Angular acceleration : From the relation τ = Iα,

$\alpha = \dfrac{\tau}{\text I} = \dfrac{12}{0.48} = 25\ \text{rad s}^{-2}$

Linear acceleration of the rope : Since there is no slipping, the rope moves with the same linear acceleration as a point on the rim of the cylinder,

$\text a = \text R \alpha = 0.40 \times 25 = 10\ \text{m s}^{-2}$

Hence, the angular acceleration of the cylinder is 25 rad s^-2 and the linear acceleration of the rope is 10 m s^-2.

To maintain a rotor at uniform angular speed of 200 rad s^-1, an engine needs to transmit a torque of 180 N m. What is the power of the engine required?

[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.

Answer

Given,

• Uniform angular speed of the rotor, ω = 200 rad s^-1
• Torque transmitted by the engine, τ = 180 N m
• The engine is 100% efficient

For a uniform angular speed of rotation no torque would be needed if there were no friction, since the angular acceleration is then zero. In practice, however, a torque has to be applied by the engine to counterbalance the frictional torque acting on the rotor, so that the net torque remains zero and the angular speed stays uniform.

The power generated by a torque is

$\text P = \tau \omega$

Substituting the values,

$\text P = 180 \times 200 = 36000\ \text W \\[1em] = 36\ \text{kW}$

Hence, the power of the engine required is 36 kW.

From a uniform disc 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.

Answer

Let O be the centre of the original disc of radius R and O′ the centre of the hole of radius R/2, so that OO′ = R/2. Let σ be the surface density, that is, the mass per unit area, of the disc.

The mass of the original complete disc is

$\text M = \pi \text R^2 \sigma$

and the mass of the portion cut out is

$\text m = \pi \left(\dfrac{\text R}{2}\right)^2 \sigma = \dfrac{1}{4}\pi \text R^2 \sigma = \dfrac{\text M}{4}$

The cut-out portion is treated as a negative mass superposed on the complete disc, since removing it is equivalent to adding a mass of − M/4 at the centre of the hole.

Taking the centre O of the original disc as the origin and the line OO′ as the X-axis, the two masses of the equivalent system are

• Mass M at x₁ = 0
• Mass − M/4 at x₂ = R/2

The centre of mass of the residual flat body with respect to O is

$\text x_{cm} = \dfrac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}$

Substituting the values,

$\text x_{cm} = \dfrac{(\text M)(0) + \left(-\dfrac{\text M}{4}\right)\left(\dfrac{\text R}{2}\right)}{\text M - \dfrac{\text M}{4}} = \dfrac{-\dfrac{\text{MR}}{8}}{\dfrac{3\text M}{4}}$

$\text x_{cm} = -\dfrac{\text{MR}}{8} \times \dfrac{4}{3\text M} = -\dfrac{\text R}{6}$

The negative sign shows that the centre of mass lies on the side opposite to the hole.

Hence, the centre of gravity of the resulting flat body is at a distance R/6 from the centre of the original disc, on the side opposite to the cut portion.

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?

Answer

Given,

• Mass of each coin = 5 g, so the mass of the two coins together = 10 g
• Position of the coins = 12.0 cm mark
• New balancing point (knife edge) K = 45.0 cm mark

Let m be the mass of the metre stick. The stick is uniform, so its weight mg acts at its centre of gravity G, which is at the 50.0 cm mark.

Taking moments of the forces about the knife edge K at the 45.0 cm mark, for the rotational equilibrium of the stick the anticlockwise moment must equal the clockwise moment,

$(10\ \text g)\text g \times (45.0 - 12.0) = \text{mg} \times (50.0 - 45.0)$

$(10\ \text g)\text g \times 33.0\ \text{cm} = \text{mg} \times 5.0\ \text{cm}$

Cancelling g from both sides,

$\text m = \dfrac{10\ \text g \times 33.0\ \text{cm}}{5.0\ \text{cm}}$

$\text m = 66.0\ \text g$

Hence, the mass of the metre stick is 66.0 g.

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 line 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.

Answer

Given,

• Mass of the oxygen molecule, M = 5.30 × 10^-26 kg
• Moment of inertia about an axis through its centre, I = 1.94 × 10^-46 kg m²
• Mean speed of the molecule, v = 500 m/s
• Kinetic energy of rotation = $\dfrac{2}{3}$ × kinetic energy of translation

Let ω be the average angular velocity of the molecule. As given,

$\dfrac{1}{2}\text I \omega^2 = \dfrac{2}{3} \times \dfrac{1}{2}\text{Mv}^2$

Cancelling $\dfrac{1}{2}$ from both sides,

$\text I \omega^2 = \dfrac{2}{3}\text{Mv}^2$

$\omega^2 = \dfrac{2}{3} \times \dfrac{\text M}{\text I} \times \text v^2$

Taking the square root,

$\omega = \sqrt{\dfrac{2}{3} \times \dfrac{\text M}{\text I}} \times \text v$

Substituting the given values,

$\omega = \sqrt{\dfrac{2}{3} \times \dfrac{5.30 \times 10^{-26}}{1.94 \times 10^{-46}}} \times 500$

$= \sqrt{\dfrac{2}{3} \times 2.732 \times 10^{20}} \times 500 = \sqrt{1.821 \times 10^{20}} \times 500$

$= (1.35 \times 10^{10}) \times 500$

$\omega = 6.75 \times 10^{12}\ \text{rad s}^{-1}$

Hence, the average angular velocity of the oxygen molecule is 6.75 × 10¹² rad s^-1.