High School

The mass of the cylinder is 118 kg and the cross-sectional area is [tex]2.29 \times 10^{-1} \, \text{m}^2[/tex]. The density of the water is [tex]1000 \, \text{kg/m}^3[/tex]. Show that the angular frequency of oscillation of the cylinder is about [tex]4.4 \, \text{rad/s}[/tex].

Answer :

The angular frequency of oscillation of the cylinder is about 8.106 rad/s.

What is Density?

Density is a physical property of matter that describes the amount of mass per unit volume of a substance. It is calculated as the mass of a substance divided by its volume. The standard unit of density is kilograms per cubic meter (kg/m^3), but it can also be expressed in grams per cubic centimeter (g/cm^3) or other units of mass and volume.

To find the angular frequency of oscillation of the cylinder, we need to use the formula for the period of oscillation of a submerged cylinder in a liquid:

T = 2π * sqrt(I / (mga))

where T is the period of oscillation, I is the moment of inertia of the cylinder about its axis of rotation, m is the mass of the cylinder, g is the acceleration due to gravity, and a is the cross-sectional area of the cylinder.

The moment of inertia of a solid cylinder about its axis of rotation is given by:

I = (1/2) * m * r^2

where r is the radius of the cylinder. In this case, the cross-sectional area of the cylinder is given, so we can find the radius using the formula:

A = π * r^2

Solving for r, we get:

r = sqrt(A/π) = sqrt((2.29 x 10^-1 m^2) / π) = 0.2706 m

So the moment of inertia of the cylinder is:

I = (1/2) * m * r^2 = (1/2) * (118 kg) * (0.2706 m)^2 = 4.378 kg*m^2

Now we can use the formula for the period of oscillation to find the angular frequency:

T = 2π * sqrt(I / (mga))

T = 2π * sqrt(4.378 kg*m^2 / ((118 kg) * (9.81 m/s^2) * (2.29 x 10^-1 m^2)))

T = 2π * sqrt(0.01915)

T = 0.775 s

The angular frequency is the reciprocal of the period:

ω = 2π / T = 2π / 0.775 s ≈ 8.106 rad/s

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