Rotational Inertia MCQ Quiz - Objective Question with Answer for Rotational Inertia - Download Free PDF

Correct option is: (3) 7 / 57

For a larger solid sphere about diameter Y-axis:

I_whole = (2 / 5) × M × (2R)² = (8 / 5) × M × R²

Density of sphere is uniform:

M / V_whole = M_smaller / V_smaller

M / ((4/3)π(2R)³) = M′ / ((4/3)πR³)

⇒ M′ = M / 8

Using parallel axis theorem for smaller sphere:

I′ = I_cm + M′ × R² = (2 / 5) × (M / 8) × R² + (M / 8) × R² = (7 / 40) × M × R²

Ratio:

Ratio = I_smaller / I_remaining = I′ / (I_whole − I′)

= ((7 / 40) × M × R²) / (((8 / 5) − (7 / 40)) × M × R²)

= 7 / 57

Calculation:
Case I: mR²/2 ω₀ = (mR²/2 + mR²/4 + mR²/4) ω ω = ω₀/2 = 3 rad/s

Case II: ω = ω₀ = 6 rad/s

Case III: mR²/2 ω₀ + mvR²/2 = (mR²)ω So, 3 + 8/2 = 7 rad/s

Case IV: mR²/2 ω₀ + mvR²(1/2) = mR²ω 3 + 8/4 = 5 rad/s

CONCEPT:

Moment of inertia:

• The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of rotation.
• The moment of inertia of a particle  is

I = mr2

Where r = the perpendicular distance of the particle from the rotational axis.

• Moment of inertia of a body made up of a number of particles (discrete distribution)

I = m1r12 + m2r22 + m3r32 + m4r42 + -------

EXPLANATION:

• The moment of inertia of a solid sphere of mass 'M' and radius 'R' about its diameter is 

\(⇒ I_s =\frac{2}{5}MR^2=0.4MR^2\)    ----- (1)

• The moment of inertia of a hollow sphere of mass 'M' and radius 'R2' about its diameter is 

\(⇒ I_h =\frac{2}{3}MR^2=0.67MR^2\)     ----- (2)

• From the above two-equation, it is clear that the moment of inertia of a hollow sphere is more than that of a hollow sphere. Therefore option 4 is correct. 

Explanation:

Moment of Inertia: 

Moment of inertia plays the same role in rotational motion as mass plays in linear motion. It is the property of a body due to which it opposes any change in its state of rest or of uniform rotation.

The moment of inertia of a particle is:

I = mr2

where r = perpendicular distance of the particle from the rotational axis.

Perpendicular axis theorem: 

This theorem states that the moment of inertia of a plane lamina 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.

Hence it can be expressed as \({I_z} = {I_y} + {I_x}\)

Here Ix, Iy and Iz are the moment of inertia of an object along X, Y and Z- axis respectively.

This is a moment of inertia of circular disc about the different axis of rotation.

The moment of inertia of a circular disc about diameter (Ix and Iy):

\(I_x=I_y=\frac{MR^2}{4}=I\)

\({I_z} = {I_y} + {I_x}\)

\({I_z} =\frac{MR^2}{4}+\frac{MR^2}{4}=\frac{MR^2}{2}=2I\)

Additional Information. This is a moment of inertia of circular disc about the different axis of rotation.

Calculation:
The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I₀. The moment of inertia about an axis passing through one of its ends and perpendicular to its length is derived using the parallel axis theorem.

According to the parallel axis theorem, the moment of inertia about an axis through one end of the rod is:

I = I₀ + M × (L/2)²

Substituting the value for I₀ (the moment of inertia through the center), we get:

I = I₀ + M × L²/4

Hence, the correct option is Option 2: I₀ + M × L²/4

CONCEPT:

• The angular momentum of a particle rotating about an axis is defined as the moment of the linear momentum of the particle about that axis.
• It is measured as the product of linear momentum and the perpendicular distance of its line of action from the axis of rotation.
• The relation between the angular momentum and moment of inertia is given by

L = Iω

Where I = moment of inertia, L = angular momentum, and ω = angular velocity.

EXPLANATION:

• If there is no external torque acting on system then initial angular momentum (L_initial) of system is equal to final momentum (L_final).
• Hence, the angular momentum in a closed system is a conserved.

∴ Iω = constant

⇒ I ∝ 1/ω

i.e. Moment of inertia is inversely proportional to the angular velocity.

• Hence if the moment of inertia of a rotating body is increased then the angular velocity decreases.

CONCEPT:

Torque (τ): 

• It is the twisting force that tends to cause rotation.
• The point where the object rotates is known as the axis of rotation. 
• Mathematically it is written as,

τ = rFsin θ 

• Angular acceleration (α): It is defined as the time rate of change of angular velocity of a particle is called its angular acceleration.
• If Δω is the change in angular velocity time Δt, then average acceleration is

\(\vec α = \frac{{{\rm{\Delta }}\omega }}{{{\rm{\Delta }}t}}\)

EXPLANATION:

• Torque is the measure of the amount of force acting on an object that can cause it to rotate.
• The torque that is needed to produce angular acceleration depends on the mass distribution of the object which is described by the moment of inertia.
• Therefore, torque (\(\tau \)) is the product of the angular acceleration (a) and the moment of inertia (I) of an object. Therefore option 2 is correct.

           \(\tau = I \times a\)

CONCEPT:

Moment of inertia:

• The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.
• The moment of inertia of a particle  is

⇒ I = mr2

Where r = the perpendicular distance of the particle from the rotational axis.

• Moment of inertia of a body made up of a number of particles (discrete distribution)

⇒ I = m1r12 + m2r22 + m3r32 + m4r42 + -------

Rotational kinetic energy: 

• The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
• A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
• Mathematically rotational kinetic energy can be written as -

⇒ KE \( = \frac{1}{2}I{\omega ^2}\)

Where I = moment of inertia and ω = angular velocity.

EXPLANATION:

• The moment of inertia of the ring about an axis passing through the center and perpendicular to its plane is given by

⇒ I_ring = MR²

• Moment of inertia of the disc about an axis passing through center and perpendicular to its plane is given by -

\(⇒ {I_{disc}} = \frac{1}{2}M{R^2}\)

• As we know that mathematically rotational kinetic energy can be written as

\(⇒ KE = \frac{1}{2}I{\omega ^2}\)

• According to the question, the angular velocity of a thin disc and a thin ring is the same. Therefore, the kinetic energy depends on the moment of inertia.
• Therefore, a body having more moments of inertia will have more kinetic energy and vice - versa.
• So, from the equation, it is clear that,

⇒ Iring > Idisc

∴ Kring > Kdisc

• The ring has higher kinetic energy.

CONCEPT:

Moment of Inertia:

• Moment of inertia plays the same role in rotational motion as mass plays in linear motion. It is the property of a body due to which it opposes any change in its state of rest or of uniform rotation.
• Moment of inertia of a particle is

I = mr2

• For a uniform rod with negligible thickness, the moment of inertia about its centre of mass is:

\({I_{cm}} = \frac{1}{{12}}M{L^2}\)

Where

M = mass of the rod

L = length of the rod

r = perpendicular distance of the particle from the rotational axis

Explanation:

From the above explanation, we can see that 

• Moment of inertia of the rod when the axis is perpendicular to it and passes through its centre is 

​\(I_c= \;\frac{{M{L^2}}}{{12}}\)           

CONCEPT:

• Parallel axis theorem: Moment of inertia of a body about a given axis I is equal to the sum of moment of inertia of the body about an axis parallel to given axis and passing through centre of mass of the body Io and Ma2, where ‘M’ is the mass of the body and ‘a’ is the perpendicular distance between the two axes.

⇒ I = I_o + Ma²

EXPLANATION:

• For a uniform rod with negligible thickness, the moment of inertia about its centre of mass is:

\({I_{cm}} = \frac{1}{{12}}M{L^2}\)

Where M = mass of the rod and L = length of the rod

∴ The moment of inertia about the end of the rod is

\(\Rightarrow {I_{end}} = {I_{cm}} + M{d^2} \)

\(\Rightarrow I_{end}= \frac{1}{{12}}M{L^2} + M{\left( {\frac{L}{2}} \right)^2} = \frac{1}{3}M{L^2}\)

CONCEPT:

Moment of inertia:

• Moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.
• The moment of inertia of a particle  is

⇒ I = mr2

Where r = the perpendicular distance of the particle from the rotational axis.

• Moment of inertia of a body made up of a number of particles (discrete distribution)

⇒ I = m1r12 + m2r22 + m3r32 + m4r42 + -------

EXPLANATION:

• Moment of inertia of the disc about an axis passing through center and perpendicular to its plane is given by -

\(⇒ {I_{disc}} = \frac{1}{2}M{R^2} = 0.5M{R^2}\)

• Moment of inertia of the solid sphere about an axis passing through center and perpendicular to its plane is given by -

\(⇒ {I_{sphere}} = \frac{2}{5}M{R^2} = 0.4M{R^2}\)

• So, from above it is very clear that the moment of inertia of the disc is greater than the moment of inertia of the solid sphere.

CONCEPT:

• Moment of Inertia: A quantity expressing a body's tendency to resist angular acceleration is called the moment of Inertia.
• For point mass Moment of inertia is simply the mass times the square of the perpendicular distance to the axis of rotation.

I = m × r​2 

where I is the Moment of Inertia, m is point mass, r is the perpendicular distance from the axis of rotation.

• For a rigid body system, the moment of inertia is the sum of the moments of inertia of all its particles taken about the same axis.

\(I=\sum m_{i}{r_{i}}^{2}\)

where I is the Moment of Inertia, m is point mass, r is the perpendicular distance from the axis of rotation.

EXPLANATION:

• If the mass is situated at a large distance from the axis Moment of Inertia will be large because the distance of mass particles from the axis of rotation will be large.
• As the mass is at a greater distance from the axis of rotation, the effective radius will be larger. 
• This is why the moment of inertia (\(I=\sum m_{i}{r_{i}}^{2}\)) will be larger.
• Hence the correct answer is option 2.

Polar moment of inertia:

Moment of inertia about an axis perpendicular to the plane of an area is known as the polar moment of inertia.

Let Oz be the axis perpendicular to the axis Ox and Oy passing through the origin O.

Moment of inertia of area of plane lamina about Oz axis is given by:

\(\rm{{I_{zz}} = \smallint {r^2}dA = \smallint \left( {{x^2} + {y^2}} \right)dA = \smallint {x^2}dA + \smallint {y^2}dA = {I_{xx}} + {I_{yy}}}\)

The moment of inertia of an area about an axis perpendicular to its plane (polar moment of inertia) at any point O is equal to the sum of moments of inertia about any two mutually perpendicular axes passing through the same point O and lying in the plane of area.

The centre of mass of a body is a point where the whole mass of the body is supposed to be concentrated.

The centre of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry.

The centre of mass may reside inside or outside the body.

CONCEPT:

• Moment of Inertia (I): Moment of inertia is the inertia of a rotating body concerning its rotation. Moment of inertia is also known as rotational inertia.

I = Σ MR2

Where I is the moment of inertia of the body, M is the mass of the body, R is the radius.

• Parallel axis theorem: The moment of inertia of a body about an axis parallel to the body passing through its center is equal to the sum of moment of inertia of the body about the axis passing through the center and product of the mass of the body times the square of the distance between the two axes.

I = Io + MR2

• Moment of Inertia of a rod about an axis through its center and perpendicular to the plane is:

\(I = \frac{1}{12}M{L^2}\)

EXPLANATION:

• Moment of Inertia of a rod about an axis through its center and perpendicular to the plane is:

\(I = \frac{1}{12}M{L^2}\)

Where M = mass of the body and L = length of the rod

• Since the length of the rod is l, the distance from the centre of a rod to one end of the rod is

⇒ I' = Io + Ma2

Here, a = l/2

 \(\Rightarrow I' = M\frac{{{l^2}}}{{12}} + M{\left( {\frac{l}{2}} \right)^2} = \frac{{M{l^2}}}{3}\)