Classical Mechanics - Harmonic Oscillator

A particle of mass m undergoes harmonic oscillation with period T₀. A force f proportional to the speed v of the particle, f = −bv, is introduced. If the particle continues to oscillate, the period with f acting is

A. Larger than T₀
B. Smaller than T₀
C. Independent of b
D. Dependent linearly on b
E. Constantly changing

(GR9677 #08)

Solution:

 f = −bv → minus sign means f is restoring force (damped oscillation).
The oscillation is getting slower (larger period) before it finally comes to stop.

Answer: A

Classical Mechanics - Torque

A solid cone hangs from a frictionless pivot at the origin O as shown above. If î, ĵ and k̂ are unit vectors, and a, b, and c are positive constants, which of the following forces F applied to the rim of the cone at point P results in a torque τ on the cone with a negative component τ_z?

A. F = ak̂, P is (0, b, −c)
B. F = −ak̂, P is (0, −b, −c)
C. F = aĵ, P is (–b, 0, −c)
D. F = aĵ, P is (b, 0, −c)
E. F = −ak̂, P is (−b, 0, −c)

(GR9277 #08)

Solution:

Torque:

Result desired: negative component of τ_z so we are looking for minus k̂ component.

For k̂ component: r_xF_y − r_yF_x

Thus, F_x and F_y cannot be zero

(A) FALSE
 F = ak̂ → F_x = 0 and F_y = 0

(B) and (E) FALSE
F = −ak̂ → F_x = 0 and F_y = 0

(C) TRUE
F = aĵ → F_x = 0, F_y = a,  r_x = −b, r_y = 0,
P is (–b, 0, −c) → r_x = −b, r_y = 0
r_xF_y − r_yF_x = −ab − 0 =  −abk̂ (negative k̂ component)

(D) FALSE
F = aĵ → F_x = 0, F_y = a,
P is (b, 0, −c) → r_x = b, r_y = 0,
→ r_xF_y − r_yF_x = ab − 0 =  abk̂ (positive k̂ component)

Answer: C

Classical Mechanics - Kinematics

A 5 kilogram stone is dropped on a nail and drives the nail 0.025 meter into a piece of wood. If stone is moving at 10 meters per second when it hits the nail, the average force exerted on the nail by the stone while the nail is going into the wood is most nearly 

A. 10 N
B. 100 N
C. 1000 N
D. 10,000 N
E. 100,000 N

(GR8677 #08)

Solution:

Force: F = ma

The acceleration when the stone hits the nail: 2ay = v² + v₀²

v₀ = 0 → 2ay = v²

a = v²/2y = 10²/2(0.025) = 2 × 10³ m/s²

F = 5 × 2 × 10³ = 10,000 N

Answer: D

Electromagnetism - Method of Image

A positive charge Q is located at a distance L above an infinite grounded conducting plane, as shown in the figure above. What is the total charge induced on the plane?

A. 2Q
B. Q
C. 0
D. –Q
E. –2Q

(GR0177 #08)

Solution:

Method of image:

• +Q attracts negative charges in the plate and repels positive ones. 
• As the positive charges want to “get away”, they succeed in doing so through the ground, leaving the negative charges behind. 
• The plate is left with a net negative charge.

Answer: D