Rotational Motion
Q. A torque of 15 N·m is applied to a wheel with a radius of 0.3 m. What is the force applied tangentially to the wheel?
A.
25 N
B.
50 N
C.
45 N
D.
30 N
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Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 15 N·m / 0.3 m = 50 N.
Correct Answer: B — 50 N
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Q. A torque of 25 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?
A.
50 N
B.
25 N
C.
75 N
D.
100 N
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Solution
Force = Torque / Radius = 25 Nm / 0.5 m = 50 N.
Correct Answer: A — 50 N
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Q. A torque of 25 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?
A.
10 N
B.
25 N
C.
50 N
D.
5 N
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Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 25 Nm / 0.5 m = 50 N.
Correct Answer: A — 10 N
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Q. A torque of 30 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied at the edge of the wheel?
A.
60 N
B.
30 N
C.
15 N
D.
75 N
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Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer: A — 60 N
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Q. A torque of 30 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially?
A.
15 N
B.
30 N
C.
60 N
D.
75 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer: C — 60 N
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Q. A torque of 30 Nm is applied to a wheel. If the radius of the wheel is 0.5 m, what is the force applied tangentially at the edge of the wheel?
A.
15 N
B.
30 N
C.
60 N
D.
75 N
Show solution
Solution
Torque (τ) = F × r, thus F = τ / r = 30 Nm / 0.5 m = 60 N.
Correct Answer: C — 60 N
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Q. A torque of 40 Nm is required to rotate a wheel. If the radius of the wheel is 0.4 m, what is the force applied tangentially?
A.
100 N
B.
80 N
C.
60 N
D.
40 N
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Solution
Force = Torque / Radius = 40 Nm / 0.4 m = 100 N.
Correct Answer: B — 80 N
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Q. A torque of 5 Nm is applied to a wheel. If the radius of the wheel is 0.25 m, what is the force applied tangentially?
A.
10 N
B.
20 N
C.
5 N
D.
15 N
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Solution
Torque (τ) = F × r, thus F = τ / r = 5 Nm / 0.25 m = 20 N.
Correct Answer: A — 10 N
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Q. A torque of 5 N·m is applied to a wheel with a moment of inertia of 2 kg·m². What is the angular acceleration of the wheel?
A.
2.5 rad/s²
B.
5 rad/s²
C.
10 rad/s²
D.
1 rad/s²
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Solution
Angular acceleration α = Torque/I = 5 N·m / 2 kg·m² = 2.5 rad/s².
Correct Answer: A — 2.5 rad/s²
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Q. A torque of 50 Nm is applied to a wheel with a radius of 0.25 m. What is the force applied at the edge of the wheel?
A.
100 N
B.
200 N
C.
250 N
D.
300 N
Show solution
Solution
Force = Torque / Radius = 50 Nm / 0.25 m = 200 N.
Correct Answer: B — 200 N
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Q. A torque of 50 Nm is applied to a wheel with a radius of 0.5 m. What is the force applied tangentially to the wheel?
A.
100 N
B.
50 N
C.
25 N
D.
75 N
Show solution
Solution
Torque (τ) = Force (F) × Radius (r) => F = τ / r = 50 Nm / 0.5 m = 100 N.
Correct Answer: A — 100 N
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Q. A torque of 50 Nm is created by a force acting at a distance of 2 m. What is the force applied?
A.
20 N
B.
25 N
C.
30 N
D.
35 N
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Solution
Force = Torque / Distance = 50 Nm / 2 m = 25 N.
Correct Answer: B — 25 N
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Q. A torque τ is applied to a rigid body with moment of inertia I. If the body starts from rest, what is the angular displacement θ after time t?
A.
(1/2)(τ/I)t^2
B.
(τ/I)t^2
C.
(1/2)(I/τ)t^2
D.
(I/τ)t^2
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Solution
Using the equation of motion for rotation, θ = (1/2)αt^2, where α = τ/I, thus θ = (1/2)(τ/I)t^2.
Correct Answer: A — (1/2)(τ/I)t^2
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Q. A torque τ is applied to a rotating object with moment of inertia I. If the object starts from rest, what is its angular acceleration α?
A.
τ/I
B.
I/τ
C.
Iτ
D.
τI
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Solution
From Newton's second law for rotation, τ = Iα, thus α = τ/I.
Correct Answer: A — τ/I
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Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity of the rod when it makes an angle θ with the vertical?
A.
√(g/L)(1-cosθ)
B.
√(2g/L)(1-cosθ)
C.
√(g/L)(1+cosθ)
D.
√(2g/L)(1+cosθ)
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Solution
Using conservation of energy, the potential energy lost equals the rotational kinetic energy gained. The angular velocity ω can be derived as ω = √(2g/L)(1-cosθ).
Correct Answer: B — √(2g/L)(1-cosθ)
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Q. A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity just before it hits the ground?
A.
√(3g/L)
B.
√(2g/L)
C.
√(g/L)
D.
√(4g/L)
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Solution
Using conservation of energy, potential energy at the top = rotational kinetic energy at the bottom. mgh = (1/2)Iω^2. For a rod, I = (1/3)ML^2, h = L/2. Solving gives ω = √(3g/L).
Correct Answer: B — √(2g/L)
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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. A wheel is rotating with an angular velocity of 10 rad/s. If it accelerates at a rate of 2 rad/s², what will be its angular velocity after 5 seconds?
A.
20 rad/s
B.
10 rad/s
C.
30 rad/s
D.
0 rad/s
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Solution
Final angular velocity = initial angular velocity + (angular acceleration × time) = 10 + (2 × 5) = 20 rad/s.
Correct Answer: A — 20 rad/s
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Q. A wheel of radius R is rolling without slipping on a horizontal surface. What is the relationship between the linear velocity v of the center of the wheel and its angular velocity ω?
A.
v = Rω
B.
v = ω/R
C.
v = 2Rω
D.
v = ω/2R
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Solution
For rolling without slipping, the linear velocity v is related to angular velocity ω by the equation v = Rω.
Correct Answer: A — v = Rω
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Q. A wheel of radius R rolls on a flat surface. If it rolls without slipping, what is the distance traveled by the center of mass after one complete rotation?
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Solution
The distance traveled by the center of mass after one complete rotation is equal to the circumference of the wheel, which is 2πR.
Correct Answer: A — 2πR
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Q. A wheel of radius R rolls without slipping on a horizontal surface. If it rotates with an angular velocity ω, what is the linear velocity of the center of the wheel?
A.
Rω
B.
2Rω
C.
ω/R
D.
R/ω
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Solution
The linear velocity v of the center of the wheel is related to the angular velocity ω by the equation v = Rω.
Correct Answer: A — Rω
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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 composite body made of two solid cylinders of mass M1 and M2 and radius R, what is the total moment of inertia about the same axis?
A.
I1 + I2
B.
I1 - I2
C.
I1 * I2
D.
I1 / I2
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Solution
The total moment of inertia of a composite body is the sum of the individual moments of inertia: I_total = I1 + I2.
Correct Answer: A — I1 + I2
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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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