Classical Mechanics - Rotational Motion

Seven pennies are arranged in a hexagonal, planar pattern so as to touch each neighbor, as shown in the figure. Each penny is a uniform disk of mass m and radius r. What is the moment of inertia of the system of seven pennies about an axis that passes through the center of the central penny and is normal to the plane of the pennies?

A. (7/2) mr²
B. (13/2) mr²
C. (29/2) mr²
D. (49/2) mr²
E. (55/2) mr²

(GR0177 #25)

Solution:

Parallel axis theorem: I = ml² + I_CM

Moment inertia of each penny (a uniform disk): I = ½ mr²

with l = 2r,

I = m(2r)² + ½ mr² = (9/2) mr²

I_total =  I_{6 outer pennies} + I_{1 central penny}

I_total = 6 × (9/2) mr²  + ½ mr² = (55/2) mr²

Answer: E

Classical Mechanics - Rotational Motion

Two uniform cylindrical disks of identical mass M, radius R, and moment inertia ½MR² collide on a frictionless, horizontal surface. Disk I, having an initial counterclockwise angular velocity ω₀ and a center-of-mass velocity v₀ =½ω₀R to the right, makes a grazing collision with disk II initially at rest. If after the collision the two disks stick together, the magnitude of the total angular momentum about the point P is

A. Zero
B. ½MR²ω₀
C. ½MR²v₀
D. MRv₀
E. Dependent on the time of the collision

(GR8677 #97)  

Solution:

L_total = L_trans + L_rot

L_trans = r × p
r = distance from the center of disk I to point P.
At point P, R = r.

L_trans = R × p = R × Mv₀ = M(R × v₀)

From the problem, ω₀  is counterclockwise and v₀ is to the right. Thus, the crossproduct (R × v₀) is negative. Also, v₀ =½ω₀R

L_trans =  −½MR²ω₀

L_rot = Iω₀

Moment inertia, I = ½MR²

L_rot = ½MR²ω₀

L_total = −½MR²ω_{0 +} ½MR²ω₀ = 0

Answer: A