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ArbitrLikvidat [17]

2 years ago

6

An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is

0.95 rad/s and the natural time period of the axial vibration is found to be 0.35 sec. What is the mass of the object.

1 answer:

devlian [24]2 years ago

4 0

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= $2\pi\sqrt{\frac{L}{g}}$

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= $2\pi\sqrt{\frac{L}{9.81}}$

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = $\frac{\textup{AE}}{\textup{L}}$

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = $\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}$

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = $\sqrt{\frac{K}{m}}$

or

mass, m = $\sqrt{\frac{K}{\omega_n^2}}$

or

mass, m = $\sqrt{\frac{20.66\times10^8}{0.95^2}}$

mass, m = 22.90 × 10⁸ kg

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