Physics Problems & Solutions: Pendulum

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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 - Pendulum

Two pendulums are attached to a massless spring, as shown above. The arms of the pendulums are of identical lengths l, but the pendulum balls have unequal masses m₁and m₂. The initial distance between the masses is the equilibrium length of the spring, which has spring contant K. What is the highest normal mode frequency of this system?

A. $\sqrt{g/l}$

B. $\sqrt{\frac{K}{m_1+m_2}}$

C. $\sqrt{\frac{K}{m_1}+\frac{K}{m_2}}$

D. $\sqrt{\frac{g}{l}+\frac{K}{m_1}+\frac{K}{m_2}}$

E. $\sqrt{\frac{2g}{l}+\frac{K}{m_1+m_2}}$

(GR9677 #84)

Solution:

A. FALSE
ω = √(g/l) is angular frequency for single pendulum

B. and C are FALSE
Both answers do not depend on g/l

D. TRUE
If there is no dependence on K → ω = √(g/l)
And if m₂→ ∞, m₁will still oscillate with spring constant K but has no dependence on m₂_,as if it were connected to a stationary object.

E. FALSE
If there is no dependence on K → ω = √(2g/l), not angular frequency for single pendulum
And if m₂→ ∞, ω = √(2g/l) has no dependence on K and m₁

Answer: D

Classical Mechanics - Pendulum

A cylindrical tube of mass M can slide on a horizontal wire. Two identical pendulum, each of mass m and length l, hang from the ends of the tube, as shown above. For small oscillations of the pendulums in the plane of the paper, the eigenfrequencies of the normal modes of oscillation of the system are 0, $\sqrt{\frac{g(M+2m)}{lM}}$ , and …

A. $\sqrt{\frac{g}{l}}$

B. $\sqrt{\frac{g}{l} \frac{M+m}{M}}$

C. $\sqrt{\frac{g}{l} \frac{m}{M}}$

D. $\sqrt{\frac{g}{l} \frac{m}{M+m}}$

E. $\sqrt{\frac{g}{l} \frac{m}{M+2m}}$

(GR9277 #07)

Solution:

2 modes in which the system can oscillate:

1. The 2 pendulum oscillate out of phase so there is torsional effect on the tube M → there is M in the equation.

2. In phase → no torsional effect on M → No M in the equation.

Answer: A

Classical Mechanics - Physical Pendulum

A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, shown above, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of configuration I?

A. (6/5)^½
B. (3/2)^½
C. 6/5
D. 3/2
E. 5/3

(GR9277 #61)

Solution:

Angular frequency for physical pendulum:

ω = (MgL/I)^1/2

Pendulum I
M = 2m
I = mr² + mr² = 2mr²
L = r + r = 2r

ω_I= (2mg2r/2mr²)^1/2= (2g/r)^1/2

Pendulum II
M = 2m
I = mr² + m(r/2)² = (⁵/₄)mr²

L = ½ r + r = (³/₂)r

ω_II=[ 2mg (³/₂)r / (⁵/₄)mr²]^1/2= (12g/5r)^1/2

The ratio:

ω_II/ ω_I= (12g/5r)^1/2/ (2g/r)^1/2= 6/5

Answer: A

Classical Mechanics - Small Oscillation

Two circular hoops, X and Y, are hanging on nails in a wall. The mass of X is four times that of Y, and the diameter of X is also four times that of Y. If the period of small oscillations of X is T, the period of small oscillations of Y is

A. T
B. T/2
C. T/4
D. T/8
E. T/16

(GR9277 #74)

Solution:

Small oscillation → simple pendulum.

Period of simple pendulum:
$T=2\pi\sqrt{\frac{L}{g}}$
→ $T\propto \sqrt{L}$

In this case,

= radius of the hoop.

$\\T\propto \sqrt{r_x}\\T_y\propto \sqrt{r_y}=\sqrt{r_x/4}=\frac{\sqrt{r_x}}{2}=\frac{T}{2}$

Answer: B

Classical Mechanics - Pendulum

Two small spheres of putty, A and B of mass M and 3M, respectively, hang from the ceiling on strings of equal length l. Sphere A is drawn aside so that it is raised to a height h₀ as shown above and then released. Sphere A collides with sphere B; they stick together and swing to a maximum height h equal to

A. (1/16) h₀
B. (1/8) h₀
C. (1/4) h₀
D. (1/3) h₀
E. (1/2) h₀

(GR8677 #5)

Solution:

Conservation of energy of A before and when it hits B:
$\begin{aligned} Mgh_0&=\frac{1}{2} Mv_A^2\\v_A&=\sqrt{2gh_0}\end{aligned}$

Conservation of momentum when and after collision:
$\begin{aligned} m_A v_A+m_B v_B&=(m_A+m_B)v_{AB}\\M\sqrt{2gh_0}+0&=4Mv_{AB}\\v_{AB}&=\frac{\sqrt{2gh_0}}{4}\end{aligned}$

Conservation of energy of A and B at h = 0 and h_max:
$\begin{aligned} \frac{1}{2} m_{AB} v_{AB}^2&=m_{AB} gh_{\max}\\\frac{1}{2} 4M\left (\frac{\sqrt{2gh_0}}{4} \right )^2&=4Mgh_{\max}\\h_{\max}&=\frac{1}{16} h_0\end{aligned}$

Answer: A

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 - Pendulum

Which of the following best illustrates the acceleration of a pendulum bob at points a through e?

(GR0177 #01)

Solution:

The acceleration of pendulum: a = a_centripetal + a_tangential
a_cent = v²/r = ω²r
a_tan = αr

At equilibrium (position B): ω = constant
α = dω/dt = 0
a = a_cent

At maximum amplitude (position A and C): v = 0
a = a_tan

At other positions:
a = a_cent + a_tan

Answer: C

Classical Mechanics - Pendulum

The figure above shown a small mass connected to a string, which is attached to a vertical post. If the mass is released when the string is horizontal as shown, the magnitude of the total acceleration of the mass as a function of the angle θ is

A. $g \sin \theta$
B. $2g \cos \theta$
C. $2g \sin \theta$
D. $g\sqrt{3\cos^2\theta+1}$
E. $g\sqrt{3\sin^2\theta+1}$

(GR9277 #93)

Solution:

Total acceleration:

$\vec{a}=\vec{a}_{\text{tangential}}+\vec{a}_{\text{centripetal}}$

At initial position where $\theta=0$ , there is no string tension $\rightarrow \vec{a}_{\text{centripetal}}=0$

$\vec{a}=\vec{a}_{\text{tangential}}+0=\vec{a}_{\text{tangential}}=g$

Using method of elimination:

A. $g \sin \theta = g \sin 0=0 \rightarrow$ FALSE

B. $2g \cos \theta=2g \cos 0=2g \rightarrow$ FALSE

C. $2g \sin \theta=2g \sin 0=0 \rightarrow$ FALSE

D. $g\sqrt{3\cos^2\theta+1}=g\sqrt{3(1)+1}=2g \rightarrow$ FALSE

E. $g\sqrt{3\sin^2\theta+1}=g\sqrt{0+1}=g \rightarrow$ TRUE

Answer: E

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