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Showing posts from August, 2021

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Mariah Carey - Hero

Hero Mariah Carey Music Pavarotti

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Katherine Johnson - A brilliant mind

Katherine Johnson STOP RACISM IN SCIENCES AND MUSIC WOMEN IN SCIENCES

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Giannis Parios & Antonis Vardis - What a musical meeting...

Antonis Vardis Giannis Parios Greek Music Music

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Shape of a hanging rope between two points - Mathematical analysis

Hanging chain Mechanics Physics

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Monika - An exceptional artist - Over The Hill

Greek Music Monika Music

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The initial acceleration of a 3-body system

Mechanics Physics

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Centripetal force of an object in an arbitrary curve

Centripetal Force Mechanics Physics

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Welcome to Science & Music

A blog created to combine the beauty of natural sciences with the magic world of music !

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" I am among those who believe science has a great beauty" - Marie Curie

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August 20217
July 202118

Popular Posts

Hanging Chain - Where is the tension greatest ?

A chain hangs between two points in the same horizontal line. We want to compare the tension of the chain in 3 different points (A, B and C). Point C is lower than B and point B is lower than A. We can answer this problem by calculating the tension at each point and using function analysis etc. But there is a much easier way to find the answer. First of all, it is essential to notice that the horizontal tension in any point is constant. Just take a small part of the chain and analyse each tension to the x-axis. Since this small part is in equilibrium, the horizontal tensions are equal and constant to a certain value. That is Tcosθ = C, where θ is the angle of the tangent and the horizontal level. So T = C/cosθ. Thus, the smaller the cosθ, the bigger the tension. Or the bigger the value of θ, the bigger the tension. The angle θ is bigger at point A than at point B, and the angle at B is bigger than the angle at C. So T(A) > T(B) > T(C). Of course if we know the exact coordina

Shape of a hanging rope between two points - Mathematical analysis

Let's assume that we take a rope (or a chain as shown in the picture) and hang it up between two distinct points. We want to find the shape of the rope. What equation describes it? Here is where physics comes. Well...everything is about physics! Let's dive in. So we see that the equation that describes the shape of the rope is of a hyperbolic form!

Mathematical analysis of the motion of a homogenous rotating hemisphere

Assume a homogenous hemisphere of radius R, that can rotate around a horizontal axis, inside the homogenous gravitational field of Earth ( g ) . If we define a coordinate system, starting from point O(0,0,0) as shown in the figure below, then the axis of rotation which is parallel to y'y and lies on the plane z=0 , has an abscissa x=ε (it passes through point A(ε,0,0)) • We want to find what should the distance ε from O be, in order for the initial angular acceleration of the hemisphere is the maximum possible (assume that initially the hemisphere is horizontal). • If θ is the angle of the axis x′x (as shown in figure) and the vertical, then prove that for small angles θ, the hemisphere oscillates harmonically. We want to find the value of ε, so that the period of oscillation is minimum.

A Falling Chain - Maximum reaction force

A homogenous chain of mass density ρ and length L is kept vertical above a horizontal level. We let the chain free. What is the maximum reaction force of the horizontal level ? Well, we can solve this problem easily by writing down Newton's II Law for the chain. That is F - ρLg = mυ' + m'υ. But if we name x the height of the chain x=x(t), then the mass m that goes into the momentum is m=ρx. Thus, F - ρLg = (ρx)x'' + ρυ^2. The acceleration x'' is equal to the gravitational acceleration (x''=-g) and thus the velocity υ is given by υ^2 = 2g(L - x). So plugging that into II Law, we get F = 3ρg( L - x). So the maximum force is when x = 0, and is equal to F=3ρgL=3Mg. The reaction force is maximum just before the chain stops, and immediately after the reaction force balances the weight of the chain (F=Mg).

Centripetal force of an object in an arbitrary curve

Let's assume that an object moves in an arbitrary curve which is described by y=y(x). We want to find the centripetal force that acts on the object in any point. Well, we know that the centripetal force for an object moving in a circle is mυ^2/R, where R is the radius of the circle, m the mass and υ the velocity. If we assume that any point of the arbitrary curve is also a point of hypothetical circle of radius R(x), then finding the function R(x) means that we can calculate the centripetal force (mυ^2/R(x)) at any point. Let's take a small part of the curve with length dS. As a part of circle with radius R(x) it has a length of dS=R.dθ. As a part of the curve y(x) it has a length of sqrt(1+y'^2)dx, where y'=dy/dx. That is, R. dθ=sqrt(1+y'^2)dx , or, R.θ' = sqrt(1+y'^2) , where θ'=dθ/dx. We know that tanθ=dy/dx=y'. Differentiating we get, θ' = cos^2(θ).y'' But we know that cos^2(θ)(1+y'2)=1. So we finally get that: R(x) = (1+y

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