SYSTEM OF PARTICLES AND ROTATORY MOTION – PART 2

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Comparison of moment of inertia with mass

In linear motion, greater in the mass of the body, greater is the force required to produce the linear acceleration in it. Thus in linear motion mass of the body is a measure of its inertia. In rotational motion a torque is applied to produce angular acceleration. Moment of inertia is a measure of rotational inertia of the body. In rotation the moment of Inertia plays a similar role as mass does in the linear motion.

Fly wheel

A fly wheel is a circular disc, whose most of the mass is concentrated on its rim and it rotates about an axel passing through its centre and perpendicular to its plane. The machines such as steam engine and automobile engine that produce rotational motion have a fly wheel.

Working of Fly wheel

Jerky motion of a vehicle can be prevented by attaching a fly wheel with its engine. Since the most of the mass of the fly wheel is concentrated at its rim, the fly wheel has large moment of inertia. Therefore fly wheel opposes or resists the rotational motion to a great extent. Whenever there is a sudden increase or decrease in the speed of vehicle, the fly wheel opposes this sudden increase or decrease in the speed of vehicle due to its large moment of Inertia.

Factors on which moment of inertia depends

Moment of inertia depends on, (i) Position of the axis of rotation with respect to the body. (ii) Orientation of the axis of rotation. (iii) Mass of the rotating body. (iv) Distance from the axis of rotation to the distribution of mass.

Radius of gyration

It the distance of a mass point from the axis of rotation, whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the whole body about the axis.

Explanation: The moment of inertia of the body about the axis of rotation is given by,

If m is the mass of the each particle,then,

𝐼 = π‘š(π‘Ÿ12 + π‘Ÿ22 + π‘Ÿ32+. … … … + π‘Ÿπ‘›2)

If the body contains n particles, then by multiplying and dividing RHS by n,

𝐼 = π‘šπ‘› [(π‘Ÿ12 + π‘Ÿ22 + π‘Ÿ32+. … … … + π‘Ÿπ‘›2)/𝑛]

𝐼 = 𝑀[(π‘Ÿ12 + π‘Ÿ22 + π‘Ÿ32+. … … … + π‘Ÿπ‘›2)/𝑛]

𝑰 = π‘΄π’ŒπŸ

k is called radius of gyration and is given by, π’Œ = [(π‘Ÿ12 + π‘Ÿ22 + π‘Ÿ32+. … … … + π‘Ÿπ‘›2)/𝑛]1/2

 SI unit of radius of gyration is metre(m).

Theorems of perpendicular and parallel axis

These are two useful theorems, which are used to find the moment of inertia of regular shaped bodies about any axis of rotation if the value of the moment of inertia of the given body is known about a certain axis of rotation of the body.

Theorem of perpendicular axis

It states that the moment of inertia of a planar body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

Explanation: Let X, Y and Z are the mutually perpendicular axes. Consider an object in X-Y plane. Let 𝐼𝑧 β†’ moment of inertia of the body about 𝑍 βˆ’ π‘Žπ‘₯𝑖𝑠

𝐼π‘₯ β†’ moment of inertia of the body about X – axis

𝐼𝑦 β†’ moment of Inertia of the body about Y-axis

Then 𝑰𝒛 = 𝑰𝒙 + π‘°π’š

Proof: Consider a particle of mass m at point P (x,y) at a distance r from O. The moment of inertia of the particle about Z-axis is 𝐼𝑧 = π‘šπ‘Ÿ2

The moment of inertia of the whole body with respect to z-axis is 𝐼𝑧 = βˆ‘π‘šπ‘–π‘Ÿπ‘–2 Similarly, the moment of inertia of the body with respect to x-axis is 𝐼π‘₯ = βˆ‘π‘šπ‘–π‘₯𝑖2 and with respect to 𝑦 βˆ’ π‘Žπ‘₯𝑖𝑠 is 𝐼𝑦 = βˆ‘π‘šπ‘–π‘¦π‘–2

Now 𝐼𝑧 = βˆ‘π‘šπ‘–π‘Ÿπ‘–2

𝐼𝑧 = βˆ‘π‘šπ‘–(π‘₯𝑖2 + 𝑦𝑖2) ∡ π‘Ÿ2 = π‘₯2 + 𝑦2

𝐼𝑧 = βˆ‘π‘šπ‘–π‘Ÿπ‘–2 +βˆ‘π‘šπ‘–π‘¦π‘–2

𝑰𝒛 = 𝑰𝒙 + π‘°π’š

Note: This theorem is applicable to the bodies which are planar or flat whose thickness is very small compared to their other dimensions.

Theorem of parallel axes

It states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

Explanation: Let Z’ be the axis about which the moment of inertia of the body is to be calculated. Z is the axis passing through the centre of mass of the body and parallel to Z’. Let a be the distance between the two axes. Then,

𝑰𝒁 β€² = 𝑰𝒛 + π‘΄π’‚πŸ

Proof: Consider a particle at P of mass m whose distance from Z axis is y. Moment of inertia of the particle about Z axis is = my2 The moment of Inertia of the body about 𝑍 is 𝐼𝑧 = βˆ‘π‘šπ‘–π‘¦π‘–2

The moment of inertia of the particle about Z’ is = π‘š(𝑦𝑖 βˆ’ π‘Ž)2

The moment of inertia of the body about 𝑍’ is given by, 𝐼𝑧′ = βˆ‘π‘šπ‘–(𝑦𝑖 βˆ’ π‘Ž)2

𝐼𝑧 = βˆ‘π‘šπ‘–(𝑦𝑖2 + π‘Ž2 βˆ’ 2π‘¦π‘–π‘Ž)

𝐼𝑧 = βˆ‘π‘šπ‘–π‘¦π‘–2 + βˆ‘π‘šπ‘–π‘Ž2 βˆ’βˆ‘π‘šπ‘–2π‘¦π‘–π‘Ž

𝐼𝑧 = βˆ‘π‘šπ‘–π‘¦π‘–2 + π‘Ž2βˆ‘π‘šπ‘– βˆ’ 2π‘Žβˆ‘π‘šπ‘–π‘¦π‘–

The centre of mass is at the origin then, π‘…π‘π‘š = 0 and π‘…π‘π‘š = (1/𝑀)βˆ‘π‘šπ‘–π‘¦π‘–

0 = (1/𝑀)βˆ‘π‘šπ‘–π‘¦π‘– Since 𝑀 β‰  0, βˆ‘π‘šπ‘–π‘¦π‘– = 0

Substituting in the above equation, 𝐼𝑧′ = 𝐼𝑧 + π‘€π‘Ž2 βˆ’ 2π‘Ž(0)

𝑰𝒛′ = 𝑰𝒛 + π‘΄π’‚πŸ

Note: This theorem is applicable to a body of any shape.

Kinematics of Rotational motion about a fixed axis

The kinematical quantities in rotational motion, angular displacement (πœƒ), angular velocity (πœ”) and anglar acceleration (𝛼) respectively correspond to kinematic quantities in linear motion displacement (x), velocity (v) and acceleration (a). They are,

πœ” = πœ”0 + 𝛼𝑑

πœƒ βˆ’ πœƒ0 = πœ”0𝑑 + (1/2)𝛼𝑑2

πœ”2 = πœ”02 + 2𝛼(πœƒ βˆ’ πœƒ0)

Dynamics of rotational motion about a fixed axis

Consider a cross–section of a rigid body rotating about a fixed axis, perpendicular the plane of the paper. Consider a particle at P in x-y plane which describes a circular path of radius r with centre at C. In time βˆ†t the particle moves to the position P’ due to application of force 𝐹. (i) Work done by the force is, π‘‘π‘Š = 𝐹 βˆ™ 𝑑𝑠

π‘‘π‘Š = 𝐹𝑑𝑠 cosπœ™

π‘‘π‘Š = 𝐹(π‘Ÿπ‘‘πœƒ) cosπœ™

π‘‘π‘Š = (π‘Ÿπ‘‘πœƒ)𝐹 cosπœ™

Now πœ™ + 𝛽 = 90Β°, because displacement vector is in the direction of tangent to the circular path.

By resolving component of 𝐹 with respect to r and with respect to displacement vector as shown in figure, we get, 𝐹 cosπœ™ = 𝐹 sin𝛽

∴ π‘‘π‘Š = (π‘Ÿπ‘‘πœƒ) 𝐹 sin𝛽

π‘‘π‘Š = ( π‘ŸπΉ sin 𝛽)π‘‘πœƒ 𝑑

π‘Š = (π‘Ÿ Γ— 𝐹 )π‘‘πœƒ (∡ π‘Ÿ Γ— 𝐹 = π‘ŸπΉ sin 𝛽)

Total work done is, 𝒅𝑾 = 𝝉 π’…πœ½

This equation is similar to the expression π‘‘π‘Š = 𝐹𝑑𝑠

(ii) Now dividing the equation π‘‘π‘Š = 𝜏 π‘‘πœƒ throughout by dt.

π‘‘π‘Š/𝑑𝑑 = 𝜏 π‘‘πœƒ/𝑑𝑑

𝑷 = π‰πŽ

Where π‘‘π‘Š/𝑑𝑑 = 𝑃 (rate of doing work = power)

(iii) The rate at which the work is done on the body is equal to rate at which kinetic energy changes.

π‘‘π‘Š/𝑑𝑑 = 𝑑/𝑑𝑑(𝐾𝐸)

𝑃 = 𝑑/𝑑𝑑(πΌπœ”2/2)

πœπœ” = 𝐼(2πœ”/2)(π‘‘πœ”/𝑑𝑑)

πœπœ” = πΌπœ”π›Ό

𝝉 = π‘°πœΆ which is similar to F = ma

Angular momentum in case of rotation about of fixed axis

Consider a rigid body of mass M rotating with an angular velocity πœ” along z-axis. The rigid body is made up of large number of elements. Consider one such element of mass mi whose position vector is π‘Ÿπ‘– and linear momentum is 𝑝𝑖. The angular momentum of this element about the axis of rotation is given by,

𝑙𝑖 = π‘Ÿπ‘– Γ— 𝑝𝑖 = (π‘Ÿπ‘ sin 90Β°)π‘˜Μ‚

𝑙𝑖 = (π‘Ÿπ‘–π‘π‘–)π‘˜Μ‚

𝑙𝑖 = (π‘Ÿπ‘–π‘šπ‘–π‘£π‘–)π‘˜Μ‚

𝑙𝑖 = (π‘Ÿπ‘–π‘šπ‘–π‘Ÿπ‘–πœ”)π‘˜Μ‚

𝑙𝑖 = (π‘šπ‘–π‘Ÿπ‘–2πœ”)π‘˜Μ‚

The total angular momentum is given by, 𝐿 = βˆ‘(π‘šπ‘–π‘Ÿπ‘–2πœ”)π‘˜Μ‚

𝐿 = (πœ”π‘˜Μ‚)βˆ‘π‘šπ‘–π‘Ÿπ‘–2

𝐿 = (πœ”π‘˜Μ‚)𝐼

𝑳 = (π‘°πŽ)π’ŒΜ‚ or 𝑳 = π‘°πŽ

Law of conservation of angular momentum

In the absence of external torque, the net angular momentum of the system is conserved.

Explanation: We have 𝐿 = πΌπœ”

Differentiating both sides with respect to t

𝑑𝐿/𝑑𝑑 = 𝑑/𝑑𝑑(πΌπœ”)

But 𝑑𝐿/𝑑𝑑 = πœπ‘›π‘’π‘‘ ∴ 𝑑/𝑑𝑑(πΌπœ”) = πœπ‘›π‘’π‘‘

If the net external torque acting on that body is zero then, 𝑑/𝑑𝑑(πΌπœ”) = 0

π‘°πŽ = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕

This is the law of conservation of angular momentum.

Note: If the M.I of the body changes from 𝐼1 to 𝐼2 due to the change of the distribution of mass of the body, then angular velocity of the body changes from πœ”1 to πœ”2 such that, 𝐼1πœ”1 = 𝐼2πœ”2 or 𝐼1πœ”1 = 𝐼2πœ”2

Illustrations for conservation of Angular momentum

1. Suppose a man is sitting on a rotating table with his arms stretched outward. When the man with draws his arms towards his chest, the moment of inertia of the man decreases. Hence his angular speed increases.

2. A ballet dancer varies her angular speed by stretching her legs and arms out or in. As she stretches her legs and arms out, her moment of inertia increases and angular speed decreases.

3. When a diver Jumps from the spring board he curls his body by rolling his arms and legs in, by doing so he decreases his moment of inertia and hence angular speed increases.

4. When a planet revolving around the sun in an elliptical orbit comes near the sun, its speed increases. This is because as the planet comes near the sun, its moment of inertia decreases and hence its angular velocity increases.

Rolling Motion

The combination of Rotational motion (without slipping) and the translational motion of a rigid body is known as rolling motion. Consider a spherical rigid body of radius R rolling over a horizontal surface. Let πœ” be the angular velocity of the rigid body about the axis of rotation. When this rigid body rolls over a smooth horizontal surface it has two types of motions simultaneously,

(i) The rotational motion about its centre of mass.

(ii) Translational motion of the centre of mass of the body.

The body will roll over the surface without slipping if the point of contact (say A) of the body with the surface is at rest at any contact with respect to the centre of mass. The magnitude of linear velocity, π‘£π‘π‘š = π‘…πœ” at A. The velocity of the point B with respect to centre of mass = πœ”π‘… + π‘£π‘π‘š = 2πœ”π‘… or = 2π‘£π‘π‘š

The velocity of the top most point on the body is maximum with respect to the centre of mass of the body.

Kinetic energy of rolling motion

The kinetic energy of a system of particles (K) can be separated into the kinetic energy of motion of the centre of mass (1/2)π‘šπ‘£π‘π‘š2 and kinetic energy of rotational motion about the centre of mass of the system of particles (1/2)πΌπœ”2  

𝐾 = (1/2)π‘šπ‘£π‘π‘š2 + (1/2)πΌπœ”2  

Where π‘š ⟢ mass of the body

π‘£π‘π‘š ⟢ velocity of centre of mass

𝐼 ⟢ moment of Inertia of the body

πœ” ⟢ angular velocity of the body

𝐾 = (1/2)π‘šπ‘£π‘π‘š2 + (1/2)(π‘šπ‘˜2)(π‘£π‘π‘š2𝑅2)

𝐾 = (1/2)π‘šπ‘£π‘π‘š2 + (1/2)π‘šπ‘£π‘π‘š2π‘˜2𝑅2

𝐾 = (1/2)π‘šπ‘£π‘π‘š2(1 + π‘˜2𝑅2)

This equation applies to any rolling body: a disc, a cylinder a ring or a sphere.

Problems

1. Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g and 200g respectively. Each side of the equilateral triangle is 0.5m long.

Co – ordinates of the mass m1 are, π‘₯1 = 0, 𝑦1 = 0.

Co-ordinates of the mass m2 are, π‘₯2 = 0.5, 𝑦2 = 0.

Co-ordinates of the mass m3 are, π‘₯3 = 0.5/2 = 0.25, 𝑦3 = (√3/2)(0.5) = 0.25√3

Let (π‘‹π‘π‘š, π‘Œπ‘π‘š) be the co-ordinates of the centre of mass

π‘‹π‘π‘š = (π‘š1π‘₯1 + π‘š2π‘₯2 + π‘š3π‘₯3)/(π‘š1 + π‘š2 + π‘š3) = (100(0) + 150(0.5) + 200(0.25))/(100 + 150 + 200𝑔) = (75 + 50)/450 = 125/450 = 5/18.

π‘Œπ‘π‘š = (π‘š1𝑦1 + π‘š2𝑦2 + π‘š3𝑦3)/(π‘š1 + π‘š2 + π‘š3) = [100(0) + 150(0) + 200(0.25√3)]/ (100 + 150 + 200𝑔) = (50√3)/450 = √3/9 = 1/3√3

(π‘‹π‘π‘š, π‘Œπ‘π‘š) = (5/18, 1/3√3 ) π‘œπ‘Ÿ 𝑅 = 5/18 𝑖+ 1/3√3𝑗

2. Find the centre of mass of a uniform L-shaped lamina with dimensions as shown. The mass of the lamina is 3kg.

Now we can divide the Lamina in to 3 squares of each length 1m and the mass of each lamina is 1kg, since it is uniform. Then C1, C2, C3 are the centre of masses of each square and the coordinates are 𝐢1(1/2, 3/2) , 𝐢2(1/2, 1/2) and 𝐢3(3/2, 1/2) because the geometric centres of the square is the centre of mass of the square. Now the centre of mass of lamina is the centre of mass of these three masses.

∴ π‘‹π‘π‘š = (π‘š1π‘₯1 + π‘š2π‘₯2 + π‘š3π‘₯3)/(π‘š1 + π‘š2 + π‘š3) = [1(1/2) + 1(1/2) + 1(3/2)]/(1 + 1 + 1)= (5/2)/3 = 5/6

π‘Œπ‘π‘š = (π‘š1y1 + π‘š2y2 + π‘š3y3)/(π‘š1 + π‘š2 + π‘š3) =  [1(3/2) + 1(1/2) + 1(1/2)]/(1 + 1 + 1) = 5/6

The co-ordinates of the centre of mass of L-shaped Lamina is,

(π‘‹π‘π‘š, π‘Œπ‘π‘š) = (5/6, 5/6)

Its position vector is 𝑅= 5/6𝑖 + 5/6𝑗

3. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27Γ…. Find the approximate location of the CM 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.

Let these two atoms lay on the x-axis. Its centre of mass lie at (π‘‹π‘π‘š, 0)

π‘š1 = π‘šπ» = 1 𝑒𝑛𝑖𝑑

π‘š2 = π‘šπΆπ‘™ = 35.5 𝑒𝑛𝑖𝑑

∴ π‘‹π‘π‘š = (π‘š1π‘₯1 + π‘š2π‘₯2)/(π‘š1 + π‘š2) = [1(0) + 35.5 (1.27 Γ— 10βˆ’10)]/(1 + 35.5) = (35.5/36.5) Γ— 1.27 Γ— 10βˆ’10 = 1.235Γ…

(π‘‹π‘π‘š, 0) = (1.235Γ…, 0)

𝑅 = 1.235 Γ— 10βˆ’10𝑖

4.  Find the scalar and vector product of two vector 𝐴 = 3π‘–βˆ’ 4𝑗+ 5π‘˜ and 𝐡 = βˆ’2𝑖+ π‘—βˆ’3π‘˜.

(i) 𝐴 βˆ™ 𝐡 = 𝐴π‘₯𝐡π‘₯ + 𝐴𝑦𝐡𝑦 + 𝐴𝑧𝐡𝑧 = [(3)(βˆ’2) + (βˆ’4)(1) + (5)(βˆ’3)] = βˆ’6 βˆ’ 4 βˆ’ 15 = βˆ’25

(ii)

= 𝑖[(βˆ’4)(βˆ’3) βˆ’ 5] βˆ’ 𝑗[(3)(βˆ’3) βˆ’ (βˆ’2)(5)] + π‘˜[(3)(1) βˆ’ (βˆ’2)(βˆ’4)] = 𝑖[12 βˆ’ 5] βˆ’ 𝑗[βˆ’9 + 10] + π‘˜[3 βˆ’ 8] = 7π‘–βˆ’ 1π‘—βˆ’ 5π‘˜

5. Show that 𝐴 = βˆ’6𝑖+ 9π‘—βˆ’ 12π‘˜ and 𝐡̂ = 2π‘–βˆ’ 3𝑗+ 4π‘˜ are parallel to each other. 𝐴 and 𝐡 will be parallel to each other if 𝐴 Γ— 𝐡 = 0.

𝐴 Γ— 𝐡 = 𝑖[(9)(4) βˆ’ (βˆ’12)(βˆ’3)] βˆ’ 𝑗[(βˆ’6)(4) βˆ’ (βˆ’12)(2)] + π‘˜[(βˆ’6)(βˆ’3) βˆ’ (9)(2)]

𝐴 Γ— 𝐡 = 𝑖[36 βˆ’ 36] βˆ’ 𝑗[βˆ’24 + 24] + π‘˜[+18 βˆ’ 18]

𝐴 Γ— 𝐡 = 0

Since 𝐴 Γ— 𝐡 = 0 therefore 𝐴 and 𝐡 are parallel to each other.

6. Magnitude of the cross product of two vectors 𝐴 and 𝐡 represents the area of the parallelogram. Prove it.

Consider a parallelogram OPQR whose adjacent sides are QP and OR are represented both in magnitude and direction by two vectors 𝐴 and 𝐡

|𝐴 Γ— 𝐡| = 𝐴𝐡 sin πœƒ = 𝐴(𝐡 sin πœƒ) = 𝑂𝑃. 𝑅𝑁 = π΅π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘

|𝐴 Γ— 𝐡| = π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Žπ‘™π‘™π‘’π‘‘π‘œπ‘”π‘Ÿπ‘Žπ‘š

Note: Since the parallelogram has two triangles OPQ and ORQ of equal areas,

Area of each triangle = (1/2)|𝐴 Γ— 𝐡|

Thus area of a triangle contained between the vectors 𝐴 and 𝐡 is one half of the magnitude of 𝐴 and 𝐡 is one half of the magnitude of 𝐴 Γ— 𝐡.

7. Find the torque of a force 7𝑖+ 3π‘—βˆ’ 5π‘˜ about the origin. The force acts on a particle whose position vector is π‘–βˆ’ 𝑗+ π‘˜. Given π‘Ÿ = π‘–βˆ’ 𝑗+ π‘˜ and 𝐹 = 7𝑖+ 3π‘—βˆ’ 5π‘˜ Torque 𝜏 = π‘Ÿ Γ— F

= 𝑖[(1 βˆ’ 1)(βˆ’5) βˆ’ (1)(3)] βˆ’ 𝑗[(1)(βˆ’5) βˆ’ (1)(7)] + π‘˜[(1)(3) βˆ’ (βˆ’1)(7)] = 𝑖[+5 βˆ’ 3] βˆ’ 𝑗[βˆ’5 βˆ’ 7] + π‘˜[3 + 7]

𝜏 = 2𝑖+ 12𝑗+ 10π‘˜

8. A meter scale is balanced on a knife edge at its centre. When two coins each of mass 5g 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 meter stick?

From the principle of moments, 𝐹1𝑑1 = 𝐹2𝑑2

π‘šπ‘” Γ— (45 βˆ’ 12) = 𝑀𝑔 Γ— (5)
π‘šπ‘” Γ— 33 = 𝑀𝑔 Γ— 5

10 Γ— 𝑔 Γ— 33 = 𝑀 Γ— 𝑔 Γ— 5

𝑀 = (10 Γ—33)/5 = 66 π‘”π‘Ÿπ‘Žπ‘š

9. A car weights 1800 kg the distance between its front and back axel is 1.8 m its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level road on each front wheel and each back wheel.

Let F1, F2 and Mg be the force exerted by the ground on front wheels, back wheels and weight of the car respectively. Now car is at rest. For translational equilibrium,

𝐹1 + 𝐹2 βˆ’ 𝑀𝑔 = 0

𝐹1 + 𝐹2 = 𝑀𝑔

and also car is in rotational equilibrium

Torque on front wheels = 0

(𝐹2 Γ— 1.8) – (𝑀𝑔 Γ— 1.05) = 0

𝐹2 = (𝑀𝑔×1.05)/1.8 = (1800Γ—9.8Γ—1.05)/1.8

𝐹2 = 10290𝑁

Force on each back wheel = 5145N

Torque on rare wheels = 0

(𝐹1 Γ— 1.8) – (𝑀𝑔 Γ— 0.75) = 0

𝐹1 = (𝑀𝑔×1.07)/1.8 = (1800Γ—9.8Γ—0.75)/1.8

𝐹2 = 7350N

Force on each Front wheel = 3675 N

10. A non-uniform bar of weight w is suspended at rest by two strings of negligible weight as shown the angles made by the strings with the vertical re 26.90 and 53.10 respectively. The bar is 2m long calculate the distance d of the centre of gravity of the bar from its left end.

Since the bar is at rest so the net external force and net external torque on the bar is zero.

βˆ‘πΉπ‘₯ = 0 and βˆ‘πΉπ‘¦ = 0

βˆ‘πΉπ‘₯ = 𝑇1 sin πœƒ βˆ’ 𝑇2 sin πœƒ = 0

𝑇1 sin(36.9) βˆ’ 𝑇2 sin(53.1) = 0

(𝑇1 Γ— 0.6004) – (𝑇2 Γ— 0.7996) = 0

𝑇2 Γ— (0.6004/0.7996)𝑇1 = 0.7508 𝑇1 βˆ’ βˆ’ βˆ’ βˆ’> (1)

βˆ‘πΉπ‘¦ = 𝑇1 cos πœƒ + 𝑇2 cos πœƒ βˆ’ 𝑀𝑔 = 0

(𝑇1 cos(36.9)) + (𝑇2 Γ— cos(53.1)) βˆ’ 𝑀𝑔 = 0

(𝑇1 Γ— 0.7996) + (𝑇2 Γ— 0.6004) βˆ’ 𝑀𝑔 = 0

0.7996𝑇1 + 0.6004𝑇2 = 𝑀𝑔 ——-Γ  (2)

Now torque acting on the left end of the bar is = 0

(𝑇2 cos πœƒ Γ— 𝑙) βˆ’ 𝑀𝑔𝑑 = 0

(𝑇2 cos(53.1) Γ— 2) – (𝑀𝑔 Γ— 𝑑) = 0

(𝑇2 Γ— 0.6004 Γ— 2) βˆ’ 𝑀𝑔𝑑 = 0

1.2008𝑇2 = 𝑀𝑔𝑑 ——-Γ  (3)

1.2008𝑇2 = (0.7996𝑇1 + 0.6004𝑇2)𝑑 from (2)

1.2008 (0.7508𝑇1) = [0.7996𝑇1 + 0.6004(0.7508)𝑇1]𝑑 from (1)

0.9015𝑇1 = (0.7996𝑇1 + 0.4507𝑇1)𝑑

0.9015𝑇1 = 1.2503𝑇1𝑑

𝑑 = 0.9015𝑇1/1.2503𝑇1 = 0.72 π‘š

11. What is the moment of a disc about one of its diameters? We have, moment of Inertia of inertia of the disc about an axis perpendicular to it is given by, 𝐼𝑧 = 𝑀𝑅22

Where 𝑀 β†’ mass of the disc

𝑅 β†’ Radius of the disc

According to the perpendicular axes theorem, 𝐼𝑧 = 𝐼π‘₯ + 𝐼𝑦

X and Y axes are along two diameters, ∴ 𝐼π‘₯ = 𝐼𝑦

𝐼𝑧 = 𝐼π‘₯ + 𝐼π‘₯

𝐼𝑧 = 2𝐼π‘₯

But 𝐼𝑧 = 𝑀𝑅2/2

𝑀𝑅2/2 = 2𝐼π‘₯

𝐼π‘₯ = 𝑀𝑅2/4

12. Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of a sphere about any of its diameter to be 2π‘šπ‘…2/5, where m is the mass of the sphere and R is the radius of the sphere.

Let AB is a diameter of the given sphere. M.I of the sphere about its diameter 𝐴𝐡 = 𝐼𝐴𝐡 = (2/5)π‘šπ‘…2

According to the theorem of parallel axis

𝐼𝐢𝐷 = 𝐼𝐴𝐡 + π‘šπ‘…2 = (2/5)π‘šπ‘…2 + π‘šπ‘…2

𝐼𝐢𝐷 = π‘šπ‘…2((2/5) + 1)

𝐼𝐢𝐷 = (7/5)π‘šπ‘…2

13. What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end?

The moment of inertia of the body about an axis passing through its centre is given by,

𝐼𝐴𝐡 = 𝑀𝑙2/12

Using the parallel axis theorem,

𝐼𝐢𝐷 = 𝐼𝐴𝐡 + 𝑀(𝑙/2)2

𝐼𝐢𝐷 = (𝑀𝑙2/12) + (𝑀𝑙2/4) = 𝑀𝑙2[(1/12) + (1/4)]

𝐼𝐢𝐷 = 𝑀𝑙2[4/12] = 𝑀𝑙2/3

14. What is the moment of inertia of a ring about a tangent to the circle of the ring? The tangent to the ring in the plane of the ring is parallel to the diameter of the ring.

According to parallel axis theorem, πΌπ‘‘π‘Žπ‘› = πΌπ‘‘π‘–π‘Ž + 𝑀𝑅2

Where 𝑀 β†’ mass of the ring

𝑅 β†’ Radius of the ring

πΌπ‘‘π‘Žπ‘› = 𝑀𝑅2/2 + 𝑀𝑅2

πΌπ‘‘π‘Žπ‘› = (3/2)𝑀𝑅2

15 Given the momentum of inertia of disc of mass m and radius R about any of its diameter to be π‘šπ‘…2/4. Find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

M. I about any diameter = (1/4)π‘šπ‘…2

𝐼π‘₯ = 𝐼𝑦 = (1/4)π‘šπ‘…2

By perpendicular axis theorem,

𝐼𝑧 = 𝐼π‘₯ + 𝐼𝑦

𝐼𝑧 = (1/4)π‘šπ‘…2 + (1/4)π‘šπ‘…2

𝐼𝑧 = (1/2)π‘šπ‘…2

Now Z and Z1 are parallel, according to parallel axis theorem,

𝐼𝑧1 = 𝐼𝑧 + π‘šπ‘…2

𝐼𝑧1 = (1/2)π‘šπ‘…2 + π‘šπ‘…2

𝐼𝑧1 = (3/2)π‘šπ‘…2

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

m = 20 kg, R = 0.25 m, πœ” = 100 rad/s, I=?, K = ?, L=?

M.I of the solid cylinder about its axis of symmetry,

𝐼 = (1/2)π‘šπ‘…2

𝐼 = (1/2) Γ— 0.625 Γ— 100 Γ— 100

𝐼 = 0.625kgm2

𝐾 = (1/2)πΌπœ”2

𝐾 = (1/2) Γ— 20 Γ— 0.25 Γ— 0.25 𝐾 = 3125𝐽

𝐿 = πΌπœ”

𝐿 = 0.625 Γ— 100 = 62.5 𝐽𝑠

17. A child stands at the centre of a turn table with his two arms out stretched. The turn table is set rotating with an angular speed of 4 over / min how much is the angular speed of the child if he folds his hands back and there by reduces his moment of inertia to 2/5 times the initial value? Assume that the turn table rotates without friction. 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?

According to the law of conservation of angular momentum,

𝐼1πœ”1 = 𝐼2πœ”2

πœ”2 = 𝐼1πœ”1/𝐼2

πœ”2 = (𝐼1 Γ— 40)/((2/5)𝐼1) = (5 Γ— 40)/2 = 100 π‘Ÿπ‘’π‘£/π‘šπ‘–π‘›

π‘˜π‘–/π‘˜π‘“ = (1/2)𝐼1πœ”12/(1/2)𝐼2πœ”22

π‘˜π‘–/π‘˜π‘“ = [𝐼1 Γ— (40)2]/[(2/5)𝐼1 Γ— (100)2]

π‘˜π‘–/π‘˜π‘“ = (5/2) Γ— (1600/10000) = (5/2) Γ— (4/25)

π‘˜π‘–/π‘˜π‘“ = 2/5

𝐾𝑖 = (2/5)π‘˜π‘“ π‘œπ‘Ÿ 𝐾𝑓 = (5/2)π‘˜π‘–

This increase in kinetic energy is obtained from the muscular energy of the child when he folds back his arms.

18. 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 30N? What is the linear acceleration of the rope? Assume that there is no slipping.

m = 3kg, R =40cm = 0.4m, F = 30N

M.I of the hollow cylinder

𝐼 = π‘šπ‘…2

𝐼 = 3 Γ— 0.4 Γ— 0.4 = 0.48π‘˜π‘”π‘š2

π‘‡π‘œπ‘Ÿπ‘žπ‘’π‘’, 𝜏 = π‘Ÿ Γ— 𝐹

𝜏 = 0.4 Γ— 30 = 12π‘π‘š

𝛼 = 𝜏/𝐼 = 12/0.48 = 25π‘Ÿπ‘Žπ‘‘/𝑠2

We have 𝑣 = π‘Ÿπœ”

𝑑𝑣/𝑑𝑑 = 𝑑/𝑑𝑑(π‘Ÿπœ”) = π‘Ÿ(π‘‘πœ”/𝑑𝑑)

π‘Ž = π‘Ÿπ›Ό = 0.4 Γ— 25 = 10π‘šπ‘ βˆ’2

19. To maintain a rotor at a uniform angular speed of 200rad/s, an engine needs to transmit a torque of 180Nm. What is the power required by the engine? Assume that the engine is 100% efficient.

πœ” = 200 π‘Ÿπ‘Žπ‘‘/𝑠 Power required, P = πœπœ”, 𝜏 = 180π‘π‘š

𝑃 = 180 Γ— 200

𝑃 = 36000𝑀

𝑃 = 36 π‘˜π‘Š

20. A hoop of radius 2m, weights 100kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20cm/s. How much work has to be done to stop it?

Work done to stop the hoop = Kinetic energy of the hoop.

π‘Š = (1/2)π‘šπ‘£π‘π‘š2 + (1/2)πΌπœ”2

π‘Š = (1/2)π‘šπ‘£π‘π‘š2 + (1/2)(π‘šπ‘…2)(π‘£π‘π‘š/𝑅)2

π‘Š = (1/2)π‘šπ‘£π‘π‘š2 + (1/2)π‘šπ‘£π‘π‘š2 = π‘šπ‘£π‘π‘š2

π‘Š = 100 Γ— 0.2 Γ— 0.2 = 4 𝐽

21. The oxygen molecule has a mass of 5.36 Γ— 10βˆ’26π‘˜π‘” and a moment of inertia of 1.94 Γ— 10βˆ’46π‘˜π‘”π‘š2 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 500m/s and that is K.E of rotation is 2/3 of its K.E of translation find the average angular velocity of the molecule.

π‘š = 5.36 Γ— 10βˆ’26π‘˜π‘”

𝐼 = 1.94 Γ— 10βˆ’46π‘˜π‘”/π‘š2

𝑣 = 500π‘š/𝑠

Given, K.E of rotation = (2/3) Γ— 𝐾. 𝐸 of translation

(1/2)πΌπœ”2 = (2/3) Γ— (1/2)π‘šπ‘£2

πœ”2 = π‘šπ‘£2/3𝐼

πœ” = √[(2 Γ— 5.30 Γ— 10βˆ’26 Γ— 500 Γ— 500)/(3 Γ— 1.94 Γ— 10βˆ’46)] = 6.7478 Γ— 102 π‘Ÿπ‘Žπ‘‘/s

Important Questions for Exam

One mark.

1. What is a rigid body?

2. What is the angle between 𝑃 Γ— 𝑄 and 𝑄 Γ— 𝑃?

3. Is 𝑃 Γ— 𝑄 = 𝑄 Γ— 𝑃?

4. Define angular displacement.

5. Mention the SI unit of angular velocity.

6. Define Torque.

7. Write the dimensional formula for torque.

8. Define angular momentum of a rigid body.

9. State the law of conservation of angular momentum.

Two marks.

1. Define Angular velocity and angular acceleration.

2. Give two general conditions of equilibrium of a rigid body. or Write the conditions for equilibrium of a rigid body.

3. What is couple? Mention effect of couple acting on a body.

4. Define moment of inertia and mention its SI unit.

5. Mention the formula for moment of inertia of a solid cylinder about its axis and explain the symbols.

6. Define radius of gyration. Is it constant for a body?

Three Marks.

1. What is the value of (a) 𝑖 Γ— 𝑖  (b) 𝑗 Γ— 𝑗 (c) π‘˜ Γ— π‘˜

2. What is the value of (a) 𝑖 Γ— 𝑗 (b) 𝑗 Γ— π‘˜ (c) π‘˜ Γ— 𝑖

3. What is the value of (a) 𝑗 Γ— 𝑖 (b) π‘˜ Γ— 𝑗 (c) 𝑖 Γ— π‘˜

4. What is centre of mass? Write the expression for its position in a two body system using a diagram and explain each term.

5. Distinguish between scalar product and vector product.

6. Define vector product of two vectors. Write an expression for it explaining each them.

7. Derive the relation between linear velocity and angular velocity.

8. Explain the principle of lever.

9. Write the expression for moment of inertia of a circular disc about an axis perpendicular to it at the centre and give the expression for its radius of gyration.

10. Compare the translation motion and rotational motion. or Compare the equations of linear and rotational motions.

Five marks.

1. State and explain theorems of perpendicular and parallel axis. or State and explain theorems of moment of inertia.

2. Define torque. Show that the rate of change of angular momentum is equal to torque acting on the system. or Deduce an expression for torque, 𝜏 = 𝑑𝑙/𝑑𝑑 𝐨𝐫 Derive the relation 𝑑𝑙/𝑑𝑑 = 𝜏 with usual notation or Derive the relation between torque and angular momentum.

3. Define torque. State and explain perpendicular and parallel axis theorem.

4. Derive an expression for Kinetic energy of a rolling body.

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