Systems of Particles and Rotational Motion
Key Concepts:
Center of Mass:
A point representing the average position of a system’s mass.
Its motion follows Newton’s laws.
Position vector: R = (Σmᵢrᵢ) / Σmᵢ
Linear Momentum:
Total linear momentum: P = Σmᵢvᵢ = MV (M: total mass, V: velocity of center of mass)
Conservation: If no external force acts, total linear momentum remains constant.
Angular Momentum:
L = r × p (r: position vector, p: linear momentum)
Conservation: If no external torque acts, total angular momentum remains constant.
Kinetic Energy:
Total KE = KE of center of mass + KE of particles relative to center of mass
Moment of Inertia (I):
Measures resistance to rotational motion.
For a point mass: I = mr²
Rotational Kinetic Energy:
KE_rotational = (1/2)Iω² (ω: angular velocity)Torque (τ):
Time rate of change of angular momentum is Torque: τ = dL/dt
Rotational analog of force: τ = Iα (α: angular acceleration)
Key Points:
The center of mass moves as if all mass were concentrated there.
Conservation laws for linear momentum and angular momentum are fundamental.
Moment of inertia is crucial for rotational motion.
Torque is the rotational equivalent of force.
Understanding these concepts helps analyze the motion of various systems, from simple objects to celestial bodies.
1. 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 ?
Ans : (i) Sphere: The center of mass of a sphere of uniform mass density is at its geometric center.
(ii) Cylinder: The center of mass of a cylinder of uniform mass density lies on its axis of symmetry, at the midpoint of the axis.
(iii) Ring: The center of mass of a ring of uniform mass density lies at its geometric center.
(iv) Cube: The center of mass of a cube of uniform mass density lies at its geometric center.
No, the center of mass of a body does not necessarily lie inside the body. For example, the center of mass of a ring lies at its center, which is outside the material of the ring.
2. 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.
Ans : Let’s denote the mass of the hydrogen atom as m and the distance between the two nuclei as r.
Then, the mass of the chlorine atom is approximately 35.5m.
Given,
(CM): center of mass of a two-particle system
r_cm = (m₁r₁ + m₂r₂) / (m₁ + m₂)
In this case:
* m₁ is mass of hydrogen atom [m]
* r₁ = 0 (assuming the hydrogen atom is at the origin)
* m₂ = 35.5m (mass of chlorine atom)
* r₂ = r = 1.27 Å = 1.27 × 10⁻¹⁰ m
Substituting these values into the equation:
r_cm = (m * 0 + 35.5m * 1.27 × 10⁻¹⁰ m) / (m + 35.5m)
= (35.5 * 1.27 × 10⁻¹⁰ m) / 36.5
≈ 1.24 × 10⁻¹⁰ m
Therefore, the center of mass of the HCl molecule is approximately 1.24 Å from the hydrogen atom, closer to the chlorine atom.
3. 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 ?
Ans : The speed of the center of mass (CM) of the (trolley + child) system remains constant at V.
This is because there is no external horizontal force acting on the system. The internal forces, like the child running on the trolley, do not affect the motion of the center of mass. Therefore, the CM continues to move with the same initial velocity V.
4. Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
Ans : Let’s consider two vectors, a and b, forming the sides of a triangle.
Find area of the triangle by using formula:
Area = (1/2) * base * height
Let’s take the magnitude of vector a as the base of the triangle. The height of the triangle can be represented by the component of vector b perpendicular to vector a. This component can be found using the cross product of vectors a and b:
|a × b| = |a| |b| sinθ
where,
θ is the angle formed by vectors a and b.
The height of the triangle, h = |b| sinθ
Therefore, the area of the triangle:
Area = (1/2) * |a| * |b| sinθ
Comparing this with the magnitude of the cross product:
|a × b| = |a| |b| sinθ
We can see that:
Area = (1/2) * |a × b|
Hence, the area of the triangle formed by vectors a and b is half the magnitude of their cross product.
5. Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c.
Ans : Let’s consider the parallelepiped formed by vectors a, b, and c. The volume of this parallelepiped can be determined by multiplying its base area by its height.
The base of the parallelepiped is a parallelogram formed by vectors a and b. The area of this parallelogram is given by the magnitude of the cross product of a and b:
Area of base = |a × b|
The height of the parallelepiped is the component of vector c perpendicular to the plane formed by vectors a and b. This component can be found by taking the dot product of vector c with the unit vector perpendicular to the plane of a and b.
This unit vector is given by
(a × b)
————
|a × b|.
Height of parallelepiped =
c . (a × b)
————-
|a × b|
Therefore, the volume of the parallelepiped is:
Volume = Base Area × Height
= |a × b| * c . (a × b) / |a × b|
= c . (a × b)
Hence, the volume of the parallelepiped is equal to the scalar triple product a.(b × c).
6. Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px , py and pz . Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Ans : in component form:
l = (yi – xj + zk) × (p_xi + p_yj + p_zk)
Using the properties of the cross product, we get:
l = (yp_z – zp_y)i + (zp_x – xp_z)j + (xp_y – yp_x)k
Therefore, the components of the angular momentum along the x, y, and z axes are:
* l_x = yp_z – zp_y
* l_y = zp_x – xp_z
* l_z = xp_y – yp_x
If the particle moves only in the x-y plane, then z = 0 and p_z = 0.
Substituting these values into the expressions for l_x, l_y, and l_z, we get:
* l_x = 0
* l_y = 0
* l_z = xp_y – yp_x
Thus, the angular momentum has only a z-component when the particle moves in the x-y plane.
7. 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.
Ans : Let’s consider three points: A, B, and C, lying on the line connecting the two particles.
Angular momentum about A:
* The particle at A has zero distance from point A and hence zero angular momentum.
* The particle at B has a distance of d from point A. Its linear momentum is mv, and its angular momentum about A is given by L_A = mvd.
Angular momentum about B:
* The particle at B has zero distance from point B and hence zero angular momentum.
* The particle at A has a distance of d from point B. Its linear momentum is mv, and its angular momentum about B is given by L_B = mvd.
Angular momentum about C:*
The particle at A has a distance of y from point C. Its linear momentum is mv, and its angular momentum about C is given by L_C1 = mvy.
* The particle at B has a distance of (d-y) from point C. Its linear momentum is mv, and its angular momentum about C is given by L_C2 = mv(d-y).
* The total angular momentum about C is L_C = L_C1 + L_C2 = mvy + mv(d-y)
= mvd.
ConclusionIn all three cases, the magnitude of the angular momentum is mvd, and the direction is perpendicular to the plane containing the two particles and the chosen point. Therefore, the angular momentum of the two-particle system is the same, regardless of the point chosen as the reference point.
8. 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.
Ans : Let’s denote the tensions in the two strings as T₁ and T₂, respectively.
For the bar to be in equilibrium, the net torque about any point must be zero.
Let’s take the left end of the bar as the reference point for calculating torques.
Torque equation:
T₁ * 2 * sin(36.9°) = T₂ * 2 * sin(53.1°) + W * d
Force balance equation:
T₁ * cos(36.9°) + T₂ * cos(53.1°) = W
By solving these two equations simultaneously, we can eliminate T₁ and T₂ to obtain an equation for d. After simplification, we find that:
d = 0.745 m
Therefore, the center of gravity of the bar is located 0.745 meters from the left end.
9. 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.
Ans : Let F₁ and F₂ be the forces exerted by the ground on each front wheel and each back wheel, respectively.
Considering rotational equilibrium about the front axle
The torque due to the weight of the car (mg) about the front axle must be balanced by the torque due to the forces on the back wheels.
1.8F₂ = 1.05mg
F₂ = (1.05/1.8)mg = (1.05/1.8) × 1800 kg × 9.8 m/s²
= 10290 N
Considering rotational equilibrium about the back axle:
Similarly, the torque due to the weight of the car about the back axle must be balanced by the torque due to the forces on the front wheels.
1.8F₁ = 0.75mg
F₁ = (0.75/1.8)mg = (0.75/1.8) × 1800 kg × 9.8 m/s² = 7350 N
Therefore, the force exerted by the ground on each front wheel is 7350/2 N = 3675 N, and the force exerted on each back wheel is 10290/2 N = 5145 N.
10. 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.
Ans : The torque applied to both objects is the same. We know that torque (τ) is related to angular acceleration (α) by the equation:
τ = Iα
where I is the moment of inertia.
For a hollow cylinder of mass M and radius R, the moment of inertia about its axis of symmetry is:
I_cylinder = MR²
For a solid sphere of mass= M
radius = R,
Given, moment of inertia about its central axis
I_sphere = (2/5)MR²
Since I_sphere < I_cylinder, for the same torque, the solid sphere will have a greater angular acceleration.
Therefore, after a given time, the solid sphere will acquire a greater angular speed.
11. 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?
Ans : Kinetic Energy of Rotation:**
Given,
kinetic energy of a rotating object
KE = (1/2)Iω²
where:
* I is the moment of inertia
* ω is the angular speed
The moment of inertia of a solid cylinder rotating about its axis is given by:
I = (1/2)MR²
where:
* M = mass of the cylinder
* R = radius of the cylinder
Substituting the given values:
I = (1/2)(20 kg)(0.25 m)²
= 0.625 kg m²
Now, calculating the kinetic energy:
KE = (1/2)(0.625 kg m²)(100 rad/s)²
= 3125 J
Angular Momentum:
Given,
rotational momentum of a rotating object
L = Iω
Substituting the values:
L = (0.625 kg m²)(100 rad/s)
= 62.5 kg m²/s
Therefore, the kinetic energy associated with the rotation of the cylinder is 3125 J, and the magnitude of its angular momentum is 62.5 kg m²/s.
12. (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?
Ans : (a) Conservation of Angular Momentum:
Since there is no external torque acting on the system (child + turntable), angular momentum is conserved.
Initial angular momentum = Final angular momentum
I₁ω₁ = I₂ω₂
Where:
* I₁ = initial moment of inertia
* ω₁ = initial angular speed (40 rev/min)
* I₂ = final moment of inertia (2/5 * I₁)
* ω₂ = final angular speed
Substituting the values:
I₁ * 40 rev/min = (2/5)I₁ * ω₂
Solving for ω₂, we get:
ω₂ = 100 rev/min
Therefore, the final angular speed of the child is 100 rev/min.
(b) Kinetic Energy:
Initial kinetic energy, KE₁ = (1/2)I₁ω₁²
Final kinetic energy, KE₂ = (1/2)I₂ω₂²
Substituting I₂ = (2/5)I₁ and ω₂ = 100 rev/min:
KE₂ = (1/2)(2/5)I₁(100)² = (2/5)I₁(100)²
Comparing KE₁ and KE₂, we see that KE₂ > KE₁.
This increase in kinetic energy comes from the work done by the child in pulling his arms inward. The child exerts a torque on the turntable, which increases the angular speed and hence the kinetic energy of the system.
13. 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.
Ans : Given:
Mass of the cylinder: 3 kg
Radius of the cylinder: 40 cm (0.4 m)
Applied force on the rope: 30 N
To find:
Angular acceleration, α
Linear acceleration, a
1. Angular Acceleration:
* Torque, τ = Force × Perpendicular distance from the axis of rotation = F × R
* Moment of inertia of a hollow cylinder, I = MR²
We know that:
τ = Iα
Substituting the values:
F × R = MR²α
Solving for α:
α = (F × R) / (MR²) = F / (MR)
Plugging in the numbers:
α = 30 N / (3 kg × 0.4 m)
= 25 rad/s²
2. Linear Acceleration:
For a point on the rim of the cylinder, linear acceleration (a) is related to angular acceleration (α) by:
a = Rα
Substituting the values:
a = 0.4 m × 25 rad/s² = 10 m/s²
Therefore, the angular acceleration of the cylinder is 25 rad/s², and the linear acceleration of the rope is 10 m/s².
14. 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.
Ans : We are given:
* Angular speed, ω = 200 rad/s
* Torque, τ = 180 Nm
Power:
Power (P) = rate of work done, the rate at which energy is transferred.
Given,
power (In rotational motion,)
P = τω
Substituting the given values:
P = 180 Nm × 200 rad/s
= 36000 W
Hence, the power required by the engine is 36,000 Watts.
15. 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.
Ans : Let’s consider a uniform disk of radius R with center O. We cut out a smaller circular hole of radius R/2 with its center at a distance of R/2 from the center of the larger disk. Let the center of mass of the remaining flat body be at point O2, and let the distance OO2 be x.
If σ is the mass per unit area of the disk, then:
* Mass of the original disk (M1) = πR²σ
* Mass of the removed portion (M2) = π(R/2)²σ = (1/4)πR²σ = M1/4
To find the position of the center of mass O2, we can use the principle of moments:
M1 * 0 = M2 * (R/2) + (M1 – M2) * x
Solving for x:
x = -(M2 * R/2) / (M1 – M2) = -(M1/4 * R/2) / (M1 – M1/4)
= -R/6
Therefore, the center of mass of the remaining flat body (O2) is located at a distance of R/6 from the center of the original disk (O) in the direction opposite to the center of the hole.
16. 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?
Ans : Let, M = mass of the meter stick.
The meter stick is balanced at 45 cm, so its center of mass is at 50 cm.
When the two coins (total mass 10 g) are placed at 12 cm, the center of mass of the system shifts to 45 cm.
To find the mass of the meter stick, we can use the principle of rotational equilibrium.
(Mass of meter stick × Distance of its center of mass from the pivot) = (Mass of coins × Distance of coins from the pivot)
M × (50 – 45) cm = 10 g × (45 – 12) cm
M × 5 cm = 10 g × 33 cm
M = 66 g
Therefore, the mass of the meter stick is 66 grams.
17. 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 lines 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.
Ans : Given
* Mass of the oxygen molecule, m = 5.30 × 10⁻²⁶ kg
* Moment of inertia,
I is 1.94 × 10⁻⁴⁶ kg m²
* Mean speed, v = 500 m/s
To find
* Average angular velocity, ω
Solution:
We are given that the rotational kinetic energy is two-thirds of the translational kinetic energy:
(1/2)Iω² = (2/3)(1/2)mv²
Simplifying and substituting the given values:
ω² = (2/3)(mv²)/I
ω = √[(2/3)(mv²)/I]
Calculating ω:
ω = √[(2/3)(5.30 × 10⁻²⁶ kg)(500 m/s)² / (1.94 × 10⁻⁴⁶ kg m²)]
ω ≈ 6.67 × 10¹² rad/s
Therefore, the average angular velocity of the oxygen molecule is approximately 6.67 × 10¹² rad/s.
