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 - Orbital Path

When it is about the same distance from the Sun as is Jupiter, a spacecraft on a mission to the outer planets has a speed that is 1.5 times the speed of Jupiter in its orbit. Which of the following describes the orbit of the spacecraft about the Sun?

A. Spiral
B. Circle
C. Ellipse
D. Parabola
E. Hyperbola

(GR9677 #66)

Solution:

The mission is to outer planets, so the path should not be bounded (no longer a close path): A, B, C are FALSE.

If v_escape = √2 v_circular → Parabola
If v_e  √2 v_c → Hyperbola
Since v_e = 1.5  √2 → Hyperbola

Answer: E

Notes:

To find escape velocity:
F_c = F_g 
mv²/ r = GMm / r² 
v_circular = √(GM / r)

KE = PE_gravity 
½ mv² = GMm / r  
v_escape = √(2GM / r) = √2 √(GM / r)
v_escape = √2 v_circular

Classical Mechanics - Circular Motion

A rigid cylinder rolls at constant speed without slipping on top of a horizontal plane surface. The acceleration of a point on the circumference of the cylinder at the moment when the point touches the plane is

A. directed forward
B. directed backward
C. directed up
D. directed down
E. zero

(GR9277 #40)

Solution:

Constant speed → no tangential acceleration → Uniform Circular Motion



When the point touches the plane,

 is directed up

Answer: C

Classical Mechanics - Circular Motion

A particle of mass M moves in a circular orbit of radius r around a fixed point under the influence of an attractive force F = ^K⁄_r³, where K is a constant. If the potential energy of the particle is zero at an infinite distance from the force center, the total energy of the particle in the circular orbit is

A. − ^K⁄_r² 
B. − ^K⁄_2r²
C. 0
D. ^K⁄_2r²
E. ^K⁄_r²

(GR9277 #87)

Solution:

Attractive force = Centripetal Force
 ^K⁄_r³ = ^mv²⁄_r
mv² = ^K⁄_r²

Kinetic energy:  T = ½mv² = ^K⁄_2r²
Potential energy: V(r) = − ∫ F dr = −K ∫ ¹⁄_r³ dr = ^K⁄_2r²
Attractive force → negative potential energy, V(r) = − ^K⁄_2r²
Total energy:  T + V = ^K⁄_2r²  −  ^K⁄_2r² = 0

Answer: C

Classical Mechanics - Satellite

A Satellite orbits the Earth in a circular orbit. An Astronaut on board perturbs the orbit slightly by briefly firing a control jet aimed toward the Earth’s center. Afterward, which of the following is true of the satellite’s path?

A. It is an ellipse
B. It is a hyperbola
C. It is a circle with larger radius
D. It is a spiral with increasing radius
E. It exhibits many radial oscillations per revolution.

(GR8677 #02)

Solution:

Initially, the object orbits the Earth in a circular orbit.

Perturbs the orbit slightly means giving some extra momentum, so the orbit won't be circular any longer, and will be elliptic, not enough to be Parabolic or Hyperbolic.

Also, logically the astronaut will not want the satellite to have v = v_escape (parabolic) or v v_escape (hyperbolic).

Answer: A


Notes:

Types of Orbits	Eccentricity	Energy	Velocity
Circular	e = 0	E = Vmin
Elliptic	0 e 1	Vmin E 0	v vescape
Parabolic	e = 1	E = 0	v = vescape It will escape the gravitational pull of the planet. If v is increased it will become a hyperbolic orbit.
Hyperbolic	e 1	E 1	v vescape It escapes the gravitational pull of the planet and continues to travel infinitely until it is acted upon by another body with sufficient gravitational force.

The orbital eccentricity, e is the amount by which its orbit deviates from a perfect circle.

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

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

Classical Mechanics - Non Uniform Circular Motion

A particle constrained to move in a circle with 10-meter radius. At one instant, the particle’s speed is 10 meters per second and is increasing at a rate of 10 meters per second squared. The angle between the particle’s velocity and acceleration vector is

A. 0^o
B. 30^o
C. 45^o
D. 60^o
E. 90^o

(GR0177 #23)

Solution:



 = acceleration vector

The speed is increasing →   ∥ 

Given:
v = 10 m/s
r = 10 m
a_tan = 10 m/s²

a_c = v²/r = (10)²/(10) = 10 m/s²

tan θ = a_tan/a_c = 10/10 = 1 → θ = 45^o

Answer: C