Classical Mechanics - Moment Inertia

The period of physical pendulum is 2π√(I/mgd), where I is the moment of inertia about the pivot point and d is the distance from the pivot to the center of mass. A circular hoop hangs from a nail on a barn wall. The mass of the hoop is 3 kilograms and its radius is 20 centimeters. If it is displaced slightly by a passing breeze, what is the period of the resulting oscillations?

A. 0.63 s
B. 1.0 s
C. 1.3 s
D. 1.8 s
E. 2.1 s

(GR9677 #21)

Solution:

T = 2π√(I/mgd)
m = 3 kg
r = 20 cm = 0.2 m
In this case d = r

To find total I:
Parallel axis theorem: I = ml² + I_CM
I_CM  = I_loop = mr² 

In this case l = r
→ I = mr² + mr² = 2mr²

T = 2π√(2mr² / mgr) 
= 2π√(2r/g) 
= 2π√[2(0.2)/(10)] 
= 2π(0.2)
= 1.25 s

Answer: C

Classical Mechanics - Rotational Kinetic Energy

Three equal masses m are rigidly connected to each other by massless rods of length l forming an equilateral triangle. The assembly is to be given an angular velocity ω about an axis perpendicular to the triangle. For fixed ω, the ratio of the kinetic energy of the assembly for an axis through B compared with that for an axis through A is equal to

A. 3
B. 2
C. 1
D. 1/2
E. 1/3

(GR9677 #32)

Solution:

KE = ½Iω²

For fix ω, the ratio KE_B/KE_A = I_B/I_A

I = ∑_i m_ir_i²

From A:
m₁= m₂ = m₃ = m
x₁= x₂ = x₃ = l/√3 (see notes)

I_A = m₁x₁² + m₂x₂² + m₃x₃²
= 3m( l/√3)² = ml²

From B:
I_B = m₁x₁² + m₂x₂² 
= ml² + ml² = 2ml²

KE_A/KE_B = 2ml²/ml² = 2

Answer: B

Notes:

Cos 30^o = ½l / x₁ 
Cos 30^o = ½√3
½l / x₁ = ½√3
x₁ = x₂ = x₃ = l/√3 

Classical Mechanics - Lagrangian

A bead is constrained to slide on a frictionless rod that is fixed at an angle θ with a vertical axis and is rotating with angular frequency ω about the axis, as shown above. Taking the distance s along the rod as the variable, the Lagrangian for the bead is equal to

A. ½ mṡ ² − mgs cos θ 
B. ½ mṡ ² + ½ m(ωs)² − mgs 
C. ½ mṡ ² + ½ m(ωs cos θ)² + mgs cos θ
D. ½ m(ṡ sin θ)² − mgs cos θ 

E. ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

(GR9677 #68)

Solution:

Lagrangian: L = T − U

Potential energy: U = mgh = mgs cos θ 
Kinetic Energy:  T_kin  =  ½ mṡ ²
Rotational kinetic energy:  T_rot =  ½ Iω²
with moment inertia: I = mr²  = m(s sin θ)²
→ T_rot = ½ m(ωs sin θ)²

L = T_kin + T_rot − U =  ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

Answer: E

Classical Mechanics - Rotational Motion

Questions 41-42

A cylinder with moment inertia 4 kgm² about a fixed axis initially rotates at 80 radians per second about this axis. A constant torque is applied to slow it down to 40 radians per second. The kinetic energy lost by the cylinder is

A. 80 J
B. 800 J
C. 4000 J
D. 9600 J
E. 19,200 J

(GR9277 #41) 

Solution:

Rotational Kinetic Energy:  



Answer: D

Classical Mechanics - Angular Speed

A thin plate of mass M, length L, and width 2d is mounted vertically on a frictionless axle along the z-axis. Initially the object is at rest. It is then tapped with a hammer to provide a torque τ, which produces an angular impulse H about the z-axis of magnitude H = ∫ τ dt. What is the angular speed ω of the plate about the z-axis after the tap? 

A. H/2Md²
B. H/Md²
C. 2H/Md²
D. 3H/Md²
E. 4H/Md²

(GR9277 #82)

Solution:

H = ∫ τ dt

Torque: τ = Iα
Angular acceleration: α = ω/t → ω = αt

H = ∫ τ dt = Iα ∫dt = Iαt = Iω

Moment inertia for the plate about the z-axis: I = ¹/₃Md²

H = ¹/₃Md²ω
ω = 3H/Md²

Answer: D

Nuclear & Particle Physics - Hydrogen

The spacing of the rotational energy levels for the hydrogen molecule H₂ is most nearly

A. 10⁻⁹ eV
B. 10⁻³ eV
C. 10 eV
D. 10 MeV
E. 100 MeV

(GR9277 #90)

Solution:

Rotational kinetic energy: E = ^L²∕_2I
Angular momentum: L² = l(l+1)ħ² with l = 0,1,2,⋯
Moment Inertia: I = mr²

For hydrogen atom:
r = 0.529 × 10⁻¹⁰ m
m =10⁻²⁷ kg (mass of proton)
→ I = 10⁻²⁷(0.529×10⁻¹⁰)² ≈ 10⁻⁴⁷ kgm²
ħ = ^h∕_2π = ^{6.63×10−34}∕ _2(3.14) ≈ 10⁻³⁴

For l = 0 → L = 0 → E = 0
For l = 1 → L = 2ħ² → E = ^ħ²∕_I

The spacing of the rotational energy levels:
∆E = ^ħ²∕_I  − 0 = ^(10−34)2∕ _10⁻⁴⁷ = 10⁻⁶⁸⁺⁴⁷ = 10⁻²¹ J

Convert to eV: 1 eV = 1.6 × 10⁻¹⁹ J
→ ∆E = ¹⁰⁻²¹∕ _{1.6 × 10⁻¹⁹} ≈ 10⁻³ eV

Answer: B

Classical Mechanics - Rotational Motion

A hoop of mass M and radius R is at rest at the top of an inclined plane. The hoop rolls down the plane without slipping. When the hoop reached the bottom, its angular momentum around its center of mass is

A. MR√(gh)
B. ½ MR√(gh)
C. M√(2gh)
D. Mgh
E. ½ Mgh

(GR8677 #76)

Solution:

Conservation of energy: Total energy at the top = Total energy at the bottom



Velocity: v_CM = ωR
Moment inertia of the hoop: I_CM = MR²



Angular Momentum: 

Answer: A

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

A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point, as shown in the figure, and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?

A. 
B. 
C. 
D. 
E. 

(GR0177 #26)

Solution:


 


Conservation of energy of the system before and after the rod falls onto the ground:
Mgy = ½Iω²

y = L/2 (center of the mass of the rod is at the center of the rod)
Moment inertia of uniform rod: I = ¹/₃ML² 
v = rω 
v_f = ωL → ω = v_f /L 

Mg(L/2) = ½(¹/₃ML²)(v_f /L)²
gL = ¹/₃ v_f ²
v_f  = (3gL)^1/2 

Answer: C

Classical Mechanics - Rotational Motion

A child is standing on the edge of a merry-go-around that has the shape of a solid disk, as shown in the figure. The mass of the child is 40 kilograms. The merry-go-around has a mass 200 kilograms and a radius of 2.5 meters and it is rotating with an angular velocity of 2.0 radians per second. The child then walk slowly toward the center of the merry-go-around. What will the final angular velocity of the merry-go-around when the child reaches the center? (The size of the child can be neglected).

A. 2.0 rad/s
B. 2.2 rad/s
C. 2.4 rad/s
D. 2.6 rad/s
E. 2.8 rad/s

(GR0177 #89)

Solution:

Angular Momentum: L = Iω

Conservation of Angular Momentum:

∑ L_i = ∑ L_f

(I_m + I_c)ω = (I_m' + I_c')ω' 
ω' = ω(I_m + I_c)/(I_m' + I_c')

Moment inertia of the uniform disk is the same before and after the child moves to the center:

I_m = I_m' = I_m = ½ m_mr² = ½ (200)(2.5)²
ω = 2 rad/s

Initally, moment inertia of the child:

I_c = m_cr² = (40)(2.5)²

When the child reaches the center, r = 0:

I_c' = 0

Thus,

ω' = ω(I_m + I_c)/(I_m' + I_c')
= (2) [½(200)(2.5)² + (40)(2.5)²] / [½ (200)(2.5)²]
= (2) [½(200) + (40)] / [½ (200)]
= 280/100
= 2.8 rad/s

Answer: E

Classical Mechanics - Rotational Motion

The cylinder shown above, with mass M and radius R, has a radially dependent density. The cylinder starts from from rest and rolls without slipping down an inclined plane of height H. At the bottom of the plane its translational speed is (8gH/7)^{1/2. Which of the following is the rotational inertia of the cylinder?}

(GR0177 #91)

Solution:

Initally, it starts from rest, from height H,



At the bottom of the plane,



Conservation of Energy,



with and



Answer: B

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