Electromagnetism - Oscillation

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

A small particle of mass m and charge –q is placed at point P and released. If R ≫ x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to:

(GR9677 #04)

Solution:

F_electric = kqQ/r²
F_centripetal = mv²/r = mω²r

F_e = F_c
kqQ/r² = mω²r
ω² = kqQ/mr³

with
k = 1/4πɛ₀
r²  = R² + x²
R ≫ x
r² ∼ R²  → r³ ∼ R³

ω = √(^qQ/_4πɛ0mR³)

Answer: A

Notes: see problem GR9277 #65

Classical Mechanics - Circular Motion

A car travels with constant speed on a circular road on level ground. In the diagram above, F_air is the force of air resistance on the car. Which of the other force shown best represents the horizontal force of the road on the car's tires?

A. F_A 
B. F_B
C. F_C
D. F_D
E. F_E

(GR9677 #05)

Solution:

F_A = Centripetal force
F_C = Frictional force of the road, equals to force exerted by tires exert in the backward direction so that the car moves in the forward direction (Newton's third law Action-Reaction).

The horizontal force on the car’s tires, F_A + F_C = F_B

Answer: B

Classical Mechanics - Pendulum

The figure above represents a point mass m attached to the ceiling by a cord of fixed length l. If the point mass moves in a horizontal circle of radius r with uniform angular velocity ω, the tension in the cord is

A. mgr / l
B. mg cos(θ/2)
C. mωr sin(θ/2)
D. m(ω²r² +g²)^½
E. m(ω⁴r² +g² )^½

(GR8677 #37)

Solution:

Horizontal components:

T sin (θ/2) = F_a = mv² / r = mω²r

Vertical components:
T cos(θ/2) = mg

The magnitude of T:
T² = [T sin(θ/2)]² + [T cos(θ/2)]² = (mω²r)² + (mg)²
T = m(ω⁴r² +g² )^½

Answer: E

Electromagnetism - Particle's Trajectory

A particle with charge q and momentum p is moving in the horizontal plane under the action of a uniform vertical magnetic field of magnitude B. Measurements are made of the particle's trajectory to determine the “sagitta” s and half-chord length l, as shown in the figure above. Which of the following expressions gives the particle's momentum in terms of q, B, s, and l? (Assume s ≪ l).

A. qBs²/2l
B. qBs²/l
C. qBl/s
D. qBl²/2s
E. qBl²/8s

(GR9677 #89)

Solution:



Answer: D

Classical Mechanics - Circular Motion

The coefficient of static friction between a small coin and the surface of a turntable is 0.30. The turntable rotates at 33.3 revolutions per minute. What is the maximum distance from the center of the turntable at which the coin will not slide?

A. 0.024 m
B. 0.048 m
C. 0.121 m
D. 0.242 m
E. 0.484 m

(GR0177 #02)

Solution:

The coin will not slide if  F_centripetal = F_friction
mω²r = μ_smg
r = μ_sg/ω² =  μ_sg/(4π²f²)

Given:
μ_s = 0.30
f = 33.3 revolutions per minute = 33.3/60 = (100/3)(1/60) = 5/9
and take π² = (3.14)² ≈ 10

Radius:
r = (0.3)(10)/[4(10)(25/81)] = (3)(81)/[4(10)(25)] = 243/1000 = 0.243

Answer: D


Notes:


F_centripetal = mv²/r = mω²r
F_friction = μ_sN = μ_smg

Classical Mechanics - Uniform Circular Motion

A satellite of mass m orbits a planet of mass M in a circular orbit of radius R. The time required for one revolution is

A. independent of M
B. proportional to √m
C. linear in R
D. proportional to R^3/2
E. proportional to R²

(GR0177 #03)

Solution:

F_c = F_G
mω²R = GmM / R²
ω² = GM / R³
(2π/T)² = GM / R³
T = (4π²R³/ GM)^1/2
T ∝ R^3/2

Answer: D

Classical Mechanics - Satellite

An astronomer observes a very small moon orbiting a planet and measures the moon’s minimum and maximum distances from the planet’s center and the moon’s maximum orbital speed. Which of the following CANNOT be calculated from these measurements?

A. Mass of the moon
B. Mass of the planet
C. Minimum speed of the moon
D. Period of the orbit
E. Semimajor axis of the orbit

(GR0177 #22)

Solution:

Let m = mass of the moon and M = mass of the planet,

F_c = F_G

mv²/r   = GMm/r²

m cancels out so, there’s no way to calculate mass of the moon.

Answer: A

Electromagnetism - Cyclotron

A nonrelativistic particle with a charge twice that of an electron moves through a uniform magnetic field. The field has a strength of π/4 tesla and is perpendicular to the velocity of the particle. What is the particle’s mass if it has a cyclotron frequency of 1,600 hertz?

A. 2.5 × 10^-23 kg
B. 1.2 × 10^-22 kg
C. 3.3 × 10^-22 kg
D. 5.0 × 10^-21 kg
E. 7.5 × 10^-21 kg

(GR0177 #62)

Solution:





Angular velocity: 





Answer: A