Rotational Motion
Q. A wheel of radius R rolls without slipping on a horizontal surface. If the wheel has an angular velocity ω, what is the linear velocity of the center of the wheel?
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Solution
The linear velocity v of the center of the wheel is related to the angular velocity by v = Rω.
Correct Answer: A — Rω
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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
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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. 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 body is rotating with an angular momentum L and its moment of inertia is halved, what will be the new angular momentum if the angular velocity remains constant?
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Solution
Angular momentum L = Iω; if I is halved and ω remains constant, L remains L.
Correct Answer: A — L
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Q. If a child sitting on a merry-go-round moves closer to the center, what happens to the angular velocity of the merry-go-round?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
As the child moves closer, the moment of inertia decreases, and to conserve angular momentum, angular velocity must increase.
Correct Answer: A — Increases
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Q. If a child sitting on a merry-go-round moves from the center to the edge, what happens to the angular momentum of the system if no external torque acts?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
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Solution
Angular momentum remains constant if no external torque acts, according to the conservation of angular momentum.
Correct Answer: C — Remains constant
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Q. If a child sitting on a merry-go-round moves towards the center, what happens to the angular velocity of the merry-go-round?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
As the child moves towards the center, the moment of inertia decreases, thus angular velocity increases to conserve angular momentum.
Correct Answer: A — Increases
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Q. If a disc rolls without slipping on a flat surface, what is the relationship between its linear velocity v and angular velocity ω?
A.
v = rω
B.
v = 2rω
C.
v = 1/2 rω
D.
v = 3rω
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Solution
For rolling without slipping, the linear velocity v is equal to the radius r times the angular velocity ω, hence v = rω.
Correct Answer: A — v = rω
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Q. If a force of 12 N is applied at an angle of 30 degrees to a lever arm of 1 m, what is the torque about the pivot?
A.
6 Nm
B.
10 Nm
C.
12 Nm
D.
15 Nm
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Solution
Torque = Force × Distance × sin(θ) = 12 N × 1 m × sin(30°) = 12 N × 1 m × 0.5 = 6 Nm.
Correct Answer: A — 6 Nm
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Q. If a force of 15 N is applied at an angle of 30 degrees to the lever arm of length 1.5 m, what is the torque about the pivot?
A.
3.75 Nm
B.
7.5 Nm
C.
11.25 Nm
D.
12.99 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 15 N × 1.5 m × sin(30°) = 15 × 1.5 × 0.5 = 11.25 Nm.
Correct Answer: B — 7.5 Nm
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Q. If a hollow cylinder and a solid cylinder of the same mass and radius roll down the same incline, which one reaches the bottom first?
A.
Hollow cylinder
B.
Solid cylinder
C.
Both reach at the same time
D.
Depends on the angle of incline
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Solution
The solid cylinder reaches the bottom first because it has a smaller moment of inertia compared to the hollow cylinder.
Correct Answer: B — Solid cylinder
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Q. If a hollow cylinder rolls down an incline, how does its acceleration compare to that of a solid cylinder?
A.
Hollow cylinder accelerates faster
B.
Solid cylinder accelerates faster
C.
Both accelerate equally
D.
Depends on the angle of incline
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Solution
The solid cylinder has a lower moment of inertia compared to the hollow cylinder, thus it accelerates faster down the incline.
Correct Answer: B — Solid cylinder accelerates faster
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Q. If a hollow sphere and a solid sphere of the same mass and radius roll down the same incline, which one reaches the bottom first?
A.
Hollow sphere
B.
Solid sphere
C.
Both reach at the same time
D.
Depends on the angle of incline
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Solution
The solid sphere has a smaller moment of inertia compared to the hollow sphere, thus it accelerates faster and reaches the bottom first.
Correct Answer: B — Solid sphere
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Q. If a planet rotates about its axis, which of the following statements is true regarding its angular momentum?
A.
It is zero
B.
It is constant
C.
It changes with time
D.
It depends on the distance from the sun
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Solution
The angular momentum of a rotating planet is constant if no external torques act on it.
Correct Answer: B — It is constant
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Q. If a rolling object has a mass m and radius r, what is the expression for its total kinetic energy?
A.
(1/2)mv^2
B.
(1/2)mv^2 + (1/2)Iω^2
C.
(1/2)mv^2 + (1/2)mr^2ω^2
D.
(1/2)mv^2 + (1/2)(2/5)mr^2(ω^2)
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Solution
The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which can be expressed as (1/2)mv^2 + (1/2)(2/5)mr^2(ω^2).
Correct Answer: D — (1/2)mv^2 + (1/2)(2/5)mr^2(ω^2)
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Q. If a rolling object has a radius of R and rolls with a speed v, what is its kinetic energy?
A.
(1/2)mv^2
B.
(1/2)mv^2 + (1/2)Iω^2
C.
(1/2)mv^2 + (1/2)(1/2)mR^2(v/R)^2
D.
None of the above
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Solution
The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which simplifies to (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.
Correct Answer: C — (1/2)mv^2 + (1/2)(1/2)mR^2(v/R)^2
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Q. If a rolling object has a radius R and rolls with an angular velocity ω, what is its linear velocity?
A.
Rω
B.
2Rω
C.
R/2ω
D.
3Rω
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Solution
The linear velocity v of a rolling object is given by v = Rω.
Correct Answer: A — Rω
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Q. If a rolling object has a translational speed of v and a rotational speed of ω, what is the relationship between them for rolling without slipping?
A.
v = ωR
B.
v = 2ωR
C.
v = ω/R
D.
v = R/ω
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Solution
For rolling without slipping, the relationship is v = ωR, where v is the translational speed and ω is the angular speed.
Correct Answer: A — v = ωR
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