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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Q. If a rotating body has an angular momentum of L and its moment of inertia is I, what is the angular velocity ω of the body?
A.
L/I
B.
I/L
C.
L^2/I
D.
I^2/L
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Solution
Angular momentum L = Iω, thus ω = L/I.
Correct Answer: A — L/I
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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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Q. If a rotating object has a moment of inertia of 5 kg·m² and is rotating with an angular velocity of 3 rad/s, what is its angular momentum?
A.
15 kg·m²/s
B.
5 kg·m²/s
C.
8 kg·m²/s
D.
10 kg·m²/s
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Solution
Angular momentum L is given by L = Iω. Thus, L = 5 kg·m² * 3 rad/s = 15 kg·m²/s.
Correct Answer: A — 15 kg·m²/s
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Q. If a rotating object has a moment of inertia of 5 kg·m² and is spinning with an angular velocity of 3 rad/s, what is its angular momentum?
A.
15 kg·m²/s
B.
5 kg·m²/s
C.
8 kg·m²/s
D.
10 kg·m²/s
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Solution
Angular momentum L = Iω = 5 kg·m² * 3 rad/s = 15 kg·m²/s.
Correct Answer: A — 15 kg·m²/s
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Q. If a rotating object has a moment of inertia of I and is rotating with an angular velocity ω, what is its rotational kinetic energy?
A.
1/2 Iω
B.
1/2 Iω^2
C.
Iω^2
D.
Iω
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Solution
The rotational kinetic energy is given by KE = 1/2 Iω^2.
Correct Answer: B — 1/2 Iω^2
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Q. If a satellite is in a geostationary orbit, what is its orbital period?
A.
24 hours
B.
12 hours
C.
6 hours
D.
1 hour
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Solution
A geostationary satellite has an orbital period equal to the Earth's rotation period, which is 24 hours.
Correct Answer: A — 24 hours
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Q. If a satellite is in a stable orbit, what can be said about the net force acting on it?
A.
It is zero
B.
It is equal to the gravitational force
C.
It is equal to the centripetal force
D.
It is equal to the sum of gravitational and centripetal forces
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Solution
In a stable orbit, the net force acting on the satellite is zero because the gravitational force provides the necessary centripetal force.
Correct Answer: A — It is zero
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Q. If a satellite is launched into a higher orbit, how does its potential energy change compared to its initial orbit?
A.
It decreases
B.
It remains the same
C.
It increases
D.
It becomes zero
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Solution
The potential energy of a satellite increases when it is moved to a higher orbit due to the increase in distance from the Earth's center.
Correct Answer: C — It increases
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Q. If a satellite is moved to a higher orbit, what happens to its orbital period?
A.
It decreases.
B.
It increases.
C.
It remains the same.
D.
It becomes zero.
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Solution
The orbital period of a satellite increases when it is moved to a higher orbit, according to Kepler's third law.
Correct Answer: B — It increases.
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Q. If a satellite is moving in a circular orbit, what is the relationship between its centripetal acceleration and gravitational acceleration?
A.
Centripetal = Gravitational
B.
Centripetal > Gravitational
C.
Centripetal < Gravitational
D.
No relationship
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Solution
For a satellite in a stable circular orbit, the centripetal acceleration is equal to the gravitational acceleration.
Correct Answer: A — Centripetal = Gravitational
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Q. If a satellite is moving in a circular orbit, what type of energy does it possess?
A.
Only kinetic energy
B.
Only potential energy
C.
Both kinetic and potential energy
D.
Neither kinetic nor potential energy
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Solution
A satellite in a circular orbit possesses both kinetic energy due to its motion and potential energy due to its position in the gravitational field.
Correct Answer: C — Both kinetic and potential energy
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Q. If a satellite's altitude is doubled, how does its orbital speed change?
A.
Increases by √2
B.
Decreases by √2
C.
Remains the same
D.
Increases by 2
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Solution
If a satellite's altitude is doubled, its orbital speed decreases by √2.
Correct Answer: B — Decreases by √2
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Q. If a satellite's altitude is increased, what happens to its orbital period?
A.
It decreases
B.
It increases
C.
It remains constant
D.
It becomes zero
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Solution
As the altitude increases, the orbital period increases due to the greater distance from the Earth's center.
Correct Answer: B — It increases
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Q. If a satellite's speed is greater than the escape velocity, what will happen?
A.
It will enter a stable orbit
B.
It will escape Earth's gravitational pull
C.
It will crash into the Earth
D.
It will remain in a circular orbit
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Solution
If a satellite's speed exceeds the escape velocity, it will escape Earth's gravitational pull and move away into space.
Correct Answer: B — It will escape Earth's gravitational pull
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Q. If a satellite's speed is less than the required orbital speed, what will happen?
A.
It will remain in orbit.
B.
It will fall back to Earth.
C.
It will escape into space.
D.
It will move to a higher orbit.
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Solution
If a satellite's speed is less than the required orbital speed, it will not have enough centripetal force to maintain its orbit and will fall back to Earth.
Correct Answer: B — It will fall back to Earth.
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Q. If a simple harmonic oscillator has a frequency of 1 Hz, what is the time period?
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
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Solution
Time period (T) is the reciprocal of frequency (f). T = 1/f = 1/1 = 1 s.
Correct Answer: B — 1 s
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Q. If a simple harmonic oscillator has a maximum displacement of 5 cm, what is the amplitude?
A.
2.5 cm
B.
5 cm
C.
10 cm
D.
0 cm
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Solution
The amplitude of a simple harmonic oscillator is defined as the maximum displacement from the equilibrium position, which is 5 cm in this case.
Correct Answer: B — 5 cm
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Q. If a simple harmonic oscillator has a total energy E, what is the kinetic energy when the displacement is half of the amplitude?
A.
E/4
B.
E/2
C.
3E/4
D.
E
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Solution
The total energy E is conserved. When the displacement is half the amplitude, the potential energy is (1/2)E, so the kinetic energy is E - (1/2)E = (1/2)E.
Correct Answer: C — 3E/4
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Q. If a soap film is formed on a wire frame, what is the effect of adding more soap to the film?
A.
Surface tension increases
B.
Surface tension decreases
C.
Surface tension remains the same
D.
Surface tension becomes zero
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Solution
Adding more soap decreases the surface tension because soap molecules disrupt the cohesive forces between water molecules.
Correct Answer: B — Surface tension decreases
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Q. If a solenoid has n turns per unit length and carries a current I, what is the magnetic field inside the solenoid?
A.
0
B.
μ₀nI
C.
μ₀I/n
D.
μ₀I
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Solution
The magnetic field inside a solenoid is given by B = μ₀nI.
Correct Answer: B — μ₀nI
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Q. If a solid cylinder is rotated about its diameter, what is its moment of inertia?
A.
1/2 MR^2
B.
1/4 MR^2
C.
1/3 MR^2
D.
MR^2
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Solution
The moment of inertia of a solid cylinder about its diameter is I = 1/4 MR^2.
Correct Answer: B — 1/4 MR^2
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Q. If a solid cylinder rolls without slipping, what fraction of its total kinetic energy is translational?
A.
1/3
B.
1/2
C.
2/3
D.
1
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Solution
For a solid cylinder, the total kinetic energy is KE_total = KE_translational + KE_rotational = (1/2)mv^2 + (1/2)(1/2)mR^2(ω^2). Since ω = v/R, the translational part is 2/3 of the total.
Correct Answer: C — 2/3
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Q. If a solid cylinder rolls without slipping, what is the ratio of its translational kinetic energy to its rotational kinetic energy?
A.
1:1
B.
2:1
C.
1:2
D.
3:1
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Solution
For a solid cylinder, the translational kinetic energy (KE_trans) is (1/2)mv² and the rotational kinetic energy (KE_rot) is (1/2)(Iω²). The ratio KE_trans:KE_rot is 1:2.
Correct Answer: C — 1:2
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Q. If a solid disk rolls without slipping, what fraction of its total energy is translational at the bottom of an incline?
A.
1/4
B.
1/3
C.
1/2
D.
2/3
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Solution
For a solid disk, the translational kinetic energy is 1/3 of the total kinetic energy when rolling without slipping.
Correct Answer: B — 1/3
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Q. If a solid sphere and a hollow sphere have the same mass and radius, which one will roll down an incline faster?
A.
Solid sphere
B.
Hollow sphere
C.
Both will roll at the same speed
D.
Depends on the angle of incline
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Solution
The solid sphere will roll down the incline faster because it has a smaller moment of inertia compared to the hollow sphere.
Correct Answer: A — Solid sphere
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Q. If a solid sphere and a solid cylinder of the same mass and radius are released from rest at the same height, which will have a greater speed at the bottom?
A.
Solid sphere
B.
Solid cylinder
C.
Both have the same speed
D.
Depends on the mass
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Solution
Both will have the same speed at the bottom due to conservation of energy, as they start from the same height.
Correct Answer: C — Both have the same speed
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