6.01e Formulate models: dimensional arguments

1

1. Write down the dimensions of velocity, acceleration and force. The force \(F\) of gravitational attraction between two objects with masses \(m _ { 1 }\) and \(m _ { 2 }\), at a distance \(r\) apart, is given by $$F = \frac { G m _ { 1 } m _ { 2 } } { r ^ { 2 } }$$ where \(G\) is the universal constant of gravitation.
2. Show that the dimensions of \(G\) are \(\mathrm { M } ^ { - 1 } \mathrm {~L} ^ { 3 } \mathrm {~T} ^ { - 2 }\).
3. In SI units (based on the kilogram, metre and second) the value of \(G\) is \(6.67 \times 10 ^ { - 11 }\). Find the value of \(G\) in imperial units based on the pound \(( 0.4536 \mathrm {~kg} )\), foot \(( 0.3048 \mathrm {~m} )\) and second.
4. For a planet of mass \(m\) and radius \(r\), the escape velocity \(v\) from the planet's surface is given by $$v = \sqrt { \frac { 2 G m } { r } }$$ Show that this formula is dimensionally consistent.
5. For a planet in circular orbit of radius \(R\) round a star of mass \(M\), the time \(t\) taken to complete one orbit is given by $$t = k G ^ { \alpha } M ^ { \beta } R ^ { \gamma }$$ where \(k\) is a dimensionless constant.
   Use dimensional analysis to find \(\alpha , \beta\) and \(\gamma\).

1

1. 1. Write down the dimensions of force and the dimensions of density. When a wire, with natural length \(l _ { 0 }\) and cross-sectional area \(A\), is stretched to a length \(l\), the tension \(F\) in the wire is given by $$F = \frac { E A \left( l - l _ { 0 } \right) } { l _ { 0 } }$$ where \(E\) is Young's modulus for the material from which the wire is made.
   2. Find the dimensions of Young's modulus \(E\). A uniform sphere of radius \(r\) is made from material with density \(\rho\) and Young's modulus \(E\). When the sphere is struck, it vibrates with periodic time \(t\) given by $$t = k r ^ { \alpha } \rho ^ { \beta } E ^ { \gamma }$$ where \(k\) is a dimensionless constant.
   3. Use dimensional analysis to find \(\alpha , \beta\) and \(\gamma\).
2. Fig. 1 shows a fixed point A that is 1.5 m vertically above a point B on a rough horizontal surface. A particle P of mass 5 kg is at rest on the surface at a distance 0.8 m from B , and is connected to A by a light elastic string with natural length 1.5 m . \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c470e80e-b346-4335-9c08-beb5a46cc506-2_405_538_1338_845} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} The coefficient of friction between P and the surface is 0.4 , and P is on the point of sliding. Find the stiffness of the string.

1

1. 1. Write down the dimensions of density, kinetic energy and power. A sphere of radius \(r\) is moved at constant velocity \(v\) through a fluid.
   2. In a viscous fluid, the power required is \(6 \pi \eta r v ^ { 2 }\), where \(\eta\) is the viscosity of the fluid. Find the dimensions of viscosity.
   3. In a non-viscous fluid, the power required is \(k \rho ^ { \alpha } r ^ { \beta } v ^ { \gamma }\), where \(\rho\) is the density of the fluid and \(k\) is a dimensionless constant. Use dimensional analysis to find \(\alpha , \beta\) and \(\gamma\).
2. A rock of mass 5.5 kg is connected to a fixed point O by a light elastic rope with natural length 1.2 m . The rock is released from rest in a position 2 m vertically below O , and it next comes to instantaneous rest when it is 1.5 m vertically above O . Find the stiffness of the rope.

1 The surface tension of a liquid enables a metal needle to be at rest on the surface of the liquid. The greatest mass \(m\) of a needle of length \(a\) which can be supported in this way by a liquid of surface tension \(S\) is given by $$m = \frac { 2 S a } { g }$$ where \(g\) is the acceleration due to gravity.

1. Show that the dimensions of surface tension are \(\mathrm { MT } ^ { - 2 }\). The surface tension of water is 0.073 when expressed in SI units (based on kilograms, metres and seconds).
2. Find the surface tension of water when expressed in a system of units based on grams, centimetres and minutes. Liquid will rise up a capillary tube to a height \(h\) given by \(h = \frac { 2 S } { \rho g r }\), where \(\rho\) is the density of the liquid and \(r\) is the radius of the capillary tube. \(r\) is the radius of the capillary tube.
3. Show that the equation \(h = \frac { 2 S } { \rho g r }\) is dimensionally consistent.
4. Find the radius of a capillary tube in which water will rise to a height of 25 cm . (The density of water is 1000 in SI units.) When liquid is poured onto a horizontal surface, it forms puddles of depth \(d\). You are given that \(d = k S ^ { \alpha } \rho ^ { \beta } g ^ { \gamma }\) where \(k\) is a dimensionless constant.
5. Use dimensional analysis to find \(\alpha , \beta\) and \(\gamma\). Water forms puddles of depth 0.44 cm . Mercury has surface tension 0.487 and density 13500 in SI units.
6. Find the depth of puddles formed by mercury on a horizontal surface.

1

1. A particle P is executing simple harmonic motion, and the centre of the oscillations is at the point O . The maximum speed of P during the motion is \(5.1 \mathrm {~ms} ^ { - 1 }\). When P is 6 m from O , its speed is \(4.5 \mathrm {~ms} ^ { - 1 }\). Find the period and the amplitude of the motion.
2. The force \(F\) of gravitational attraction between two objects of masses \(m _ { 1 }\) and \(m _ { 2 }\) at a distance \(d\) apart is given by \(F = \frac { G m _ { 1 } m _ { 2 } } { d ^ { 2 } }\), where \(G\) is the universal gravitational constant.
   1. Find the dimensions of \(G\). Three objects, each of mass \(m\), are moving in deep space under mutual gravitational attraction. They move round a single circle with constant angular speed \(\omega\), and are always at the three vertices of an equilateral triangle of side \(R\). You are given that \(\omega = k G ^ { \alpha } m ^ { \beta } R ^ { \gamma }\), where \(k\) is a dimensionless constant.
   2. Find \(\alpha , \beta\) and \(\gamma\). For three objects of mass 2500 kg at the vertices of an equilateral triangle of side 50 m , the angular speed is \(2.0 \times 10 ^ { - 6 } \mathrm { rad } \mathrm { s } ^ { - 1 }\).
   3. Find the angular speed for three objects of mass \(4.86 \times 10 ^ { 14 } \mathrm {~kg}\) at the vertices of an equilateral triangle of side 30000 m .

2 The surface tension of a liquid allows a metal needle to be at rest on the surface of the liquid.
The greatest mass \(m\) of a needle of length \(l\) which can be supported in this way by a liquid of surface tension \(S\) is given by the formula \(m = \frac { 2 S l } { g }\) where \(g\) is the acceleration due to gravity.

1. Determine the dimensions of surface tension. Surface tension also allows liquids to rise up capillary tubes. Molly is experimenting with liquids in capillary tubes and she arrives at the formula \(h = \frac { 2 S } { \rho g r }\), where \(h\) is the height to which a liquid of surface tension \(S\) rises, \(\rho\) is the density of the liquid, and \(r\) is the radius of the capillary tube.
2. Show that the equation for \(h\) is dimensionally consistent. In SI units, the surface tension of mercury is \(0.475 \mathrm {~kg} \mathrm {~s} ^ { - 2 }\) and its density is \(13500 \mathrm {~kg} \mathrm {~m} ^ { - 3 }\).
3. Find the diameter of a capillary tube in which mercury will rise to a height of 10 cm . In another experiment, Molly finds that when liquid of surface tension \(S\) is poured onto a horizontal surface, puddles of depth \(d\) are formed. For this experiment she finds that \(d = k S ^ { \alpha } \rho ^ { \beta } g ^ { \gamma }\) where \(k\) is a dimensionless constant.
4. Determine the values of \(\alpha , \beta\) and \(\gamma\).

3 The time period \(T\) of a satellite in circular orbit around a planet satisfies the equation \(G M T ^ { 2 } = 4 \pi ^ { 2 } R ^ { 3 }\),
where

• \(G\) is the universal gravitational constant,
• \(M\) is the mass of the planet,
• \(\quad R\) is the radius of the orbital circle.
  1. Find the dimensions of \(G\).

A student suggests the following formula to model the approach speed between two orbiting bodies. \(v = k G { } ^ { \alpha } { } ^ { \beta } { } _ { r } \gamma _ { m _ { 1 } } m _ { 2 } \left( m _ { 1 } + m _ { 2 } \right)\),
where

2

1. Find the dimensions of energy. The moment of inertia, \(I\), of a rigid body rotating about a fixed axis is measured in \(\mathrm { kg } \mathrm { m } ^ { 2 }\).
2. State the dimensions of \(I\). The kinetic energy, \(E\), of a rigid body rotating about a fixed axis is given by the formula \(\mathrm { E } = \frac { 1 } { 2 } \mathrm { I } \omega ^ { 2 }\),
   where \(\omega\) is the angular velocity (angle per unit time) of the rigid body.
3. Show that the formula for \(E\) is dimensionally consistent. When a rigid body is pivoted from one of its end points and allowed to swing freely, it forms a pendulum. The period, \(t\), of the pendulum is the time taken for it to complete one oscillation. A student conjectures the formula \(\mathrm { t } = \left. \mathrm { k } ( \mathrm { mg } ) ^ { \alpha } \mathrm { r } ^ { \beta } \right| ^ { \gamma }\),
   where
   The moment of inertia of a thin uniform rigid rod of mass 1.5 kg and length 0.8 m , rotating about one of its endpoints, is \(0.32 \mathrm {~kg} \mathrm {~m} ^ { 2 }\). The student suspends such a rod from one of its endpoints and allows it to swing freely. The student measures the period of this pendulum and finds that it is 1.47 seconds.
4. Using the formula conjectured by the student, determine the value of \(k\).

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1. Write down the dimensions of force. The force \(F\) of gravitational attraction between two objects with masses \(m _ { 1 }\) and \(m _ { 2 }\), at a distance \(d\) apart, is given by $$F = \frac { G m _ { 1 } m _ { 2 } } { d ^ { 2 } }$$ where \(G\) is the universal gravitational constant.
   In SI units the value of \(G\) is \(6.67 \times 10 ^ { - 11 } \mathrm {~kg} ^ { - 1 } \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 2 }\).
2. Write down the dimensions of \(G\).
3. Determine the value of \(G\) in imperial units based on pounds, feet, and seconds. Use the facts that 1 pound \(= 0.454 \mathrm {~kg}\) and 1 foot \(= 0.305 \mathrm {~m}\). For a planet of mass \(M\) and radius \(r\), it is suggested that the velocity \(v\) needed for an object to escape the gravitational pull of the planet, the 'escape velocity', is given by the following formula. \(\mathrm { v } = \sqrt { \frac { \mathrm { kGM } } { \mathrm { r } } }\),
   where \(k\) is a dimensionless constant.
4. Show that this formula is dimensionally consistent. Information regarding the planets Earth and Mars can be found in the table below.

   	Earth	Mars
   Radius (m)	6371000	3389500
   Mass (kg)	\(5.97 \times 10 ^ { 24 }\)	\(6.39 \times 10 ^ { 23 }\)
   Escape velocity ( \(\mathrm { m } \mathrm { s } ^ { - 1 }\) )	11186

5. Using the formula \(\mathrm { v } = \sqrt { \frac { \mathrm { kGM } } { \mathrm { r } } }\), determine the escape velocity for planet Mars.