Moment of Inertia
Q. A composite body consists of a solid cylinder and a solid sphere, both of mass M and radius R. What is the total moment of inertia about the same axis?
A.
(7/10) MR^2
B.
(9/10) MR^2
C.
(11/10) MR^2
D.
(13/10) MR^2
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Solution
The total moment of inertia is I_cylinder + I_sphere = (1/2 MR^2) + (2/5 MR^2) = (7/10) MR^2.
Correct Answer: A — (7/10) MR^2
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Q. A disk and a ring of the same mass and radius are rolling down an incline. Which will reach the bottom first?
A.
Disk
B.
Ring
C.
Both will reach at the same time
D.
Depends on the angle of incline
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Solution
The disk has a smaller moment of inertia compared to the ring, hence it will reach the bottom first.
Correct Answer: A — Disk
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Q. A particle of mass m is located at a distance r from the axis of rotation. What is the moment of inertia of this particle?
A.
mr
B.
mr^2
C.
m/r
D.
m/r^2
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Solution
The moment of inertia of a point mass is given by I = mr^2.
Correct Answer: B — mr^2
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Q. A particle of mass m is located at a distance r from the axis of rotation. What is the moment of inertia of this particle about the axis?
A.
mr
B.
mr^2
C.
m/r
D.
m/r^2
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Solution
The moment of inertia of a point mass about an axis is given by I = mr^2.
Correct Answer: B — mr^2
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Q. A rectangular plate of mass M and dimensions a x b is rotated about an axis along one of its edges. What is its moment of inertia?
A.
1/3 Ma^2
B.
1/12 Ma^2 + 1/3 Mb^2
C.
1/2 Ma^2
D.
1/4 Ma^2 + 1/3 Mb^2
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Solution
The moment of inertia of a rectangular plate about an edge is I = 1/12 Ma^2 + 1/3 Mb^2.
Correct Answer: B — 1/12 Ma^2 + 1/3 Mb^2
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Q. A solid sphere and a hollow sphere of the same mass and radius are released from rest at the same height. Which one will have a greater linear speed when they reach the ground?
A.
Solid sphere
B.
Hollow sphere
C.
Both have the same speed
D.
Depends on the mass
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Solution
The solid sphere will have a greater linear speed because it has a smaller moment of inertia, allowing it to convert more potential energy into translational kinetic energy.
Correct Answer: A — Solid sphere
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Q. A solid sphere and a hollow sphere of the same mass and radius are released from rest at the same height. Which one will hit the ground first?
A.
Solid sphere
B.
Hollow sphere
C.
Both hit at the same time
D.
Depends on the mass
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Solution
The solid sphere will hit the ground first because it has a smaller moment of inertia, allowing it to roll down faster.
Correct Answer: A — Solid sphere
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Q. A solid sphere of mass M and radius R is rolling without slipping. What is its moment of inertia about an axis through its center?
A.
2/5 MR^2
B.
3/5 MR^2
C.
1/2 MR^2
D.
MR^2
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Solution
The moment of inertia of a solid sphere about its center is I = 2/5 MR^2.
Correct Answer: A — 2/5 MR^2
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Q. A solid sphere of mass M and radius R is rotating about an axis through its center. What is its moment of inertia?
A.
2/5 MR^2
B.
3/5 MR^2
C.
1/2 MR^2
D.
1/3 MR^2
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Solution
The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer: A — 2/5 MR^2
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Q. A solid sphere rolls without slipping down an incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.
1:2
B.
2:3
C.
1:1
D.
1:3
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Solution
For a solid sphere, the ratio of translational kinetic energy to total kinetic energy is 2:3.
Correct Answer: B — 2:3
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Q. A thin rod of length L and mass M is rotated about an axis perpendicular to its length through one end. What is its moment of inertia?
A.
1/3 ML^2
B.
1/12 ML^2
C.
1/2 ML^2
D.
ML^2
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Solution
The moment of inertia of a thin rod about an end is I = 1/3 ML^2.
Correct Answer: A — 1/3 ML^2
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Q. A thin rod of length L and mass M is rotated about an axis perpendicular to its length and passing through one end. What is its moment of inertia?
A.
1/3 ML^2
B.
1/12 ML^2
C.
1/2 ML^2
D.
ML^2
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Solution
The moment of inertia of a thin rod about an end is I = 1/3 ML^2.
Correct Answer: A — 1/3 ML^2
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Q. A uniform rod of length L and mass M is rotated about its center. What is its moment of inertia?
A.
1/3 ML^2
B.
1/12 ML^2
C.
1/2 ML^2
D.
ML^2
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Solution
The moment of inertia of a uniform rod about its center is I = 1/12 ML^2.
Correct Answer: C — 1/2 ML^2
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Q. A uniform rod of length L is pivoted at one end. If it is allowed to fall freely, what is its angular acceleration just after it is released?
A.
g/L
B.
2g/L
C.
g/2L
D.
3g/2L
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Solution
The angular acceleration α = τ/I = (MgL/2)/(1/3 ML^2) = 3g/2L.
Correct Answer: B — 2g/L
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Q. A uniform thin circular ring of mass M and radius R is rotated about an axis through its center. What is its moment of inertia?
A.
MR^2
B.
1/2 MR^2
C.
1/3 MR^2
D.
2/5 MR^2
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Solution
The moment of inertia of a thin circular ring about an axis through its center is I = MR^2.
Correct Answer: A — MR^2
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Q. Calculate the moment of inertia of a hollow sphere of mass M and radius R about an axis through its center.
A.
2/5 MR^2
B.
3/5 MR^2
C.
2/3 MR^2
D.
MR^2
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Solution
The moment of inertia of a hollow sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer: B — 3/5 MR^2
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Q. Determine the moment of inertia of a solid sphere of mass M and radius R about an axis through its center.
A.
2/5 MR^2
B.
3/5 MR^2
C.
4/5 MR^2
D.
MR^2
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Solution
The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer: A — 2/5 MR^2
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Q. For a composite body made of a solid cylinder and a solid sphere, how do you calculate the total moment of inertia about the same axis?
A.
Add the individual moments
B.
Multiply the individual moments
C.
Subtract the individual moments
D.
Divide the individual moments
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Solution
The total moment of inertia of a composite body about the same axis is the sum of the individual moments of inertia.
Correct Answer: A — Add the individual moments
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Q. For a given mass, which of the following configurations will have the smallest moment of inertia?
A.
All mass at the center
B.
Mass distributed evenly
C.
Mass at the edge
D.
Mass concentrated at one end
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Solution
The moment of inertia is smallest when all mass is concentrated at the center, as it minimizes the distance from the axis of rotation.
Correct Answer: A — All mass at the center
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Q. For a hollow sphere of mass M and radius R, what is the moment of inertia about an axis through its center?
A.
2/5 MR^2
B.
3/5 MR^2
C.
2/3 MR^2
D.
MR^2
Show solution
Solution
The moment of inertia of a hollow sphere about an axis through its center is I = 2/5 MR^2.
Correct Answer: B — 3/5 MR^2
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Q. For a rectangular plate of mass M and dimensions a x b, what is the moment of inertia about an axis through its center and parallel to side a?
A.
1/12 Mb^2
B.
1/3 Mb^2
C.
1/4 Mb^2
D.
1/6 Mb^2
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Solution
The moment of inertia of a rectangular plate about an axis through its center parallel to side a is I = 1/12 Mb^2.
Correct Answer: A — 1/12 Mb^2
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Q. For a solid disk of mass M and radius R, what is the moment of inertia about an axis through its center and perpendicular to its plane?
A.
1/2 MR^2
B.
1/4 MR^2
C.
MR^2
D.
3/4 MR^2
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Solution
The moment of inertia of a solid disk about an axis through its center is I = 1/2 MR^2.
Correct Answer: A — 1/2 MR^2
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Q. For a system of particles, how is the moment of inertia calculated?
A.
Sum of individual moments
B.
Product of mass and distance squared
C.
Sum of mass times distance squared
D.
Average of all moments
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Solution
The moment of inertia for a system of particles is calculated as I = Σ(m_i * r_i^2), where m_i is the mass and r_i is the distance from the axis.
Correct Answer: C — Sum of mass times distance squared
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Q. For a system of particles, the moment of inertia is calculated as the sum of the products of mass and the square of the distance from the axis of rotation. This is known as:
A.
Parallel Axis Theorem
B.
Perpendicular Axis Theorem
C.
Rotational Dynamics
D.
Angular Momentum
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Solution
This is known as the Parallel Axis Theorem, which states that I = Σ(m_i * r_i^2).
Correct Answer: A — Parallel Axis Theorem
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Q. For a system of particles, the moment of inertia is calculated by summing which of the following?
A.
Masses only
B.
Distances only
C.
Mass times distance squared
D.
Mass times distance
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Solution
The moment of inertia is calculated by summing the products of mass and the square of the distance from the axis of rotation.
Correct Answer: C — Mass times distance squared
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Q. For a system of particles, the total moment of inertia is calculated by which of the following?
A.
Sum of individual moments
B.
Product of mass and distance
C.
Sum of mass times distance squared
D.
Average of individual moments
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Solution
The total moment of inertia for a system of particles is the sum of each particle's moment of inertia, I_total = Σ(m_i * r_i^2).
Correct Answer: C — Sum of mass times distance squared
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Q. For a system of particles, the total moment of inertia is calculated by which of the following methods?
A.
Adding individual moments of inertia
B.
Multiplying total mass by average distance
C.
Using the parallel axis theorem
D.
Using the perpendicular axis theorem
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Solution
The total moment of inertia for a system of particles is calculated by adding the individual moments of inertia.
Correct Answer: A — Adding individual moments of inertia
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Q. For a thin circular ring of mass M and radius R, what is the moment of inertia about an axis perpendicular to its plane through its center?
A.
MR^2
B.
1/2 MR^2
C.
2/3 MR^2
D.
1/3 MR^2
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Solution
The moment of inertia of a thin circular ring about an axis through its center and perpendicular to its plane is I = MR^2.
Correct Answer: A — MR^2
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Q. If a body has a moment of inertia of 15 kg m² and is subjected to a torque of 5 N m, what is its angular acceleration?
A.
0.33 rad/s²
B.
0.5 rad/s²
C.
1 rad/s²
D.
3 rad/s²
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Solution
Angular acceleration α = τ/I = 5 N m / 15 kg m² = 0.33 rad/s².
Correct Answer: A — 0.33 rad/s²
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Q. If a rotating object has a moment of inertia of 4 kg·m² and is spinning with an angular velocity of 3 rad/s, what is its angular momentum?
A.
12 kg·m²/s
B.
4 kg·m²/s
C.
1 kg·m²/s
D.
7 kg·m²/s
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Solution
Angular momentum L = Iω = 4 kg·m² * 3 rad/s = 12 kg·m²/s.
Correct Answer: A — 12 kg·m²/s
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