NEET UG - Parallel and perpendicular axes theorems and applications, equilibrium of rigid bodies E-Book Study Material

Parallel and Perpendicular Axes Theorems and Applications, Equilibrium of Rigid Bodies

Parallel and Perpendicular Axes Theorems

In rotational motion, understanding the moment of inertia (I) is crucial. The moment of inertia depends on the mass distribution of the body and the axis of rotation. The Parallel and Perpendicular Axes Theorems are fundamental tools that help us calculate the moment of inertia of a rigid body about different axes, especially when the moment of inertia about a specific axis is already known.

Parallel Axes Theorem

The Parallel Axes Theorem relates the moment of inertia of a body about an axis passing through its center of mass to the moment of inertia about any other axis parallel to the first one. Let I_cm be the moment of inertia of a rigid body about an axis passing through its center of mass. If we want to find the moment of inertia (I) about an axis parallel to the first axis and at a distance 'd' from it, the theorem states:

I = I_cm + Md²

Where:

I is the moment of inertia about the new axis.
I_cm is the moment of inertia about the parallel axis passing through the center of mass.
M is the total mass of the body.
d is the perpendicular distance between the two parallel axes.

This theorem is extremely useful because it allows us to shift the axis of rotation without having to re-calculate the entire moment of inertia from scratch using integration. We only need to know the moment of inertia about the center of mass axis and the distance to the new parallel axis.

Example: Consider a solid cylinder of mass M and radius R. Its moment of inertia about its central longitudinal axis is I_cm = (1/2)MR². If we want to find the moment of inertia about a parallel axis at a distance R from the center (e.g., the axis of one of its flat surfaces), we can use the parallel axes theorem:

I = I_cm + Md²

Here, d = R.

So, I = (1/2)MR² + M(R)² = (3/2)MR².

Mnemonic for Parallel Axes Theorem: Think of 'P' for Parallel. The formula is I = I_cm + M*d². The 'M' and 'd' are added to the original I_cm because moving the axis away from the center of mass generally increases the moment of inertia, as the mass elements are, on average, further from the axis.

Perpendicular Axes Theorem

The Perpendicular Axes Theorem is applicable only to planar (2D) bodies, meaning bodies lying entirely in a single plane. It relates the moment of inertia of a planar body about an axis perpendicular to its plane to the moments of inertia about two perpendicular axes lying in the plane and intersecting at the same point.

Let I_x be the moment of inertia about an axis X lying in the plane of the body. Let I_y be the moment of inertia about an axis Y lying in the plane, perpendicular to X, and intersecting X at some point. Let I_z be the moment of inertia about an axis Z perpendicular to the plane of the body and passing through the point of intersection of X and Y.

The theorem states:

I_z = I_x + I_y

This theorem is particularly useful for calculating the moment of inertia of thin, flat objects like discs, rings, and rectangular plates about axes perpendicular to their plane, provided we know their moments of inertia about axes within the plane.

Example: Consider a thin uniform disc of mass M and radius R. Its moment of inertia about a diameter (an axis in the plane of the disc) is I_diameter = (1/4)MR². Since any two perpendicular diameters are equivalent due to symmetry, we have I_x = I_y = I_diameter.

Using the perpendicular axes theorem to find the moment of inertia about an axis perpendicular to the plane and passing through the center:

I_z = I_x + I_y

I_z = (1/4)MR² + (1/4)MR² = (1/2)MR².

This matches the known formula for the moment of inertia of a disc about its central axis perpendicular to its plane.

Key Point for Perpendicular Axes Theorem: Remember this theorem only applies to planar objects. The 'Z' axis is perpendicular to the 'X' and 'Y' axes, and all three axes intersect at a single point. The formula is additive: I_z = I_x + I_y.

Applications of the Theorems

These theorems simplify moment of inertia calculations for various common shapes:

Thin Rod:

Moment of inertia about an axis through the center and perpendicular to its length: I_cm = (1/12)ML².
Moment of inertia about an axis through one end and perpendicular to its length (using parallel axes theorem with d=L/2): I_end = I_cm + M(L/2)² = (1/12)ML² + (1/4)ML² = (1/3)ML².

Annular Disc (Ring):

Moment of inertia about an axis through the center and perpendicular to its plane: I_cm = MR².
Moment of inertia about a diameter (an axis in the plane) by perpendicular axes theorem: I_diameter = I_x + I_y. Since I_x = I_y due to symmetry, 2*I_diameter = MR², so I_diameter = (1/2)MR².

Solid Disc:

Moment of inertia about an axis through the center and perpendicular to its plane: I_cm = (1/2)MR².
Moment of inertia about a diameter: I_diameter = (1/4)MR² (derived using perpendicular axes theorem as shown previously).

Rectangular Lamina:

Let the dimensions be length L and width B. For an axis through the center and parallel to side B (along the length): I_x = (1/12)ML².
For an axis through the center and parallel to side L (along the width): I_y = (1/12)MB².
For an axis through the center and perpendicular to the plane: I_z = I_x + I_y = (1/12)M(L² + B²).

These calculations would be far more complex if we had to derive them from the basic definition of moment of inertia (∫r²dm) every time.

Equilibrium of Rigid Bodies

A rigid body is an idealized body where the distance between any two constituent particles remains constant. In reality, no body is perfectly rigid, but for many mechanical problems, we can treat them as such. A rigid body can undergo two types of motion: translation and rotation. Therefore, for a rigid body to be in equilibrium, both its translational motion and rotational motion must be absent or unchanging.

Conditions for Equilibrium

There are two main conditions for a rigid body to be in equilibrium:

First Condition for Equilibrium (Translational Equilibrium): The net external force acting on the rigid body must be zero. This means the vector sum of all forces acting on the body is zero.
ΣF = 0

This implies that the net force in each of the x, y, and z directions is zero:

ΣF_x = 0, ΣF_y = 0, ΣF_z = 0

This condition ensures that the center of mass of the rigid body is either at rest or moving with a constant velocity (no acceleration). For static equilibrium, the center of mass must be at rest.
Second Condition for Equilibrium (Rotational Equilibrium): The net external torque acting on the rigid body about any point must be zero. Torque is the rotational equivalent of force.
Στ = 0

This implies that the net torque in each of the rotational directions (e.g., clockwise and counter-clockwise) is zero:

Στ_clockwise = Στ_{counter-clockwise}

This condition ensures that the angular acceleration of the rigid body is zero. For static equilibrium, the body must not be rotating, or if it is, it must be rotating with a constant angular velocity.

A body satisfying both conditions is said to be in mechanical equilibrium. If a body is in translational equilibrium and also has zero angular acceleration, it is in dynamic equilibrium. If it is also at rest, it is in static equilibrium.

Important Note: For the second condition (Στ = 0), the torques must be calculated about the *same* point. If the net force is also zero (ΣF = 0), then the net torque will be zero about *any* point. This is a very useful property in solving problems.

Types of Equilibrium

Based on how a rigid body returns to its original position after a small displacement, equilibrium can be classified into three types:

Stable Equilibrium: If, after a small displacement, the body experiences a net force or torque that tends to restore it to its original position, it is in stable equilibrium. In this case, the potential energy of the body is at a minimum.
Example: A cone resting on its base, a pendulum bob hanging freely.
Unstable Equilibrium: If, after a small displacement, the body experiences a net force or torque that tends to move it further away from its original position, it is in unstable equilibrium. In this case, the potential energy of the body is at a maximum.
Example: A cone balanced on its vertex, a pencil balanced on its tip.
Neutral Equilibrium: If, after a small displacement, the body neither tends to return to its original position nor move further away, it remains in its new position. In this case, the potential energy of the body remains constant.
Example: A cone lying on its side, a ball on a flat horizontal surface.

Applications and Problem Solving in Equilibrium

Problems involving rigid body equilibrium often involve forces like gravity, tension, normal forces, and friction. To solve these problems, we typically follow these steps:

Draw a Free-Body Diagram (FBD): This is the most critical step. Draw the rigid body and represent all external forces acting on it as vectors originating from their points of application. Also, indicate the pivot or point about which torques might be calculated.
Choose a Coordinate System: Usually, horizontal (x) and vertical (y) axes are convenient.
Apply the First Condition for Equilibrium (ΣF = 0): Resolve all forces into their x and y components and set the sum of components in each direction to zero. This gives two scalar equations.
Choose a Pivot Point: Select a point to calculate torques. Often, choosing a point where one or more unknown forces act simplifies the calculation because the torque due to those forces will be zero.
Apply the Second Condition for Equilibrium (Στ = 0): Calculate the torque due to each force about the chosen pivot point. Remember that torque (τ) is given by τ = rFsinθ, where r is the distance from the pivot to the point of force application, F is the magnitude of the force, and θ is the angle between r and F. Alternatively, calculate the perpendicular distance from the pivot to the line of action of the force. Sum the torques, considering clockwise torques as negative and counter-clockwise torques as positive (or vice versa), and set the sum to zero. This gives a third scalar equation.
Solve the Equations: Solve the system of linear equations obtained from the force and torque conditions to find the unknown forces, distances, or angles.

Example Problem: A uniform rod of length 2m and mass 1kg is supported horizontally by two vertical strings attached at its ends. A load of mass 2kg is placed at a distance of 0.5m from one end. Find the tensions in the two strings.

Solution:

Let the rod be AB, with length L=2m. Mass of rod M=1kg. Load m=2kg. Load is at distance x=0.5m from end A.
Let T_A be the tension in string at end A, and T_B be the tension in string at end B.
Weight of the rod W = Mg = 1kg * 9.8 m/s² = 9.8 N. This acts at the center of the rod (at 1m from either end).
Weight of the load w = mg = 2kg * 9.8 m/s² = 19.6 N. This acts at 0.5m from end A.
First Condition (ΣF = 0): The rod is in vertical equilibrium.

T_A + T_B - W - w = 0

T_A + T_B = W + w = 9.8 N + 19.6 N = 29.4 N ...(1)

Second Condition (Στ = 0): Let's choose end A as the pivot point. The torque due to T_A is zero.

Torque due to W: W acts at 1m from A. Torque = W * 1m (counter-clockwise).

Torque due to w: w acts at 0.5m from A. Torque = w * 0.5m (counter-clockwise).

Torque due to T_B: T_B acts at 2m from A. Torque = T_B * 2m (clockwise).

Στ_{about A} = 0

(W * 1m) + (w * 0.5m) - (T_B * 2m) = 0

(9.8 N * 1m) + (19.6 N * 0.5m) = T_B * 2m

9.8 Nm + 9.8 Nm = 2 * T_B Nm

19.6 Nm = 2 * T_B Nm

T_B = 19.6 / 2 = 9.8 N

Solve for T_A using equation (1):

T_A + 9.8 N = 29.4 N

T_A = 29.4 N - 9.8 N = 19.6 N

So, the tension in the string at the end where the load is placed is 19.6 N, and the tension at the other end is 9.8 N.

Equilibrium Shortcut: For problems involving a horizontal beam supported at two points, if you choose the pivot at one of the support points, the tension in that string becomes zero in the torque equation, making it much simpler to solve for the tension in the other string. Always draw the FBD carefully!