Rotational Motion
Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot?
A.
5 Nm
B.
10 Nm
C.
20 Nm
D.
15 Nm
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Solution
Torque = Force × Distance = 10 N × 2 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 10 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
A.
10 Nm
B.
17.32 Nm
C.
20 Nm
D.
5 Nm
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Solution
Torque = Force × Distance × sin(60°) = 10 N × 2 m × (√3/2) = 17.32 Nm.
Correct Answer: B — 17.32 Nm
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Q. A force of 15 N is applied at an angle of 60 degrees to a lever arm of 1 m. What is the torque?
A.
7.5 Nm
B.
12.5 Nm
C.
15 Nm
D.
25 Nm
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Solution
Torque = Force × Distance × sin(θ) = 15 N × 1 m × sin(60°) = 15 N × 1 m × (√3/2) = 12.5 Nm.
Correct Answer: B — 12.5 Nm
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Q. A force of 20 N is applied at an angle of 30 degrees to the lever arm of 1 m. What is the torque about the pivot?
A.
10 Nm
B.
17.32 Nm
C.
20 Nm
D.
5 Nm
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Solution
Torque = Force × Distance × sin(30°) = 20 N × 1 m × 0.5 = 10 Nm.
Correct Answer: B — 17.32 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?
A.
10 Nm
B.
20 Nm
C.
30 Nm
D.
40 Nm
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Solution
Torque = Force × Distance × sin(θ) = 20 N × 2 m × sin(60°) = 20 N × 2 m × (√3/2) = 20√3 Nm.
Correct Answer: B — 20 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of length 0.5 m. What is the torque about the pivot?
A.
5 Nm
B.
10 Nm
C.
8.66 Nm
D.
17.32 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 20 N × 0.5 m × sin(60°) = 20 N × 0.5 m × (√3/2) = 8.66 Nm.
Correct Answer: C — 8.66 Nm
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Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
10 Nm
B.
20 Nm
C.
17.32 Nm
D.
34.64 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 17.32 Nm
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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?
A.
15 Nm
B.
30 Nm
C.
60 Nm
D.
52 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.
Correct Answer: D — 52 Nm
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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?
A.
10 Nm
B.
15 Nm
C.
20 Nm
D.
25 Nm
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Solution
Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.
Correct Answer: C — 20 Nm
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Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
20 Nm
B.
40 Nm
C.
34.64 Nm
D.
69.28 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.
Correct Answer: C — 34.64 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
86.6 Nm
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Solution
Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 N·m
B.
50 N·m
C.
86.6 N·m
D.
100 N·m
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Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.
Correct Answer: C — 86.6 N·m
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 1 m. What is the torque about the pivot?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.
Correct Answer: B — 43.3 Nm
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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?
A.
25 Nm
B.
50 Nm
C.
43.3 Nm
D.
100 Nm
Show solution
Solution
Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.
Correct Answer: C — 43.3 Nm
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Q. A hollow sphere rolls down a slope of height h. What fraction of its potential energy is converted into translational kinetic energy at the bottom?
A.
1/3
B.
1/2
C.
2/3
D.
1
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Solution
For a hollow sphere, I = (2/3)mr^2. Using energy conservation, the translational kinetic energy is 2/3 of the potential energy at the top.
Correct Answer: C — 2/3
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Q. A hollow sphere rolls down an incline. If it starts from rest, what fraction of its total energy is translational at the bottom?
A.
1/3
B.
2/3
C.
1/2
D.
1/4
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Solution
For a hollow sphere, the translational kinetic energy at the bottom is 2/3 of the total energy, hence the fraction is 2/3.
Correct Answer: B — 2/3
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Q. A hollow sphere rolls down an incline. If its mass is m and radius is R, what is its moment of inertia?
A.
(2/5)mR^2
B.
(1/2)mR^2
C.
(2/3)mR^2
D.
(3/5)mR^2
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Solution
The moment of inertia of a hollow sphere about its center is I = (2/3)mR^2.
Correct Answer: C — (2/3)mR^2
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Q. A particle is moving in a circular path of radius r with a constant angular speed ω. What is the tangential speed of the particle?
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Solution
The tangential speed v is given by v = rω.
Correct Answer: A — rω
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Q. A particle is moving in a circular path of radius r with a constant angular speed ω. What is the tangential speed v of the particle?
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Solution
The tangential speed v is given by v = rω.
Correct Answer: A — rω
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Q. A particle is moving in a circular path of radius R with a constant speed v. What is the centripetal acceleration of the particle?
A.
v²/R
B.
Rv
C.
v/R
D.
R²/v
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Solution
Centripetal acceleration a_c = v²/R.
Correct Answer: A — v²/R
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Q. A particle is moving in a circular path with a radius of 2 m and a speed of 3 m/s. What is the angular momentum of the particle if its mass is 4 kg?
A.
24 kg·m²/s
B.
12 kg·m²/s
C.
6 kg·m²/s
D.
9 kg·m²/s
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Solution
L = mvr = 4 kg * 3 m/s * 2 m = 24 kg·m²/s.
Correct Answer: A — 24 kg·m²/s
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Q. A particle is moving in a circular path with radius r. If the speed of the particle is doubled, how does its angular momentum change?
A.
Remains the same
B.
Doubles
C.
Quadruples
D.
Halves
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Solution
Angular momentum L = mvr, if v is doubled, L also doubles.
Correct Answer: B — Doubles
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Q. A particle is moving in a straight line with a constant velocity. What is its angular momentum about a point that is not on the line of motion?
A.
Zero
B.
Constant
C.
Increasing
D.
Decreasing
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Solution
The angular momentum is constant because the particle's velocity and distance from the point are constant.
Correct Answer: B — Constant
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Q. A particle is moving in a straight line with a velocity v. If it suddenly starts moving in a circular path of radius r, what will be its angular momentum about the center of the circular path?
A.
0
B.
mv
C.
mvr
D.
mv^2/r
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Solution
Angular momentum L = mvr, where v is the linear speed and r is the radius of the circular path.
Correct Answer: C — mvr
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Q. A particle is moving in a straight line with a velocity v. What is its angular momentum about a point O located at a distance r from the line of motion?
A.
0
B.
mv
C.
mvr
D.
mv^2
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Solution
Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point O.
Correct Answer: B — mv
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Q. A particle moves in a circular path of radius R with a constant speed v. What is the centripetal acceleration of the particle?
A.
v^2/R
B.
Rv^2
C.
v/R
D.
R/v^2
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Solution
Centripetal acceleration a_c = v^2/R.
Correct Answer: A — v^2/R
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Q. A particle moves in a circular path with a radius r and a constant speed v. If the speed is doubled, what happens to the angular momentum of the particle?
A.
It remains the same
B.
It doubles
C.
It quadruples
D.
It halves
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Solution
Angular momentum L = mvr; if v is doubled, L also doubles.
Correct Answer: B — It doubles
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Q. A particle moves in a straight line with a constant velocity. What is its angular momentum about a point not on the line of motion?
A.
Zero
B.
Constant
C.
Varies with time
D.
Depends on distance
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Solution
Angular momentum is constant as the particle moves with constant velocity.
Correct Answer: B — Constant
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Q. A particle moves in a straight line with a constant velocity. What is the angular momentum of the particle about a point not on the line of motion?
A.
Zero
B.
Depends on the distance from the point
C.
Infinite
D.
Constant
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Solution
The angular momentum is non-zero and depends on the distance from the point to the line of motion.
Correct Answer: B — Depends on the distance from the point
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Q. A particle moves in a straight line with a velocity v. What is its angular momentum about a point P located at a distance d from the line of motion?
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Solution
Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point P.
Correct Answer: B — mvd
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