The disc of a torsional pendulum has a moment of inertia of 600 k g ⋅ c m 2 kg \sdot cm^2 k g ⋅ c m 2 and is immersed in viscous fluid. The brass shaft attached to it is of 10 c m 10 \ cm 10 c m diameter and 40 c m 40 \ cm 40 c m long. When the pendulum is vibrating the observed amplitudes on the same side of the rest position for successive cycle are 9 ° 9 \degree 9° , 6 ° 6 \degree 6° , 4 ° 4 \degree 4° . Determine the following:
Logarithmic Decrement
Damping torque at unit velocity
The periodic time of vibration
Assume for the brass shaft, G = 4.4 ⨉ 1 0 10 N / m 2 G = 4.4⨉10^{10} N/m^2 G = 4.4⨉1 0 10 N / m 2 . What would be the frequency if the disc is removed from the viscous fluid?
Solution
Given data:
Moment of inertia of disc, I = 600 k g ⋅ c m 2 = 600 × 1 0 − 4 k g ⋅ m 2 = 0.06 k g ⋅ m 2 Diameter of the shaft, d = 10 c m = 0.1 m Length of the shaft, L = 40 c m = 0.4 m Initial amplitude, x 0 = 9 ° Second amplitude, x 1 = 6 ° Third amplitude, x 2 = 4 ° \begin{aligned}
\text{Moment of inertia of disc,} \ \ I &= 600 \ kg \sdot cm^2 \\
&= 600 \times 10^{-4} \ kg \sdot m^2 \\
&= 0.06 \ kg \sdot m^2 \\
\\
\text{Diameter of the shaft,} \ \ d &= 10 \ cm = 0.1 \ m \\
\\
\text{Length of the shaft,} \ \ L &= 40 \ cm = 0.4 \ m \\
\\
\text{Initial amplitude,} \ \ x_0 &= 9 \degree \\
\\
\text{Second amplitude,} \ \ x_1 &= 6 \degree \\
\\
\text{Third amplitude,} \ \ x_2 &= 4 \degree \\
\end{aligned} Moment of inertia of disc, I Diameter of the shaft, d Length of the shaft, L Initial amplitude, x 0 Second amplitude, x 1 Third amplitude, x 2 = 600 k g ⋅ c m 2 = 600 × 1 0 − 4 k g ⋅ m 2 = 0.06 k g ⋅ m 2 = 10 c m = 0.1 m = 40 c m = 0.4 m = 9° = 6° = 4°
1. The logarithmic decrement:
δ = log e ( x 0 x 1 ) = log e ( 9 6 ) = log e ( 1.5 ) = 0.405 \begin{aligned}
\delta &= {\log}_e \bigg( {x_0 \over x_1}\bigg) \\
\\
&= {\log}_e \bigg( {9 \over 6}\bigg) \\
\\
&= {\log}_e (1.5) \\
\\
&= 0.405
\end{aligned} δ = log e ( x 1 x 0 ) = log e ( 6 9 ) = log e ( 1.5 ) = 0.405
2. Damping torque at unit velocity:
ξ = δ 4 π 2 + δ 2 = 0.405 4 π 2 + ( 0.405 ) 2 = 0.0642 \begin{aligned}
\xi &= {\delta \over \sqrt{4 {\pi}^2 + {\delta}^2}} \\
&= {0.405 \over \sqrt{4 {\pi}^2 + (0.405)^2}} \\
&= 0.0642
\end{aligned} ξ = 4 π 2 + δ 2 δ = 4 π 2 + ( 0.405 ) 2 0.405 = 0.0642
The torsional stiffness of shaft is,
K t = G J L = G ⋅ π 32 ⋅ ( d ) 4 L = 4.4 × 1 0 10 × π 32 ⋅ ( 0.1 ) 4 0.4 = 1.07 × 1 0 6 N m / r a d \begin{aligned}
K_t &= {GJ \over L} \\
\\
&= {G \sdot {\pi \over 32} \sdot (d)^4 \over L} \\
\\
&= {4.4 \times 10^{10} \times {\pi \over 32} \sdot (0.1)^4 \over 0.4} \\
\\
&= 1.07 \times 10^6 \ Nm/rad
\end{aligned} K t = L G J = L G ⋅ 32 π ⋅ ( d ) 4 = 0.4 4.4 × 1 0 10 × 32 π ⋅ ( 0.1 ) 4 = 1.07 × 1 0 6 N m / r a d
The natural circular frequency of system is,
ω n = K t I = 1.07 × 1 0 6 0.06 = 4242.5 r a d / s \begin{aligned}
{\omega}_n &= \sqrt{K_t \over I} \\
\\
&= \sqrt{1.07 \times 10^6 \over 0.06} \\
\\
&= 4242.5 \ \ rad/s
\end{aligned} ω n = I K t = 0.06 1.07 × 1 0 6 = 4242.5 r a d / s
Now, Damping torque,
F t = c t ⋅ x ˚ F_t = c_t \sdot \mathring{x} F t = c t ⋅ x ˚
∴ \therefore ∴ Damping torque at unit velocity, F t x ˚ = c t {F_t \over \mathring{x}} = c_t x ˚ F t = c t
Now, we know that;
ξ = c t c c \xi = {c_t \over c_c} ξ = c c c t
∴ c t = ξ ⋅ c c = ξ ( 2 ⋅ I ⋅ ω n ) = 0.0642 ( 2 × 0.06 × 4242.5 ) = 32.74 N ⋅ m ⋅ s / r a d \begin{aligned}
\therefore c_t &= \xi \sdot c_c \\
&= \xi (2 \sdot I \sdot {\omega}_n) \\
&= 0.0642(2 \times 0.06 \times 4242.5) \\
&= 32.74 \ \ N \sdot m \sdot s / rad
\end{aligned} ∴ c t = ξ ⋅ c c = ξ ( 2 ⋅ I ⋅ ω n ) = 0.0642 ( 2 × 0.06 × 4242.5 ) = 32.74 N ⋅ m ⋅ s / r a d
3. The time period of damped vibration:
t p = 2 π ω d = 2 π ω n 1 − ξ 2 = 2 π ( 4242.5 ) 1 − ( 0.0642 ) 2 = 0.00148 s e c o n d \begin{aligned}
t_p &= {2\pi \over {\omega}_d} \\
\\
&= {2\pi \over {\omega}_n \sqrt{1 - {\xi}^2}} \\
\\
&= {2\pi \over (4242.5) \sqrt{1 - (0.0642)^2}} \\
\\
&= 0.00148 \ \ second
\end{aligned} t p = ω d 2 π = ω n 1 − ξ 2 2 π = ( 4242.5 ) 1 − ( 0.0642 ) 2 2 π = 0.00148 seco n d
4. Frequency, if disc is removed from viscous fluid:
f n = ω n 2 π = 4242.5 2 π = 675.21 H z \begin{aligned}
f_n &= {{\omega}_n \over 2\pi} \\
\\
&= {4242.5 \over 2\pi} \\
\\
&= 675.21 \ Hz
\end{aligned} f n = 2 π ω n = 2 π 4242.5 = 675.21 Hz
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