Academia Aeternum
तमसो मा ज्योतिर्गमय
Work through every question with crystal‑clear steps, stress–strain visuals and exam‑oriented hints – all in a sleek deep‑space layout.
Q1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
The centre of mass (COM) of a body is the point at which the entire mass of the system may be considered to be concentrated for describing translational motion. For bodies with uniform mass density, symmetry plays a crucial role in locating the COM.
If a body has geometric symmetry in all directions, the mass is distributed equally about the centre. Therefore the centre of mass coincides with the geometric centre.
However, the centre of mass need not always lie within the material of the body. For hollow or ring-shaped objects, the COM may lie at a point where no material exists.
(i) Solid Sphere
A sphere possesses complete spherical symmetry. The mass distribution is identical in every direction. Therefore the centre of mass lies at the geometric centre of the sphere.
(ii) Cylinder
A uniform cylinder has symmetry about its central axis and about a plane through its middle. Hence the centre of mass lies at the midpoint of the axis of the cylinder.
(iii) Ring (Thin Circular Ring)
A thin ring has circular symmetry. Each mass element on the ring has an opposite element that balances it. Thus the centre of mass lies at the geometric centre of the ring.
Note that this point lies at the centre of the ring where there is no material present.
(iv) Cube
A cube possesses symmetry along all three mutually perpendicular axes. Therefore the centre of mass lies at the geometric centre of the cube, which is the intersection point of its body diagonals.
The centre of mass of a body does not necessarily lie within the material of the body. For example:
Thus, the centre of mass depends only on mass distribution, not on whether the point lies within the physical material.
Q2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
For a system of particles lying along a straight line, the centre of mass is given by
\( x_{CM} = \dfrac{\sum m_i x_i}{\sum m_i} \)
In a diatomic molecule such as HCl, nearly all the mass is concentrated in the nuclei. Therefore, the molecule can be treated as a two-particle system consisting of the hydrogen nucleus and chlorine nucleus separated by a fixed distance.
The centre of mass always lies closer to the heavier particle.
Let the mass of hydrogen atom be
\( m_H = m \)
Mass of chlorine atom
\( m_{Cl} = 35.5m \)
Distance between the nuclei
\( d = 1.27 \, Å \)
Choose the hydrogen nucleus as the origin of coordinates.
Therefore
Using the centre of mass formula:
$$ \begin{aligned} x_{CM} &= \frac{m_H x_H + m_{Cl} x_{Cl}}{m_H + m_{Cl}} \end{aligned} $$ Substitute values: $$ \begin{aligned} x_{CM} &= \frac{m(0) + 35.5m(1.27)}{m + 35.5m} \end{aligned} $$ $$ \begin{aligned} x_{CM} &= \frac{35.5m \times 1.27}{36.5m} \end{aligned} $$ Mass \(m\) cancels: $$ \begin{aligned} x_{CM} &= \frac{35.5 \times 1.27}{36.5} \end{aligned} $$ $$ x_{CM} \approx 1.24 \, Å $$The centre of mass of the HCl molecule lies
\(1.24 \, Å\) from the hydrogen nucleus towards the chlorine nucleus.
Q3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?
The velocity of the centre of mass of a system depends only on the net external force acting on the system.
\( \vec{F}_{ext} = (M_{total})\dfrac{d\vec{v}_{CM}}{dt} \)
If the net external force on a system is zero, then the centre of mass moves with constant velocity.
In this problem the floor is smooth, meaning there is no external horizontal force on the system consisting of the child and trolley.
Let
Initially both the trolley and child move with velocity
\( V \)
Total momentum of the system initially:
\( P_{initial} = (M + m)V \)
Since the floor is smooth, there is no external horizontal force. Therefore total momentum of the system remains conserved.
Velocity of the centre of mass is
$$ \begin{aligned} v_{CM} &= \frac{\text{Total Momentum}}{\text{Total Mass}} \end{aligned} $$ Substituting the values: $$ \begin{aligned} v_{CM} &= \frac{(M+m)V}{M+m} \end{aligned} $$ $$ v_{CM} = V $$The centre of mass of the (child + trolley) system continues to move with uniform speed \( V \).
Hence the centre of mass moves with constant velocity regardless of how the child moves inside the system.
Q4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
The cross product of two vectors has an important geometric meaning. If two vectors a and b originate from the same point, the magnitude of their cross product is
\( | \vec{a} \times \vec{b} | = |\vec{a}|\,|\vec{b}| \sin \theta \)
where \( \theta \) is the angle between the vectors.
This quantity represents the area of the parallelogram formed by the two vectors. Since a triangle occupies exactly half the area of that parallelogram, its area becomes
\( \frac{1}{2} |\vec{a} \times \vec{b}| \)
Consider vectors \( \vec{a} \) and \( \vec{b} \) originating from the same point. These vectors form a triangle.
Let the magnitude of vector \( \vec{a} \) be the base of the triangle:
Base = \( |\vec{a}| \)
The perpendicular height from the tip of vector \( \vec{b} \) to the base is
Height = \( |\vec{b}| \sin \theta \)
Area of the triangle:
$$ \begin{aligned} \Delta &= \frac{1}{2} \times \text{base} \times \text{height} \end{aligned} $$ Substituting values: $$ \begin{aligned} \Delta &= \frac{1}{2} |\vec{a}| (|\vec{b}| \sin \theta) \end{aligned} $$ $$ \Delta = \frac{1}{2} |\vec{a}| |\vec{b}| \sin \theta $$ But from vector algebra: $$ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta $$ Therefore, $$ \boxed{\Delta = \frac{1}{2} |\vec{a} \times \vec{b}|} $$Area of the triangle formed by vectors \( \vec{a} \) and \( \vec{b} \) is
\( \displaystyle \frac{1}{2} |\vec{a} \times \vec{b}| \)
Q5 Show that \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) is equal in magnitude to the volume of the parallelepiped formed on the three vectors \( \vec{a}, \vec{b}, \vec{c} \).
The quantity \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) is called the scalar triple product.
The vector \( \vec{b} \times \vec{c} \) is perpendicular to the plane containing \( \vec{b} \) and \( \vec{c} \), and its magnitude equals
\( |\vec{b} \times \vec{c}| = |\vec{b}|\,|\vec{c}| \sin \theta \)
which represents the area of the parallelogram formed by \( \vec{b} \) and \( \vec{c} \).
Thus the scalar triple product gives the product of
Consider vectors \( \vec{b} \) and \( \vec{c} \) forming the base of a parallelepiped.
Area of the base parallelogram:
\( A = |\vec{b} \times \vec{c}| \)
Let \( \theta \) be the angle between vector \( \vec{a} \) and the normal to the base. The perpendicular height of the parallelepiped is
\( h = |\vec{a}| \cos \theta \)
Therefore the volume becomes
$$ \begin{aligned} V &= (\text{base area}) \times (\text{height}) \end{aligned} $$ Substituting values: $$ \begin{aligned} V &= |\vec{b} \times \vec{c}| \, (|\vec{a}| \cos\theta) \end{aligned} $$ Rearranging: $$ V = |\vec{a}|\,|\vec{b} \times \vec{c}| \cos\theta $$ But from vector algebra $$ \vec{a} \cdot (\vec{b} \times \vec{c}) = |\vec{a}|\,|\vec{b} \times \vec{c}| \cos\theta $$ Hence, $$ \boxed{V = |\vec{a} \cdot (\vec{b} \times \vec{c})|} $$Magnitude of the scalar triple product equals the volume of the parallelepiped formed by the vectors.
\( \displaystyle V = |\vec{a} \cdot (\vec{b} \times \vec{c})| \)
Q6 Find the components along the x, y, z axes of the angular momentum \( \vec L \) of a particle, whose position vector is \( \vec r \) with components \(x,y,z\) and momentum is \( \vec p \) with components \(p_x, p_y, p_z\). Show that if the particle moves only in the x–y plane the angular momentum has only a z-component.
The angular momentum of a particle about the origin is defined as the cross product of its position vector and momentum:
\( \vec L = \vec r \times \vec p \)
where
The cross product determines both the magnitude and direction of angular momentum. The direction is given by the right-hand rule.
Position vector of the particle:
\( \vec r = x\hat i + y\hat j + z\hat k \)
Momentum vector:
\( \vec p = p_x\hat i + p_y\hat j + p_z\hat k \)
Angular momentum is given by the cross product
\( \vec L = \vec r \times \vec p \)
Using the determinant form: $$ \vec L = \begin{vmatrix} \hat i & \hat j & \hat k \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix} $$ Expanding the determinant: $$ \vec L = (yp_z - zp_y)\hat i + (zp_x - xp_z)\hat j + (xp_y - yp_x)\hat k $$ Hence the components are $$ \begin{aligned} L_x &= yp_z - zp_y \\ L_y &= zp_x - xp_z \\ L_z &= xp_y - yp_x \end{aligned} $$If the particle moves only in the x-y plane:
\( \vec L = (xp_y - yp_x)\hat k \)
If a particle moves in the x–y plane, the angular momentum has only a z-component.
This principle is widely used in rotational motion problems, planetary motion, and circular dynamics.
Q7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.
Angular momentum of a system of particles about a point O is
\( \vec L = \sum \vec r_i \times \vec p_i \)
If the total linear momentum of the system is zero, then the angular momentum of the system becomes independent of the origin.
In this problem, the two particles move with equal speeds in opposite directions, so their momenta cancel.
\( \vec p_1 + \vec p_2 = 0 \)
Let the particles move along two parallel lines separated by distance \(d\).
Momentum of particle 1:
\( \vec p_1 = m v \hat i \)
Momentum of particle 2:
\( \vec p_2 = -m v \hat i \)
Total linear momentum:
\( \vec P = \vec p_1 + \vec p_2 = 0 \)
Angular momentum about an arbitrary origin \(O\):
\( \vec L = \vec r_1 \times \vec p_1 + \vec r_2 \times \vec p_2 \)
Choose coordinates such that the two particles lie at
\( y = +\frac{d}{2} \) and \( y = -\frac{d}{2} \)
Their position vectors are
\( \vec r_1 = x_1\hat i + \frac{d}{2}\hat j \)
\( \vec r_2 = x_2\hat i - \frac{d}{2}\hat j \)
Compute angular momenta. For particle 1: $$ \vec L_1 = \vec r_1 \times \vec p_1 = \left(x_1\hat i + \frac{d}{2}\hat j\right) \times (mv\hat i) $$ Since \( \hat i \times \hat i = 0 \), $$ \vec L_1 = \frac{d}{2} mv (\hat j \times \hat i) $$ $$ \vec L_1 = -\frac{mvd}{2}\hat k $$ For particle 2: $$ \vec L_2 = \left(x_2\hat i - \frac{d}{2}\hat j\right) \times (-mv\hat i) $$ $$ \vec L_2 = \frac{mvd}{2}\hat k $$ Adding the two: $$ \vec L = \vec L_1 + \vec L_2 $$ $$ \vec L = mvd \hat k $$The angular momentum of the system is
\( \vec L = mvd\,\hat k \)
This value is independent of the choice of origin.
Q8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. 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.
Since the bar is at rest, it is in static equilibrium. Therefore two conditions must be satisfied:
Thus we apply three equations:
\( \sum F_x =0,\qquad \sum F_y =0,\qquad \sum \tau =0 \)
The weight \(W\) acts at the centre of gravity located at a distance \(d\) from the left end.
The bar has length \(2\,\text{m}\) and weight \(W\). Let the tensions in the left and right strings be \(T_1\) and \(T_2\).
Because the bar is in equilibrium, the horizontal components of tension must cancel.
\(T_1\sin36.9^\circ = T_2\sin53.1^\circ\)
Using
\(\sin36.9^\circ = 0.60\), \(\sin53.1^\circ = 0.80\)
\[ \begin{aligned} T_2 &= T_1\frac{\sin36.9^\circ}{\sin53.1^\circ} \\ T_2 &= \frac{0.60}{0.80}T_1 \\ T_2 &= 0.75T_1 \end{aligned} \]
Now apply vertical equilibrium. The vertical components of the tensions must balance the weight.
\[ T_1\cos36.9^\circ + T_2\cos53.1^\circ = W \]
Using
\(\cos36.9^\circ =0.80\), \(\cos53.1^\circ =0.60\)
\[ \begin{aligned} T_1(0.80) + (0.75T_1)(0.60) &= W \\ T_1(0.80 +0.45) &= W \\ 1.25T_1 &= W \end{aligned} \]
Let the centre of gravity be at distance \(d\) from the left end.
Taking moments about the left end eliminates the torque due to \(T_1\).
Clockwise moment = Anticlockwise moment
\[ T_2\cos53.1^\circ \times 2 = W \times d \]
Substituting values:\[ \begin{aligned} (0.75T_1)(0.60)\times2 &= (1.25T_1)d \\ 0.90T_1 &= 1.25T_1 d \end{aligned} \]
\[ d = \frac{0.90}{1.25} \]
\[ d = 0.72\,\text{m} \]
The centre of gravity of the bar lies
\( \boxed{0.72\ \text{m}} \) from the left end.
Q9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
When a rigid body rests on a horizontal surface, it must satisfy the conditions of static equilibrium.
The reactions of the ground act upward at the axles. Since the centre of gravity is not exactly midway between the axles, the reactions on the front and rear wheels will be different.
Mass of car:
\( m = 1800\,\text{kg} \)
Weight of car:\[ W = mg = 1800 \times 9.8 \]
\[ W = 17640\,\text{N} \]
Distance between axles = \(1.8\,\text{m}\)
Distance of centre of gravity from front axle:
\(1.05\,\text{m}\)
Distance from rear axle:\[ 1.8 - 1.05 = 0.75\,\text{m} \]
\(R_f\) = reaction at front axle \(R_r\) = reaction at rear axle
\[ R_f \times 1.05 = R_r \times 0.75 \]
\[ \frac{R_f}{R_r} = \frac{0.75}{1.05} \]
\[ \frac{R_f}{R_r} = \frac{5}{7} \]
Thus\[ R_f = \frac{5}{7}R_r \]
\[ R_f + R_r = W \]
Substitute \(R_f\):\[ \frac{5}{7}R_r + R_r = 17640 \]
\[ \frac{12}{7}R_r = 17640 \]
\[ R_r = 10290\,\text{N} \]
Therefore\[ R_f = \frac{5}{7} \times 10290 \]
\[ R_f = 7350\,\text{N} \]
Rear wheel reaction: \[ \frac{10290}{2} = 5145\,\text{N} \]
Front wheel reaction: \[ \frac{7350}{2} = 3675\,\text{N} \]
Force on each front wheel = \(3675\,\text{N}\)
Force on each rear wheel = \(5145\,\text{N}\)
Q10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
The rotational analogue of Newton's second law states:
\( \tau = I\alpha \)
whereThus
\( \alpha = \dfrac{\tau}{I} \)
For the same applied torque, the body with the smaller moment of inertia will have the greater angular acceleration.
| Body | Moment of Inertia |
|---|---|
| Hollow cylinder | \( I = MR^2 \) |
| Solid sphere | \( I = \frac{2}{5}MR^2 \) |
Let the same constant torque \( \tau \) act on both bodies.
Angular acceleration of hollow cylinder:\[ \alpha_{cylinder} = \frac{\tau}{MR^2} \]
Angular acceleration of solid sphere:\[ \alpha_{sphere} = \frac{\tau}{\tfrac{2}{5}MR^2} \]
\[ \alpha_{sphere} = \frac{5\tau}{2MR^2} \]
Since\[ \frac{5}{2MR^2} > \frac{1}{MR^2} \]
therefore\[ \alpha_{sphere} > \alpha_{cylinder} \]
If the torque acts for the same time \(t\), the angular speed becomes
\[ \omega = \alpha t \]
Hence\[ \omega_{sphere} > \omega_{cylinder} \]
The solid sphere acquires a greater angular speed than the hollow cylinder.
Mass distributed farther from the axis → larger moment of inertia → smaller angular acceleration.
Q11 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?
For a rigid body rotating about a fixed axis:
Rotational kinetic energy is given by
K = \frac{1}{2}I\omega^2Angular momentum of the rotating body is
L = I\omegaFor a solid cylinder rotating about its axis, the moment of inertia is
\( I = \frac{1}{2}MR^2 \)
Given:
\( M = 20\,\text{kg} \)
\( R = 0.25\,\text{m} \)
\( \omega = 100\,\text{rad s}^{-1} \)
\[ I = \frac{1}{2}MR^2 \]
Substituting values:\[ I = \frac{1}{2}\times 20 \times (0.25)^2 \]
\[ I = 10 \times 0.0625 \]
\[ I = 0.625\,\text{kg m}^2 \]
\[ K = \frac{1}{2}I\omega^2 \]
Substitute values:\[ K = \frac{1}{2}\times 0.625 \times (100)^2 \]
\[ K = 0.3125 \times 10000 \]
\[ K = 3125\,\text{J} \]
\[ L = I\omega \]
Substitute values:\[ L = 0.625 \times 100 \]
\[ L = 62.5\,\text{kg m}^2\text{s}^{-1} \]
Rotational kinetic energy = \(3125\,\text{J}\)
Angular momentum = \(62.5\,\text{kg m}^2\text{s}^{-1}\)
Q12
(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 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?
Since the turntable rotates without friction, no external torque acts on the system. Therefore, the angular momentum of the system remains conserved.
I_1 \omega_1 = I_2 \omega_2The rotational kinetic energy of a rotating body is
K = \frac{1}{2} I \omega^2Initial angular speed:
\( \omega_1 = 40 \,\text{rev min}^{-1} \)
Initial moment of inertia:
\( I_1 \)
When the child folds his arms, the moment of inertia becomes
\( I_2 = \frac{2}{5} I_1 \)
Applying conservation of angular momentum:\[ I_1 \omega_1 = I_2 \omega_2 \]
Substitute \(I_2\):\[ I_1 \omega_1 = \frac{2}{5} I_1 \omega_2 \]
Cancel \(I_1\):\[ \omega_2 = \frac{5}{2} \omega_1 \]
Substitute \(\omega_1 = 40\):\[ \omega_2 = \frac{5}{2} \times 40 \]
\[ \omega_2 = 100 \,\text{rev min}^{-1} \]
\[ K_1 = \frac{1}{2} I_1 \omega_1^2 \]
New kinetic energy:\[ K_2 = \frac{1}{2} I_2 \omega_2^2 \]
Substitute values:\[ K_2 = \frac{1}{2} \left(\frac{2}{5}I_1\right) \left(\frac{5}{2}\omega_1\right)^2 \]
\[ K_2 = \frac{1}{2}I_1\omega_1^2 \times \frac{5}{2} \]
Since\[ K_1 = \frac{1}{2}I_1\omega_1^2 \]
we obtain\[ K_2 = \frac{5}{2} K_1 \]
Thus\(K_2 > K_1\)
New angular speed = \(100\,\text{rev min}^{-1}\)
New kinetic energy = \( \dfrac{5}{2} \) times the initial kinetic energy
Hence the increase in rotational kinetic energy comes from the internal work done by the child.
Q13 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.
When a force is applied tangentially to the rim of a rotating body, it produces a torque about the axis of rotation.
\( \tau = F R \)
The rotational analogue of Newton’s second law is
\tau = I\alphaFor a hollow cylinder rotating about its central axis,
\( I = MR^2 \)
If there is no slipping between the rope and the cylinder, the linear acceleration of the rope is related to the angular acceleration by
\( a = \alpha R \)
Given:
\( M = 3\,\text{kg} \)
\( R = 0.40\,\text{m} \)
\( F = 30\,\text{N} \)
\[ \tau = FR \]
\[ \tau = 30 \times 0.40 \]
\[ \tau = 12\,\text{N m} \]
\[ I = MR^2 \]
\[ I = 3 \times (0.40)^2 \]
\[ I = 3 \times 0.16 \]
\[ I = 0.48\,\text{kg m}^2 \]
\[ \alpha = \frac{\tau}{I} \]
\[ \alpha = \frac{12}{0.48} \]
\[ \alpha = 25\,\text{rad s}^{-2} \]
\[ a = \alpha R \]
\[ a = 25 \times 0.40 \]
\[ a = 10\,\text{m s}^{-2} \]
Angular acceleration of cylinder = \(25\,\text{rad s}^{-2}\)
Linear acceleration of rope = \(10\,\text{m s}^{-2}\)
Q14 To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
In rotational motion, the power delivered by a rotating machine is equal to the product of the torque applied and the angular speed.
This is the rotational analogue of the linear power relation \(P = Fv\).
\( P = \tau \omega \)
whereIf a rotor moves with constant angular speed, the applied torque is only required to balance the opposing frictional torque.
Given:
\( \tau = 180\,\text{N m} \)
\( \omega = 200\,\text{rad s}^{-1} \)
The power required is\[ P = \tau \omega \]
Substitute the given values:\[ P = 180 \times 200 \]
\[ P = 36000\,\text{W} \]
Power required = \(3.6 \times 10^{4}\,\text{W}\)
\(= 36\,\text{kW}\)
Q15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
For composite bodies, the centre of mass can be found using the principle
\( x_{CM} = \dfrac{\sum m_i x_i}{\sum m_i} \)
When a portion of a body is removed, the removed part can be treated as negative mass. This method simplifies centre of mass calculations for bodies with holes.
Since the disc is uniform, mass is proportional to area.
Let the centre of the original disc be the origin \(O\).
Surface mass density = \( \sigma \)
Mass of original disc:\[ M_1 = \sigma \pi R^2 \]
Mass of removed hole:\[ M_2 = \sigma \pi \left(\frac{R}{2}\right)^2 \]
\[ M_2 = \frac{1}{4}\sigma \pi R^2 \]
Remaining mass:
\[ M = M_1 - M_2 \]
\[ M = \sigma \pi R^2 - \frac{1}{4}\sigma \pi R^2 \]
\[ M = \frac{3}{4}\sigma \pi R^2 \]
The centre of the original disc lies at
\( x_1 = 0 \)
The centre of the removed hole lies at
\( x_2 = \frac{R}{2} \)
Treat the removed hole as negative mass.\[ x_{CM} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2} \]
Substitute values:\[ x = \frac{(0)(\pi R^2) - \left(\frac{\pi R^2}{4}\right)\left(\frac{R}{2}\right)} {\frac{3}{4}\pi R^2} \]
\[ x = -\frac{\frac{\pi R^3}{8}}{\frac{3}{4}\pi R^2} \]
\[ x = -\frac{R}{6} \]
The centre of gravity of the remaining body lies
\( \dfrac{R}{6} \)
from the centre of the original disc, along the line joining the centres and away from the hole.
Q16 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?
A body remains balanced when the net torque about the pivot is zero.
\( \sum \tau = 0 \)
Thus, clockwise moments must equal anticlockwise moments about the pivot point.
For a uniform metre stick, the centre of gravity lies at the
\(50\,\text{cm}\) mark.
Mass of each coin:
\(5\,\text{g}\)
Total mass of two coins:
\(10\,\text{g}\)
Let the mass of the metre stick be
\(M\)
Pivot point:
\(45\,\text{cm}\)
\[ 45 - 12 = 33\,\text{cm} \]
### Distance of Stick's Centre of Gravity\[ 50 - 45 = 5\,\text{cm} \]
Clockwise moment = Anticlockwise moment
\[ 10 \times 33 = M \times 5 \]
\[ 330 = 5M \]
\[ M = 66\,\text{g} \]
Mass of the metre stick = \(66\,\text{g}\)
Q17 The oxygen molecule has a mass of \(5.30 \times 10^{-26}\,\text{kg}\) and a moment of inertia of \(1.94 \times 10^{-46}\,\text{kg m}^2\) 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 \(500\,\text{m s}^{-1}\) and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
The translational kinetic energy of a molecule moving with speed \(v\) is
\(K_{\text{trans}} = \dfrac{1}{2}mv^2\)
The rotational kinetic energy of a rotating molecule is
\(K_{\text{rot}} = \dfrac{1}{2}I\omega^2\)
The problem states that
\(K_{\text{rot}} = \dfrac{2}{3}K_{\text{trans}}\)
Given:
\( m = 5.30 \times 10^{-26}\,\text{kg} \)
\( I = 1.94 \times 10^{-46}\,\text{kg m}^2 \)
\( v = 500\,\text{m s}^{-1} \)
\[ K_{\text{trans}} = \frac{1}{2}mv^2 \]
Since\[ K_{\text{rot}} = \frac{2}{3}K_{\text{trans}} \]
then\[ K_{\text{rot}} = \frac{2}{3}\left(\frac{1}{2}mv^2\right) \]
\[ K_{\text{rot}} = \frac{1}{3}mv^2 \]
\[ K_{\text{rot}} = \frac{1}{2}I\omega^2 \]
Equating the two expressions:\[ \frac{1}{2}I\omega^2 = \frac{1}{3}mv^2 \]
Multiply both sides by 2:\[ I\omega^2 = \frac{2}{3}mv^2 \]
\[ \omega^2 = \frac{2mv^2}{3I} \]
\[ \omega^2 = \frac{2 \times 5.30 \times 10^{-26} \times (500)^2} {3 \times 1.94 \times 10^{-46}} \]
\[ \omega^2 = \frac{2 \times 5.30 \times 10^{-26} \times 2.5 \times 10^{5}} {5.82 \times 10^{-46}} \]
\[ \omega^2 \approx 4.55 \times 10^{25} \]
\[ \omega = \sqrt{4.55 \times 10^{25}} \]
\[ \omega \approx 6.7 \times 10^{12}\,\text{rad s}^{-1} \]
Average angular velocity of the oxygen molecule
\( \boxed{\omega \approx 6.7 \times 10^{12}\,\text{rad s}^{-1}} \)
This quick revision sheet summarizes all the most important formulas from the chapter System of Particles and Rotational Motion. These formulas are frequently used in JEE, NEET, Olympiads, and board examinations.
\( x_{CM} = \frac{\sum m_i x_i}{\sum m_i} \)
\( y_{CM} = \frac{\sum m_i y_i}{\sum m_i} \)
\( \vec{R}_{CM} = \frac{\sum m_i \vec{r_i}}{\sum m_i} \)
\( M\vec{a}_{CM} = \vec{F}_{ext} \)
\( \vec{L} = \vec{r} \times \vec{p} \)
\( \vec{L} = I\vec{\omega} \)
\( \vec{\tau} = \frac{d\vec{L}}{dt} \)
\( \tau = I\alpha \)
\( \omega = \omega_0 + \alpha t \)
\( \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \)
\( \omega^2 = \omega_0^2 + 2\alpha\theta \)
\( v = r\omega \)
\( a_t = r\alpha \)
\( K = \frac{1}{2}I\omega^2 \)
\( W = \tau \theta \)
\( P = \tau \omega \)
\( v = R\omega \)
\( a = R\alpha \)
\( K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)
Ring: \(I = MR^2\)
Disc/Cylinder: \(I = \frac{1}{2}MR^2\)
Solid Sphere: \(I = \frac{2}{5}MR^2\)
Thin Rod (centre): \(I = \frac{1}{12}ML^2\)
Thin Rod (end): \(I = \frac{1}{3}ML^2\)
Even strong students lose marks in rotational motion because of a few common conceptual mistakes. Avoiding the following traps can significantly improve performance in JEE, NEET, Olympiads and board exams.
This visual mind map summarizes the most important ideas from System of Particles and Rotational Motion. Use it for quick revision before exams.
Review the most important concepts from System of Particles & Rotational Motion in just 30 seconds. Perfect for quick revision before tests and competitive exams.
\( \tau = rF\sin\theta \)
\( \tau = I\alpha \)
\( \vec{L} = \vec{r} \times \vec{p} \)
\( \vec{L} = I\omega \)
\( K = \frac12 I\omega^2 \)
\( P = \tau \omega \)
\( W = \tau \theta \)
\( v = R\omega \)
\( a = R\alpha \)
\( K = \frac12 mv^2 + \frac12 I\omega^2 \)
Ring
\(I = MR^2\)
Disc
\(I = \frac12 MR^2\)
Solid Sphere
\(I = \frac25 MR^2\)
Rod (centre)
\(I = \frac{1}{12}ML^2\)
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