1. Define torque and angular momentum.
Torque is defined as the turning effect of a force about a fixed
point or axis. It is equal to the product of the force and the perpendicular
distance of the line of action of the force from the axis of rotation.
Torque (τ) = Force × Perpendicular distance
Angular momentum is defined as the moment of linear momentum of a
particle about a given point or axis. It is equal to the product of the
linear momentum and the perpendicular distance of its line of motion from
the axis.
Angular momentum (L) = r × p
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2. What is the relation between the torque and angular momentum?
The relation between torque and angular momentum
is given by:
Torque (τ) = Rate of change of angular momentum
Mathematically,
τ = dL / dt
This means that when the net external torque acting on a body is zero, its
angular momentum remains constant.
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3. State and explain the principle of conservation of angular momentum.
Principle of conservation of angular momentum states that, when
no external torque acts on a system, the total angular momentum of the system
remains constant.
In other words, the angular momentum of a system before interaction is equal
to the angular momentum of the system after interaction, provided the net
external torque is zero.
Mathematically,
Initial angular momentum = Final angular momentum
L = constant
This principle explains many physical phenomena such as the increase in
angular speed of a spinning skater when pulling in her arms and the
stability of rotating celestial bodies.
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4. Why is it easier to open a door by applying force at the edge rather than near the hinge?
It is easier to open a door by applying force at the edge rather than near
the hinge because the turning effect of force (torque)
depends on the perpendicular distance from the axis of rotation.
Torque is given by:
Torque = Force × Perpendicular distance from the hinge
When force is applied at the edge of the door, the perpendicular distance
from the hinge is maximum, producing a larger torque for the same applied
force. Hence, the door opens more easily.
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5. State parallel axis theorem and perpendicular axes theorem.
Parallel Axis Theorem:
Mathematically,
I = ICM + Md2
where I = moment of inertia about the given axis,
Perpendicular Axes Theorem:
Mathematically,
Iz = Ix + Iy
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The moment of inertia of a body about any axis parallel to an axis passing
through its center of mass is equal to the sum of the moment of inertia about
the center of mass axis and the product of the mass of the body and the square
of the distance between the two axes.
ICM = moment of inertia about the parallel axis through center of mass,
M = mass of the body, d = distance between the axes.
For a flat lamina lying in the XY-plane, the moment of inertia about an axis
perpendicular to the plane (Z-axis) is equal to the sum of the moments of
inertia about two mutually perpendicular axes (X and Y) lying in the plane
and intersecting at a point.
6. Moment of inertia of different Objects.
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Object
Axis of Rotation
Moment of Inertia (I) with Theorem Used
Rod of length L, mass M
About center of mass (axis perpendicular to length)
ICM = (1/12) M L²
Rod of length L, mass M
About an edge (axis perpendicular to length)
I = ICM + M(L/2)² = (1/12) M L² + (1/4) M L² = (1/3) M L²
(Parallel Axis Theorem)
Ring of radius R, mass M
Axis perpendicular to plane through center
I = M R²
Ring of radius R, mass M
Diameter in plane of ring
I = (1/2) M R²
(Perpendicular Axes Theorem)
Ring of radius R, mass M
Tangent in plane of ring
I = ICM + M R² = (1/2) M R² + M R² = (3/2) M R²
(Parallel Axis Theorem)
Ring of radius R, mass M
Tangent perpendicular to plane
I = Iperpendicular center + M R² = M R² + M R² = 2 M R²
(Parallel Axis + Perpendicular Axes)
Disk of radius R, mass M
Axis perpendicular to plane through center
I = (1/2) M R²
Disk of radius R, mass M
Diameter in plane of disk
I = (1/4) M R²
(Perpendicular Axes Theorem)
Solid sphere of radius R, mass M
Diameter through center
I = (2/5) M R²
Solid sphere of radius R, mass M
Tangent through surface
I = ICM + M R² = (2/5) M R² + M R² = (7/5) M R²
(Parallel Axis Theorem)
Hollow sphere (thin shell) radius R, mass M
Diameter through center
I = (2/3) M R²
Hollow sphere (thin shell) radius R, mass M
Tangent through surface
I = ICM + M R² = (2/3) M R² + M R² = (5/3) M R²
(Parallel Axis Theorem)
Numerical on Rotational Motion-Important Questions and Numericals | Engineering Physics (Polytechnic 1st Year)
1. A torque of 20 N·m acts on a wheel of moment of inertia 4 kg·m². Find the angular acceleration of the wheel.
A torque of 20 N·m acts on a wheel of moment of inertia
4 kg·m². Find the angular acceleration of the wheel.
Given:
Using the relation between torque and angular acceleration:
τ = I × α
Solve for angular acceleration:
α = τ / I = 20 / 4 = 5 rad/s²
Answer: Angular acceleration of the wheel is 5 rad/s².
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Torque, τ = 20 N·m
Moment of inertia, I = 4 kg·m²
2. A skater of mass 50 kg spinning with arms extended at 2 rad/s pulls her arms inwards reducing her moment of inertia from 4 kg·m² to 1.5 kg·m². Find her new angular speed.
A skater of mass 50 kg spinning with arms extended at
2 rad/s pulls her arms inwards, reducing her moment of inertia
from 4 kg·m² to 1.5 kg·m². Find her new angular speed.
Given:
Using the principle of conservation of angular momentum:
Linitial = Lfinal ⇒ I1 ω1 = I2 ω2
Substitute the values:
4 × 2 = 1.5 × ω2
ω2 = 8 / 1.5 ≈ 5.33 rad/s
Answer: The new angular speed of the skater is approximately 5.33 rad/s.
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Mass of skater, m = 50 kg
Initial angular velocity, ω1 = 2 rad/s
Initial moment of inertia, I1 = 4 kg·m²
Final moment of inertia, I2 = 1.5 kg·m²
3.A wheel of moment of inertia 5 kg·m² rotates with angular velocity 10 rad/s. Find its angular momentum.
Given:
Angular momentum is given by:
L = I × ω
Substitute the values:
L = 5 × 10 = 50 kg·m²/s
Answer: The angular momentum of the wheel is 50 kg·m²/s.
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Moment of inertia, I = 5 kg·m²
Angular velocity, ω = 10 rad/s
4. A solid sphere (mass 5 kg, radius 0.3 m) rotates about a tangent. Find its moment of inertia.
Given:
Using the Parallel Axis Theorem:
I = ICM + M R²
Moment of inertia of solid sphere about diameter (CM axis):
ICM = (2/5) M R² = (2/5) × 5 × 0.3² = 0.18 kg·m²
Moment of inertia about tangent:
I = ICM + M R² = 0.18 + 5 × 0.3² = 0.18 + 0.45 = 0.63 kg·m²
Answer: I = 0.63 kg·m²Show Answer
Mass, M = 5 kg
Radius, R = 0.3 m
Axis: Tangent to sphere
5. A solid disk of mass 4 kg and radius 0.3 m rotates about an axis tangent to the disk in its plane. Find I.
Given:
Using the Parallel Axis Theorem:
I = ICM + M R²
Moment of inertia about center of mass (disk perpendicular to plane):
ICM = (1/2) M R² = (1/2) × 4 × 0.3² = 0.18 kg·m²
Moment of inertia about tangent:
I = ICM + M R² = 0.18 + 4 × 0.3² = 0.18 + 0.36 = 0.54 kg·m²
Answer: I = 0.54 kg·m²Show Answer
Mass, M = 4 kg
Radius, R = 0.3 m
Axis: Tangent to disk in plane
6. A uniform rod of length 2 m and mass 3 kg is pivoted about its center. Find its moment of inertia.
Given:
The moment of inertia of a uniform rod about its center is:
I = (1/12) M L²
Substitute the values:
I = (1/12) × 3 × 2² = (1/12) × 3 × 4 = 12 / 12 = 1 kg·m²
Answer: The moment of inertia of the rod about its center is 1 kg·m².
Show Answer
Mass, M = 3 kg
Length, L = 2 m
Axis: Through center, perpendicular to length
