Month: July 2019

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201907280623 Solution to 1982-AL-PHY-I-7

(Non-relativistic approach.)

Set up a 2-D Cartesian coordinate system, the origin being in the position of body X at time t=0, and at the point (6,0) there being body Y.

Then the position of body X and of body Y can each be given by a function of time t:

\begin{aligned} \mathbf{r}_X(t) & = 3t\, \hat{\mathbf{i}} \\ \mathbf{r}_Y(t) & =6\, \hat{\mathbf{i}} + 4t\, \hat{\mathbf{j}} \end{aligned}

where t\in [0,\infty ).

The separation \mathbf{r}_{YX} of body Y from body X by time t is:

\mathbf{r}_{YX}(t)= \mathbf{r}_Y - \mathbf{r}_X =(6-3t)\, \hat{\mathbf{i}} + (4t)\, \hat{\mathbf{j}}.

The velocity \mathbf{v}_{YX} of body Y from body X is:

\begin{aligned} \mathbf{v}_{YX} & =\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\Big(\mathbf{r}_{YX}\Big) \\ & = -3\, \hat{\mathbf{i}} + 4\, \hat{\mathbf{j}} \end{aligned}

The magnitude v_{YX} of the velocity is

v_{YX}=|\mathbf{v}_{YX}|=\sqrt{(-3)^2+(4)^2}=5\quad (\mathrm{m\, s^{-1}}).

And the answer is B.

All above is overkill.

Notice \mathbf{v}_{YX}=\mathbf{v}_Y-\mathbf{v}_X = (0,4) - (3,0) = (-3,4) and v_{YX}=|(-3,4)|=5

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201907260745 Homework 1 (Q1)

Consider a probability density of the Gaussian distribution

|\Psi |^2=\rho (x)=Ae^{-\lambda(x-a)^2}

where A, a, and \lambda are constants.

You probably wish to know \displaystyle{\int_{-\infty}^{+\infty}}e^{-u^2}\mathrm{d}u=\sqrt{\pi}.

  1. Determine A according to the normalizing rules.
  2. Find \langle x\rangle, \langle x^2\rangle, and \sigma, the standard deviation of x.
  3. Sketch the graph of \rho (x).

Solution:

  1. Roughwork.

    \begin{aligned}1 & = \int_{-\infty}^{+\infty} |\Psi |^2 \mathrm{d}x \\1& = \int_{-\infty}^{+\infty} Ae^{-\lambda (x-a)^2}\mathrm{d}x \\\frac{1}{A} & = \int_{-\infty}^{+\infty} e^{-\big(\lambda^{\frac{1}{2}}(x-a)\big)^2} \mathrm{d}x \\\dots & \Bigg( \because \enspace \frac{\mathrm{d}\big( \lambda^{\frac{1}{2}}(x-a) \big)}{\mathrm{d}x} = \sqrt{\lambda} \Bigg) \dots \\\frac{1}{A} & = \frac{1}{\sqrt{\lambda}} \int_{-\infty}^{+\infty} e^{-\big(\lambda^{\frac{1}{2}}(x-a)\big)^2} \mathrm{d}\big( \lambda^{\frac{1}{2}}(x-a) \big) \\\frac{1}{A} & =\sqrt{\frac{\pi}{\lambda}} \\A & = \sqrt{\frac{\lambda}{\pi}}\end{aligned}

  2. Recall \langle x\rangle = \displaystyle{\int_{-\infty}^{+\infty}}x\rho (x)\,\mathrm{d}x, Eq. (1.17) in Griffith's Introduction to Quantum Mechanics

    First,

    \begin{aligned} \langle x\rangle & = A\int_{-\infty}^{+\infty} xe^{-\lambda (x-a)^2}\mathrm{d}x \\ & = A\int_{-\infty}^{+\infty} (u+a)e^{-\lambda u^2}\mathrm{d}u \\ & = A \int_{-\infty}^{+\infty} u\,\mathrm{d}u + Aa\int_{-\infty}^{+\infty}e^{-\lambda u^2}\,\mathrm{d}u \\ & = 0+Aa\sqrt{\frac{\pi}{\lambda}} \\ & = a \end{aligned}

    The section below sets a bad example of computation, the second equality sign being wishful thinking, and what follows thence is incorrect.

    \begin{aligned} \langle x^2\rangle & = A\int_{-\infty}^{+\infty} x^2e^{-\lambda (x-a)^2}\mathrm{d}x \\ & = A \int_{-\infty}^{+\infty} x^2e^{-\lambda x^2}\mathrm{d}x + A \int_{-\infty}^{+\infty} x^2e^{2\lambda ax}\mathrm{d}x + A \int_{-\infty}^{+\infty} x^2e^{-\lambda a^2}\mathrm{d}x \end{aligned}
    Recall the formula for integration by parts is \int uv'\mathrm{d}x =uv-\int vu'\mathrm{d}x, or, \displaystyle{\int u\bigg( \frac{\mathrm{d}v}{\mathrm{d}x}\bigg) \mathrm{d}x} =uv-\displaystyle{\int v\bigg( \frac{\mathrm{d}u}{\mathrm{d}x} \bigg) \mathrm{d}x}.

Evaluate term-by-term, the first term is

\begin{aligned} &\quad A\int_{-\infty}^{+\infty} x^2e^{-\lambda x^2}\,\mathrm{d}x\\ & = \frac{A}{2} \int_{-\infty}^{+\infty} xe^{-\lambda x^2}\,\mathrm{d}(x^2) \\ & =\frac{A}{2} \Bigg\{ \bigg[ \frac{-xe^{-\lambda x^2}}{\lambda}\bigg]_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty} \frac{-e^{-\lambda x^2}}{\lambda}\,\mathrm{d}(x^2) \Bigg\} \\ \dots & \Bigg( \quad \mathrm{d}(\sqrt{\lambda}x^2) = \frac{1}{2\sqrt{\lambda}}\,\mathrm{d}(x^2) \quad \Bigg) \dots \\ & =\frac{A}{2} \Bigg\{ \bigg[ \frac{-xe^{-\lambda x^2}}{\lambda}\bigg]_{-\infty}^{+\infty} +\frac{2\sqrt{\lambda}}{\lambda}\int_{-\infty}^{+\infty}e^{-(\sqrt{\lambda} x)^2} \mathrm{d}(\sqrt{\lambda}x^2) \Bigg\} \\ & = \frac{A}{2} \bigg( 0+2\sqrt{\frac{\pi}{\lambda}}\bigg) \\ & = \frac{\sqrt{\frac{\lambda}{\pi}}}{2} \bigg( 2\sqrt{\frac{\pi}{\lambda}}\bigg) \\ & = 1 \end{aligned}

The terrible blunder ends here.

Correction.

\begin{aligned} \langle x^2 \rangle & = \int_{-\infty}^{+\infty} x^2Ae^{-\lambda (x-a)^2}\mathrm{d}x \\ & = A\int_{-\infty}^{+\infty} (u+a)^2 e^{\lambda u^2}\mathrm{d}u \\ & = A\int_{-\infty}^{+\infty} u^2e^{-\lambda u^2}\mathrm{d}u + 2Aa\int_{-\infty}^{+\infty} ue^{-\lambda u^2}\mathrm{d}u + Aa^2\int_{-\infty}^{+\infty}e^{-\lambda u^2}\mathrm{d}u \end{aligned}

Step back to look closer, the third term is the easiest to compute:

\begin{aligned} & \quad Aa^2\int_{-\infty}^{+\infty}e^{-\lambda u^2}\mathrm{d}u \\ & = \frac{Aa^2}{\sqrt{\lambda}}\int_{-\infty}^{+\infty}e^{-(\sqrt{\lambda}u)^2}\mathrm{d}(\sqrt{\lambda}u) \\ & =Aa^2\sqrt{\frac{\pi}{\lambda}}\\ & = \sqrt{\frac{\lambda}{\pi}} a^2\sqrt{\frac{\pi}{\lambda}}\\ & = a^2 \end{aligned}

The first two terms might need to be evaluated using integration by parts. On the other hand, from the angle of parity, in the second term

2Aa\displaystyle{\int_{-\infty}^{+\infty}} f(u)\,\mathrm{d}u

where f(u)\stackrel{\mathrm{def}}{=}ue^{-\lambda u^2}

f(-u)=(-u)e^{-\lambda(-u)^2}=-f(u)

is an odd function. The definite integral upon evaluation will be nought:

2Aa\displaystyle{\int_{-\infty}^{+\infty}} ue^{-\lambda u^2}\,\mathrm{d}u \equiv 0.

The first term can be checked

A \displaystyle{\int}g(u)\,\mathrm{d}u

where g(u) \stackrel{\mathrm{def}}{=}u^2e^{-\lambda u^2}

that g(-u)=(-u)^2e^{-\lambda (-u)^2}=g(u) is an even function.

Upon evaluation the definite integral will have the property that

A \displaystyle{\int_{-\infty}^{+\infty}}u^2e^{-\lambda u^2}\,\mathrm{d}u = 2A \displaystyle{\int_{0}^{+\infty}}u^2e^{-\lambda u^2}\,\mathrm{d}u,

though it seems not useful here. Doing integration by parts,

\begin{aligned} & \quad A \displaystyle{\int_{-\infty}^{+\infty}}u^2e^{-\lambda u^2}\mathrm{d}u \\ & = \frac{A}{\sqrt{\lambda}} \int_{-\infty}^{+\infty} u^2e^{-(\sqrt{\lambda}u)^2}\mathrm{d}(\sqrt{\lambda}u) \\ & = \frac{A}{\sqrt{\lambda}} \Bigg\{ \bigg[ \frac{u^2e^{-(\sqrt{\lambda}u)^2}}{-2(\sqrt{\lambda}u)} \bigg]\bigg|_{-\infty}^{+\infty} -\int_{-\infty}^{+\infty} \frac{e^{-(\sqrt{\lambda}u)^2}}{-2(\sqrt{\lambda}u)} \Big( \frac{2u}{\sqrt{\lambda}} \Big) \mathrm{d}(\sqrt{\lambda}u) \Bigg\} \\ & = \frac{A}{\sqrt{\lambda}} \Bigg\{ 0+ \frac{1}{\lambda}\int_{-\infty}^{+\infty} e^{-(\sqrt{\lambda}u)^2} \mathrm{d}(\sqrt{\lambda}u) \Bigg\} \\ & = A\frac{1}{\sqrt{\lambda}}\frac{1}{\lambda}\sqrt{\pi} \\ & = \sqrt{\frac{\lambda}{\pi}} \frac{1}{\sqrt{\lambda}}\frac{1}{\lambda}\sqrt{\pi} \\ & = \frac{1}{\lambda} \end{aligned}

Thus \langle x^2\rangle = \frac{1}{\lambda} + a^2

As \sigma^2=\langle x^2\rangle -\langle x\rangle^2,

\sigma^2 = \frac{1}{\lambda} + a^2 - a^2 = \frac{1}{\lambda}.

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201907251758 Solution to 1980-CE-PHY-II-5

The kinetic energy E_\mathrm{k} of an object of mass m and speed v is given by the formula

E_\mathrm{k}=\displaystyle{\frac{1}{2}}mv^2.

In the situation that the object is thrown upwards with initial speed u, and subjected only to gravity \mathbf{g}, it can be expected that after some time of flight T, the object will return to its initial position, its downward speed in which is equal to the initial upward speed u.

Define a piecewise scalar function v(t) of time t:

v(t) = \begin{cases} u-gt & \quad \textrm{when }0\leq t\leq \displaystyle{\frac{T}{2}} \\ -u+gt & \quad \textrm{when } \displaystyle{\frac{T}{2}}\leq t\leq T \end{cases}

or simply

v:[0,T]\subset \mathbb{R} \rightarrow [0,u]\subset \mathbb{R} given by t\mapsto \big|u-g(T-t)\big|.

Then

\begin{aligned} v^2 & =(u-gt)^2\quad \big( =(-u+gt)^2\big) \\ & = u^2-2ugt+g^2t^2 \quad \big(\forall\, t\in [0,T] \big) \end{aligned}.

Thus the kinetic energy E_\mathrm{k}(t) is

\begin{aligned} E_\mathrm{k}(t) & =\displaystyle{\frac{1}{2}}m(u^2-2ugt+g^2t^2) \\ & = \bigg( \displaystyle{\frac{1}{2}}mg^2 \bigg) t^2 + ( -mug ) t + \bigg( \displaystyle{\frac{1}{2}}mu^2 \bigg) \end{aligned}

where m, u, g are constants.

The kinetic energy E_\mathrm{k}(t) is set to zero at time t':

\begin{aligned} t' & = \displaystyle{\frac{-(-mug)\pm\sqrt{\big( -mug\big)^2-4\big(\frac{1}{2}mg^2\big)\big(\frac{1}{2}mu^2\big)}}{2\big(\frac{1}{2}mg^2\big)}} \\ & = \displaystyle{\frac{mug\pm\sqrt{m^2u^2g^2-m^2u^2g^2}}{mg^2}} \\ & =\displaystyle{\frac{u}{g}} \end{aligned}

Substituting t'=\displaystyle{\frac{u}{g}} for t in v=u-gt:

v=u-g\bigg( \displaystyle{\frac{u}{g}} \bigg) =0,

as checked.

And the answer is B.

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201907251221 Short Review (Projectile Motion)

When projected in the air and subjected only to gravity, a projectile performs projectile motion. Its path/trajectory is a parabola. If the projectile is projected at an angle of projection, it is said to be in general projectile motion; else, it is said to be in horizontal projectile motion.

The duration of time from projection to landing is called the time of flight. The maximum height the projectile can reach is usually measured from the launch level. The range of projection is the horizontal distance the projectile has travelled from projection to landing.

Horizontally Projected Motion

When air resistance is negligible, a projectile moves at a uniform horizontal velocity and at a uniform vertical acceleration due to gravity. As its motion in the horizontal and in the vertical are independent of each other, so we separate the projectile motion into two perpendicular directions—the horizontal and the vertical—in order to resolve and analyse it.

i. v=u+at; ii. \displaystyle{s=\frac{(u+v)}{2}t}; iii. \displaystyle{s=ut+\frac{1}{2}at^2}; iv. v^2=u^2+2as are what you need.

At any instant t:

horizontal motion (u_x=u, a_x=0):

v_x=u_x=u;

s_x=u_xt=ut OR t=\displaystyle{\frac{s_x}{u}}

vertical motion (u_y=0, a_y=-g):

v_y=u_y+a_yt=0-gt=-gt;

s_y=u_yt+\frac{1}{2}a_yt^2=0+\frac{1}{2}(-g)t^2=-\frac{1}{2}gt^2;

v_y^2=u_y^2+2a_ys_y=0+2(-g)s_y=-2gs_y;

the velocity v can be found by

\sqrt{v_x^2+v_y^2}=\sqrt{u^2-2gs_y}\enspace (=\sqrt{u^2+g^2t^2})

its trajectory is parabolic: s_y=\displaystyle{-\frac{g}{2u^2}s_x^2}.

General Projectile Motion

A projectile is now projected with an initial velocity u at an angle \theta. As usual, we separate the motion into horizontal and vertical directions. At any instant t, the magnitude of its velocity is v=\sqrt{v_x^2+v_y^2}, and its direction makes an angle \phi with the level, where \tan\phi =\displaystyle{\frac{v_y}{v_x}}.

Horizontal motion: The initial horizontal velocity is u_x=u\cos\theta. After some time t, its horizontal displacement is s_x=u_xt+\frac{1}{2}a_xt^2 OR s_x=u\cos\theta t and its velocity remains to be v_x=u_x+a_xt OR v_x=u_x=u\cos\theta.

(Neglecting air friction, we assume zero horizontal acceleration, i.e., a_x=0.)

Vertical motion: The initial vertical velocity is u_y=u\sin\theta. After some time t, its vertical displacement is s_y=u_yt+\frac{1}{2}a_yt^2 OR s_y=u\sin\theta t-\frac{1}{2}gt^2 and its velocity changes to v_y=u_y+a_yt OR v_y=u\sin\theta -gt.

(Upward taken to be +ve, the vertical acceleration due to gravity is a_y=-g)

Better to know the equations of motion well in deriving, than to simply memorize the formulae in solving, the unknowns, lest it be wrong in some cases, say, on a slant.

Time of flight t: Consider only the vertical motion, by ①: s_y=u_yt+\displaystyle{\frac{1}{2}}a_yt^2

\begin{aligned} 0&=(u\sin\theta)t-\frac{1}{2}gt^2 \\ 0&=t(u\sin\theta -\frac{1}{2}gt) \\ t&= 0\quad \mathrm{or}\quad \boxed{t=\frac{2u\sin\theta}{g}} \end{aligned}

Or, by ②: v=u+at, and that the object lands with the same speed as is launched, we have -u\sin\theta =u\sin\theta -gt, which also gives t=2u\sin\theta /g.

Range R: Consider only the horizontal motion, by s_x=u_xt (\because a_x=0)

\begin{aligned} R&=(u\cos\theta )\bigg( \frac{2u\sin\theta}{g}\bigg) \\ &=\frac{2u^2\sin\theta\cos\theta}{g} \end{aligned}

\boxed{R=\frac{u^2\sin 2\theta}{g}}

^\dagger R is maximum if \theta =45^\circ (\sin 2\theta =1). ^{\dagger\dagger} \sin 2\theta=2\sin\theta\cos\theta. ^{\dagger\dagger\dagger} Both angles \theta and 90^\circ -\theta give the same range.

Maximum height H: Consider only the vertical motion, by ①: v_y^2=u_y^2+2a_ys_y

0^2=(u\sin\theta )^2-2gH

\boxed{H=\frac{u^2\sin^2\theta}{2g}}

Or, by ②: s_y=u_yt+\frac{1}{2}a_yt^2 and that it takes half of the time of flight to reach H, we have H=(u\sin\theta)\bigg( \displaystyle{\frac{u\sin\theta}{g}}\bigg)+\displaystyle{\frac{1}{2}}(-g)\bigg( \displaystyle{\frac{u\sin\theta}{g}}\bigg)^2 and thus the same result.

CONCEPT TEST

  1. As shown in the figure, three objects a, b, and c are projected horizontally and travelled along their respective trajectories. Objects b and c are projected at the same height. Neglecting air resistance, which of the following statements are true?
    1. The time of flight of a is longer than that of b.
    2. The time of flight of b is equal to that of c.
    3. The horizontal velocity of a is less than that of b.
    4. The initial velocity of b is greater than that of c.
    1. (I) and (II)
    2. (I) and (IV)
    3. (II) and (III)
    4. (II) and (IV)
  2. As shown in the figure, two objects A and B are in general projectile motion. They reach the same maximum height. Neglecting air resistance, which of the following statements are wrong?
    1. The acceleration of B is greater than that of A.
    2. The time of flight of B is equal to that of A.
    3. The velocity of B is equal to that of A when they are at the maximum height.
    4. The velocity of B is greater than that of A at the time of landing.
    1. (I) and (II)
    2. (I) and (III)
    3. (II) and (III)
    4. (III) and (IV)
  3. An object is now in general projectile motion. Which of the following graphs are correct?
    1. (I) and (II)
    2. (I) and (III)
    3. (II) and (IV)
    4. (III) and (IV)

Answers:

  1. D
  2. B
  3. B

Explanations:

  1. Since h=\displaystyle{\frac{1}{2}}gt^2, time of flight is given by t=\displaystyle{\sqrt{\frac{2h}{g}}}. From h_b=h_c>h_a, we get t_a<t_b=t_c. Thus (I) is wrong and (II) correct. The horizontal velocity is given by v=\displaystyle{\frac{x}{t}}. From x_a>x_b>x_c, we get v_a>v_b>v_c. Thus (III) is wrong and (IV) correct.
  2. The acceleration of A and of B is due to gravity, and is equal to -g (Upward taken to be +ve). Thus (I) is wrong. Let the vertical component of initial velocity be u_y. From 0=u_y^2-2gh, we know that A and B have the same magnitude in their vertical component of initial velocity. Thus they have the same time of flight and (II) is correct. From v_x=\displaystyle{\frac{x}{t}}, where x_B>x_A, we know that the horizontal velocity of B is greater than that of A. The velocity at maximum height is v=\sqrt{v_x^2+v_y^2}=\sqrt{v_x^2}=v_x. Thus v_B>v_A and (III) is wrong. Again by resolving components, the velocity at the time of landing is \sqrt{v_x^2+v_y^2}. From v_y being equal and v_x of B greater than that of A, it follows that (IV) is correct.
  3. By the conservation of mechanical energy, \Delta \mathrm{KE}+\Delta\mathrm{PE}=0, i.e., \bigg( \displaystyle{\frac{1}{2}}mv^2-\displaystyle{\frac{1}{2}}mu^2\bigg) +(mgh-0)=0. Arranging it into the form y=mx+c, we have \mathrm{KE}(=\frac{1}{2}mv^2)=-\mathrm{PE}(=mgh)+\frac{1}{2}mu^2. Thus (I) is correct. (II) is wrong because during the flight, there must be a non-zero horizontal component of velocity and thus \mathrm{KE}\not\equiv 0. The maximum height H=\displaystyle{\frac{u^2\sin^2\theta}{2g}}. We can readily fit it into the form y=ax^2, where y=H, x=u, and a=\displaystyle{\frac{\sin^2\theta}{2g}}. Hence (III) is correct. From s_y=u_yt+\frac{1}{2}a_yt^2, it follows that H=(u\sin\theta )t-\displaystyle{\frac{gt^2}{2}}. We can likewise fit it into a parabola y=ax^2+bx+c where y=H, x=t, a=-g/2, b=u\sin\theta, and c=0. The graph (IV) is far from correct.
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201907241013 Short Review (Work, Energy, and Power)

Q & A

Work and energy transfer

Q. What does mechanical energy include?

Ans. Kinetic energy, gravitational potential energy, and elastic potential energy.

Q. What is the unit of energy?

Ans. Joule (J).

Q. A worker is pushing a trolley loaded with goods from one place to another. What is he doing?

Ans. Work.

Q. What is the definition of work?

Ans. Work is the product of force F_\| parallel to displacement and displacement s, i.e. W=\vec{F}\cdot \vec{s}=Fs\cos\theta, where \theta is the angle between F and s.

Q. What is the unit(s) of work?

Ans. Joule (\mathrm{J}), or Newton metre (\mathrm{N\,m}).

Q. Are energy and work vectors or scalars?

Ans. Both are scalars.

Q. Is work W done always positive?

Ans. No, work W can be negative. From W=Fs\cos\theta, where F,s>0 are the magnitudes of force and of displacement, W<0\Rightarrow \cos\theta <0 \Rightarrow 90^\circ <\theta <180^\circ. Negative work is done when F and s are in opposite direction, or when they make an obtuse angle.

Q. Can you give an example of negative work done on an object?

Ans. Yes, friction f of a rough surface acts opposite to the displacement s. So work done on an object due to friction is W=fs\cos 180^\circ =-fs<0.

Q. Is work W either positive or negative?

Ans. Not really, it can be zero. From W=Fs\cos\theta, assuming there is a force and a displacement, i.e., F,s>0, still, when \theta =90^\circ, \cos\theta =0. Hence, if the force and the displacement are perpendicular to each other (F\perp s), work done W=Fs\cos\theta =0.

Kinetic energy and potential energy

Q. What is the formula for the kinetic energy of a body?

Ans. \mathrm{KE}=\displaystyle{\frac{1}{2}}mv^2.

Q. What is the formula for the gravitational potential energy of a body?

Ans. \mathrm{PE}=mgh.

Q. Is work W equivalent to kinetic energy (KE), gravitational potential energy (GPE), or elastic potential energy (EPE)?

Ans. None of them. Work W is equivalent to mechanical energy (\mathrm{KE} +\mathrm{GPE} +\mathrm{EPE}).

Q. On a level surface, an object of mass m is acted on by a force F such that it accelerates from an initial velocity u to a final velocity v after a displacement s. Show that in this instance, work W done by the force is equivalent solely to the change in kinetic energy (\mathrm{\Delta KE}).

Ans. Substitute F=ma and s=\displaystyle{\frac{v^2-u^2}{2a}} into W=Fs, we get W=(ma)\bigg( \displaystyle{\frac{v^2-u^2}{2a}}\bigg)=\displaystyle{\frac{1}{2}mv^2-\frac{1}{2}mu^2}=\mathrm{KE}_\mathrm{final}-\mathrm{KE}_\mathrm{initial}=\mathrm{\Delta KE}. Notice that the downward gravitational force mg is perpendicular to the displacement s, so GPE does not contribute to the work done.

Q. On a cliff, an object of mass m which is initially at rest is dropped vertically from a height h_1 to a height h_2. Show that in this instance, work W done by gravity is equivalent solely to the change in gravitational potential energy (\mathrm{\Delta GPE}).

Ans. W=Fs=(mg)(h_1-h_2)=\mathrm{GPE_1}-\mathrm{GPE_2}=\Delta \mathrm{GPE}.

Q. Prove that the dimensions of work and of kinetic energy are the same.

Ans. [W]=[Fs]=[ma\cdot s]=\mathrm{kg\cdot\,ms^{-2}\cdot m}=\mathrm{kg\,m^2\,s^{-2}}.

\mathrm{[KE]}=\bigg[\displaystyle{\frac{1}{2}mv^2}\bigg]=\mathrm{kg\cdot (m\,s^{-1})^2}=\mathrm{kg\,m^2\,s^{-2}}.

Q. Prove that the dimensions of work and of potential energy are the same.

Ans. [W]=\mathrm{kg\,m^2\,s^{-2}}; \mathrm{[PE]}=[mgh]=\mathrm{kg\cdot (m\,s^{-2})\cdot m}=\mathrm{kg\,m^2\,s^{-2}}.

Energy changes and conservation of energy

Q. What is the law of conservation of energy?

Ans. Energy can be changed from one form into another, but it cannot be created or destroyed.

Q. When will mechanical energy (i.e., the sum of KE and PE) not be conserved?

Ans. Mechanical energy is not conserved if frictional force is present, or if mechanical energy is converted to other forms of energy, e.g., electrical energy, thermal energy, sound energy, chemical energy, etc.

Power

Q. What is power? Its unit?

Ans. Power P is the rate at which energy E is transferred, P=\displaystyle{\frac{E}{t}}. For energy due only to heat transfer, P=\displaystyle{\frac{Q}{t}}. For energy due only to doing work, P=\displaystyle{\frac{W}{t}}. Its unit is watt (W).

Q. P=\boxed{\displaystyle{\frac{W}{t}}}_{\,\spadesuit}=\displaystyle{\frac{Fs}{t}}=\boxed{Fv}_{\,\clubsuit}. When to use \spadesuit, when to use \clubsuit?

Ans. \spadesuit: average power; \clubsuit: instantaneous power.

Example 1 (Energy conversion and conservation)

An object of mass m begins to move up with initial velocity u along a smooth (frictionless) inclined plane of slope angle \theta.

(a) What is the initial kinetic energy of the object?

Answer: (a) \mathrm{KE}_\mathrm{initial}=\displaystyle{\frac{1}{2}mu^2}

(b) Use the force approach and the energy approach to find the maximum height h that the object can reach.

Solution:

(Force approach.) As always, we first draw a free-body diagram of the object. Since the normal reaction N does not have a component along the plane, we need only to consider the component of weight mg along the plane, which is mg\sin\theta.

The acceleration along the plane is a=\displaystyle{\frac{F}{m}=\frac{-mg\sin\theta}{m}}=-g\sin\theta. The object will stop at the maximum height, i.e. final velocity v=0. By v^2=u^2+2as, we get 0=u^2+2(-g\sin\theta)s. It follows that the displacement travelled along the plane is s=\displaystyle{\frac{u^2}{2g\sin\theta}}. Since height h is related to displacement s by h=s\sin\theta, the maximum height should be h_\mathrm{max.}=\bigg( \displaystyle{\frac{u^2}{2g\sin\theta}}\bigg) (\sin\theta)=\displaystyle{\frac{u^2}{2g}}.

(Energy approach.) Since there is no friction on the inclined plane, we can apply the principle of conservation of mechanical energy. And we ignore the elastic potential energy because the object is rigid.

The final kinetic energy is zero because the final velocity is zero (\Leftarrow it stops momentarily at the maximum height h). Take the ground as the reference level, i.e., h=0.

Then,

\begin{aligned} \mathrm{\Delta\, mechanical\,energy}&=0\\ \Delta \mathrm{KE}+\Delta \mathrm{PE} &=0\\ (\mathrm{KE}_\mathrm{final}-\mathrm{KE}_\mathrm{initial})+(\mathrm{PE}_\mathrm{final}-\mathrm{PE}_\mathrm{initial})&=0\\ (0-\frac{1}{2}mu^2)+(mgh_\mathrm{max.}-0)&=0\\ mgh_\mathrm{max.}&=\frac{1}{2}mu^2\\ h_\mathrm{max.}&=\frac{u^2}{2g}\\ \end{aligned}

Example 2 (Work done by which force?)

An object of mass m is at rest on a rough surface of a wedge of slope angle \theta. Then the wedge makes a displacement s with a constant velocity to the left, while the object remains in the same position on the wedge. Find the work done on the object by i. friction f, ii. normal force N, and iii. the weight mg of the object.

Solution:

As always, we first draw a free-body diagram. From the figure, we know friction f=mg\sin\theta and normal force N=mg\cos\theta.

i. Work done by friction W_f=fs\cos (180^\circ -\theta)=(mg\sin\theta )(s)(-\cos\theta)=-mgs\sin\theta\cos\theta.

ii. Work done by normal force W_N=Ns\cos (180^\circ -90^\circ -\theta )=(mg\cos\theta)(s)(\sin\theta)=mgs\sin\theta\cos\theta.

iii. Work done by gravity W_{mg}=0 (\because mg\perp s).

Example 3 (instantaneous power \neq average power)

An object of mass m is initially at rest on the top of a rough inclined plane of height H and slope angle \theta. It then accelerates and slides down the plane. When it reaches the ground, its velocity is v and it has travelled a displacement s. Find i. the instantaneous power of friction at the instant when the object reaches the ground, and ii. the average power of friction during the whole process of sliding.

i. Use force approach: 0\neq F_\mathrm{net}=ma=mg\sin\theta -f. Substituting a=\frac{v^2-u^2}{2s}=\frac{v^2}{2s} (\because u=0), we have f=mg\sin\theta -\frac{mv^2}{2s}. Then the instantaneous power of friction is given by \boxed{P_\mathrm{ins}=fv}=\bigg(mg\sin\theta -\displaystyle{\frac{mv^2}{2s}}\bigg) v.

ii. Use energy approach: Work done against friction is \Delta \mathrm{KE}+\Delta \mathrm{PE}=(\frac{1}{2}mv^2-0)+(0-mgH)=\frac{1}{2}mv^2-mgH. Hence, work done by friction is W_f=mgH-\frac{1}{2}mv^2. Average power of friction is given by \boxed{P_\mathrm{avg}=\displaystyle{\frac{W_f}{t}}}=\displaystyle{\frac{mgH-\frac{1}{2}mv^2}{t}}.

If m=1\,\mathrm{kg}, v=4\,\mathrm{m\,s^{-1}}, t=1\,\mathrm{s}, f=1\,\mathrm{N}, \theta =30^\circ, H=1\,\mathrm{m}, s=2\,\mathrm{m}, and take g=10\,\mathrm{m\,s^{-2}}, check that 4\,\mathrm{W}=\boxed{P_\mathrm{ins}\neq P_\mathrm{avg}}=2\,\mathrm{W}.

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201907181500 Homework 2 (Q3)

Two mass points, m_1 and m_2, move under the influence of a mutual central force, where the central force potential is given by U(\mathbf{r}_1,\mathbf{r}_2)=U(|\mathbf{r}_1-\mathbf{r}_2|). Assume the center of mass is at the rest, please find the equivalent one-body problem and show that the corresponding Lagrangian can be written as

\mathcal{L}=\displaystyle{\frac{1}{2}}\mu\dot{r}^2-U_{\textrm{eff}}

where r=|\mathbf{r}_1-\mathbf{r}_2| is the relative distance between the two mass points and \mu=\displaystyle{\frac{m_1m_2}{m_1+m_2}} is the reduced mass.

Solution.

(Reference: https://www.physics.rutgers.edu/~shapiro/507/book4(DOT)pdf)

Let \mathbf{R}\stackrel{\textrm{def}}{=}\displaystyle{\frac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2}} and \mathbf{r}\stackrel{\textrm{def}}{=}\mathbf{r}_2-\mathbf{r}_1. Expressing \mathbf{r}_1 and \mathbf{r}_2 in terms of \mathbf{R} and \mathbf{r}, we write

\mathbf{r}_1=\mathbf{R}-\displaystyle{\frac{m_2}{M}}\mathbf{r},

\mathbf{r}_2=\mathbf{R}+\displaystyle{\frac{m_1}{M}}\mathbf{r}

where M=m_1+m_2.

The kinetic energy T is computed as follows:

\begin{aligned} T & = \displaystyle{\frac{1}{2}m_1\dot{r}_1^2}+\displaystyle{\frac{1}{2}m_2\dot{r}_2^2} \\ & = \displaystyle{\frac{1}{2}m_1\Bigg[ \displaystyle{\frac{\mathrm{d} }{\mathrm{d}t}} \bigg( \mathbf{R}-\displaystyle{\frac{m_2}{M}}\mathbf{r} \bigg)\Bigg]^2}+\displaystyle{\frac{1}{2}m_2\Bigg[ \displaystyle{\frac{\mathrm{d} }{\mathrm{d}t}} \bigg( \mathbf{R}+\displaystyle{\frac{m_1}{M}}\mathbf{r}\bigg)\Bigg]^2} \\ & = \displaystyle{\frac{1}{2}}(m_1+m_2)\dot{R}^2+\displaystyle{\frac{1}{2}\frac{m_1m_2}{M}}\dot{r}^2 \\ & = \displaystyle{\frac{1}{2}}M\dot{R}^2+\displaystyle{\frac{1}{2}\mu\dot{r}^2} \\ \end{aligned}

where \mu is the reduced mass \displaystyle{\frac{m_1m_2}{m_1+m_2}}.

From its formula above, T can be seen as the sum of the kinetic energy of the motion of the centre of mass, i.e., \displaystyle{\frac{1}{2}}M\dot{R}^2, and the kinetic energy of motion about the centre of mass, i.e., \displaystyle{\frac{1}{2}\mu\dot{r}^2}.

And from the fact that the Lagrangian \mathcal{L}=\displaystyle{\frac{1}{2}}M\dot{R}^2+\displaystyle{\frac{1}{2}\mu\dot{r}^2}-U(r) is cyclic on R, the centre of mass is either at rest or in uniform motion.

Thus the equation of motion for r will not contain terms involving \mathbf{R} or \dot{\mathbf{R}}. We may hence ignore the first term, and what remains in the Lagrangian is

\mathcal{L}=\displaystyle{\frac{1}{2}\mu\dot{r}^2}-U(r).

Now that we introduce a spherical coordinate system given by its equation of transformation from the Cartesian as:

\begin{aligned} x & =r\sin\theta\cos\phi \\ y & =r\sin\theta\sin\phi \\ z & =r\cos\theta \\ \end{aligned}

we can write the kinetic energy as

\begin{aligned} T & =\displaystyle{\frac{1}{2}}\mu (\dot{x}^2+\dot{y}^2+\dot{z}^2) \\ & = \displaystyle{\frac{1}{2}}\mu[(\dot{r}\sin\theta\cos\phi+\dot{\theta}r\cos\theta\cos\phi -\dot{\phi}r\sin\theta\sin\phi)^2\\ & \qquad +(\dot{r}\sin\theta\sin\phi+\dot{\theta}r\cos\theta\sin\phi+\dot{\phi}r\sin\theta\cos\phi)^2\\ & \qquad +(\dot{r}\cos\theta -\dot{\theta}r\sin\theta)^2]\\ & = \displaystyle{\frac{1}{2}}\mu [\dot{r}^2+r^2\dot{\theta}^2+r^2\sin^2\theta\dot{\phi}^2]\\ \end{aligned}

(Note that the kinetic energy is cyclic on the coordinate \phi and the conjugate momentum P_\phi=\displaystyle{\frac{\partial\mathcal{L}}{\partial\dot{\phi}}}=\mu r^2\sin^2\theta\dot{\phi}=\textrm{constant}. Observe that r\sin\theta is the distance between the particle and the z-axis, and it can be easily seen that P_\phi is the z-component of the angular momentum \mathbf{L}.)

To simplify things, we choose the direction of angular momentum \mathbf{L} as the z-direction. It follows that \theta=\pi/2, \dot{\theta}=0, and L=\mu r^2\dot{\phi}.

I wish to obtain the Euler-Lagrange equation for r. Hence I compute the following:

\begin{aligned} \displaystyle{\frac{\partial \mathcal{L}}{\partial r}} & = \mu r\dot{\phi}^2- \partial_r U\\ \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{r}}} & = \mu\dot{r}\\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial \mathcal{L}}{\partial \dot{r}}}\bigg) & = \mu\ddot{r}\\ \end{aligned}

Then one can express the one-body problem as:

\mu\ddot{r}-\mu r\dot{\phi}^2+\displaystyle{\frac{\mathrm{d}U}{\mathrm{d}r}}=0,

or,

\mu\ddot{r}-\displaystyle{\frac{L^2}{\mu r^3}}+\displaystyle{\frac{\mathrm{d}U}{\mathrm{d}r}}=0,

or,

\mu\ddot{r}+\displaystyle{\frac{\mathrm{d}}{\mathrm{d}r}U_{\textrm{eff}}(r)}=0,

where U_{\textrm{eff}}(r)=U(r)+\displaystyle{\frac{L^2}{2\mu r^2}} is the effective potential.

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201907181434 Homework 1 (Q3)

Obtain the equation of motion for a particle falling vertically under the influence of gravity when the frictional forces obtainable from a dissipation function kv^2/2 are present. Integrate the equation to obtain the velocity as a function of time and show that maximum possible velocity for a fall from rest is v=mg/k.

Solution.

Write the Lagrangian \mathcal{L}=T-V by noting

T=\displaystyle{\frac{1}{2}m\dot{\mathbf{y}}^2} and V=-mg|\mathbf{y}|,

where the upward direction is taken to be positive. The frictional force is

\mathcal{F}=\displaystyle{\frac{k\dot{\mathbf{y}}^2}{2}}.

I wish to obtain the Euler-Lagrange equation, by computing the derivatives below:

\begin{aligned} \displaystyle{\frac{\partial \mathcal{L}}{\partial y}} & = mg \\ \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}} & =m\dot{y} \\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}}\bigg) & = m\ddot{y} \\ \displaystyle{\frac{\partial \mathcal{F}}{\partial \dot{y}}} & = k\dot{y}\\ \end{aligned}

Hence I obtain the E-L equation (with dissipation):

\begin{aligned} \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}\bigg( \displaystyle{\frac{\partial \mathcal{L}}{\partial \dot{y}}}\bigg) -\displaystyle{\frac{\partial \mathcal{L}}{\partial y}}+\displaystyle{\frac{\partial \mathcal{F}}{\partial \dot{y}}} & =0\\ m\ddot{y}-mg+k\dot{y} & =0\\ \ddot{y}+\displaystyle{\frac{k}{m}}\dot{y} & =g\\ \end{aligned}

Treating u=\dot{y} as variable, I may obtain a first-order differential equation:

\dot{u}+\displaystyle{\frac{k}{m}u}-g=0

Solving it,

\begin{aligned} v(t) & =e^{\int \frac{k}{m}\mathrm{d}t}=e^{kt/m}\\ e^{kt/m}\dot{u}+\frac{k}{m}ue^{kt/m} & =ge^{kt/m}\\ \displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(ue^{kt/m}) & = ge^{kt/m}\\ ue^{kt/m} & = \displaystyle{\frac{mg}{k}}e^{kt/m}+\textrm{constant }C\\ u(t)& =\displaystyle{\frac{mg}{k}}+Ce^{-kt/m}\\ v_{\textrm{max.}}=\dot{y} & =\displaystyle{\frac{mg}{k}}\qquad\qquad (e^{-\frac{kt}{m}}\rightarrow 0\enspace \textrm{as}\enspace t\rightarrow \infty )\\ \end{aligned}

In conclusion, it is proven that the maximum possible speed for a fall from rest is v=mg/k.