Respuesta :
Answer:
L = 0.882 m
Explanation:
This problem can approximate the bell and the clapper as two pendulums, one a simple pendulum and the other a physical pendulum.
Let's start by analyzing the keystone that is a simple pendulum, with angular velocity
w₁ = √ g / L
The angular velocity is related to the frequency and period
w = 2π f = 2π / T
T₁ = 2π √ L / g
We approach the bell to a physical pendulum, the angular velocity is
w₂ = √ mg d / I
Where m is the mass of the field, d the distance to the pivot point and I the moment of inertia
T₂ = 2π √ I / Mgd
In this case, they indicate that the two bodies have the same period
T₁ = T₂
2π √ L / g = 2π √ I / M g d
L / g = I / M g d
L = I / M d
Let's calculate
L = 18.0 / (34.0 0.60)
L = 0.882 m
The length L of the clapper rod for the bell to ring silently is 0.88 m.
The given parameter:
- Center mass of the bell, r = 0.6 m
- Moment of inertia of the bell, I = 18 kgm²
- Mass of the bell, m = 34 kg
- Mass of the clapper, = 1.8 kg
- Length of the rod, = L
The period of oscillation of pendulum for small amplitude is calculated as follows;
[tex]T= 2\pi \sqrt{\frac{I}{mgd} } \\\\T_b = 2\pi \times \sqrt{\frac{18}{34 \times 9.8 \times 0.6} } \\\\T_b= 1.885 \ s[/tex]
The length of the clapper rod is calculated as follows;
[tex]T_c = 2\pi \sqrt{\frac{L}{g} } \\\\\frac{T_c}{2\pi} = \sqrt{\frac{L}{g} }\\\\\frac{T_c^2}{4\pi^2 } = \frac{L}{g} \\\\L = \frac{gT_c^2}{4\pi^2 }\\\\T_c = T_b\\\\L = \frac{gT_c^2}{4\pi^2 }\\\\L = \frac{9.8 \times (1.885)^2}{4\pi^2 } \\\\L = 0.88 \ m[/tex]
Thus, the length L of the clapper rod for the bell to ring silently is 0.88 m.
Learn more about period of oscillation here: https://brainly.com/question/20070798
