Rotational Motion
Q. A flywheel is rotating with an angular speed of 20 rad/s. If it experiences a torque of 5 Nm, what is the time taken to stop it?
A.
8 s
B.
4 s
C.
10 s
D.
5 s
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Solution
Using τ = Iα, we find α = τ/I. Assuming I = 1 kg·m², α = 5 rad/s². Time to stop = ω/α = 20 rad/s / 5 rad/s² = 4 s.
Correct Answer: B — 4 s
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Q. A flywheel is rotating with an angular velocity of 10 rad/s. If it is subjected to a torque of 5 Nm, what is the angular acceleration?
A.
0.5 rad/s²
B.
2 rad/s²
C.
0.2 rad/s²
D.
1 rad/s²
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Solution
Using τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration. Assuming I = 25 kg·m², α = τ/I = 5/25 = 0.2 rad/s².
Correct Answer: B — 2 rad/s²
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Q. A flywheel is rotating with an angular velocity of 15 rad/s. If it comes to rest in 3 seconds, what is the angular deceleration?
A.
5 rad/s²
B.
10 rad/s²
C.
15 rad/s²
D.
20 rad/s²
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Solution
Angular deceleration = (final angular velocity - initial angular velocity) / time = (0 - 15) / 3 = -5 rad/s², so the magnitude is 5 rad/s².
Correct Answer: B — 10 rad/s²
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Q. A flywheel is rotating with an angular velocity of 15 rad/s. If it experiences a torque of 3 N·m, what is the angular acceleration?
A.
0.2 rad/s²
B.
0.5 rad/s²
C.
1 rad/s²
D.
5 rad/s²
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Solution
Angular acceleration α = Torque/I. Assuming I = 6 kg·m², α = 3 N·m / 6 kg·m² = 0.5 rad/s².
Correct Answer: B — 0.5 rad/s²
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Q. A flywheel is rotating with an angular velocity of 20 rad/s. If it comes to rest in 5 seconds, what is the angular deceleration?
A.
4 rad/s²
B.
5 rad/s²
C.
20 rad/s²
D.
0 rad/s²
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Solution
Angular deceleration α = (ω_f - ω_i) / t = (0 - 20 rad/s) / 5 s = -4 rad/s², so the magnitude is 4 rad/s².
Correct Answer: B — 5 rad/s²
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Q. A flywheel is rotating with an angular velocity of 20 rad/s. If it experiences a constant torque that reduces its angular velocity to 10 rad/s in 5 seconds, what is the magnitude of the torque if the moment of inertia is 4 kg·m²?
A.
8 N·m
B.
4 N·m
C.
2 N·m
D.
10 N·m
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Solution
The angular deceleration α = (ω_final - ω_initial) / time = (10 - 20) / 5 = -2 rad/s². Torque τ = Iα = 4 kg·m² * (-2 rad/s²) = -8 N·m, so the magnitude is 8 N·m.
Correct Answer: B — 4 N·m
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Q. A force of 10 N is applied at a distance of 0.5 m from the pivot point. What is the torque about the pivot?
A.
2.0 Nm
B.
5.0 Nm
C.
10.0 Nm
D.
20.0 Nm
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Solution
Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.
Correct Answer: A — 2.0 Nm
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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 point?
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 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 a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?
A.
25 Nm
B.
43.3 Nm
C.
50 Nm
D.
0 Nm
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Solution
Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 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 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 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 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
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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
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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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