Mechanics Question (1 Viewer)

Voldemath · Jul 9, 2015

Hey,
Can someone please show me their working out on how to solve this question... Been stuck on it for the last hour.

2. A particle of mass m moves under a retarding force that is proportional to the cube of the speed. Find how long it takes to travel a distance d from the instant speed was u.

The answer is:

d/u + kd^2/ 2m

Thanks

InteGrand · Jul 9, 2015

Voldemath said:
Hey,
Can someone please show me their working out on how to solve this question... Been stuck on it for the last hour. Don't really know how to integrate 1/(mg-kv^3)

2. A particle of mass m moves under a retarding force that is proportional to the cube of the speed. Find how long it takes to travel a distance d from the instant speed was u.

The answer is:

d/u + kd^2/ 2m

Thanks

$There's no mention of gravity here.$

$First of all, define $v=u$, $x=0$ at $t=0$. In other words, start timing from the time when $v=u$, and call the position of the particle at this time the origin.$

$We are given that $\sum \vec{F}=-kv^3$, where $k$ is a positive constant.$

$So by Newton's Second Law of Motion, $m\frac{\mathrm{d}v}{\mathrm{d}t}=-kv^3$$

$$\Rightarrow \frac{\mathrm{d}v}{v^3} = -\frac{k}{m}\text{ d}t$.$

$If $v$ is the velocity at time $t$, then integrating between the appropriate bounds,$

$$\int _u ^v \frac{\mathrm{d}\nu}{\nu ^3}=\int_0 ^t -\frac{k}{m}\text{ d}\tau$ (since at $t=0$, $v=u$)$

$\Rightarrow \left[\frac{1}{-2}\nu^{-2} \right]_u ^v = -\frac{k}{m}t$

$\Rightarrow -\frac{1}{2} \left(\frac{1}{v^2}-\frac{1}{u^2} \right) = -\frac{k}{m}t$

$\Rightarrow \frac{1}{v^2}-\frac{1}{u^2} = \frac{2k}{m}t$

$\Rightarrow \frac{1}{v^2} = \frac{1}{u^2} + \frac{2k}{m}t=\frac{m+2ku^2 t}{u^2 m}$

$\Rightarrow v^2 = \frac{u^2 m}{m+2ku^2 t}$

$$\Rightarrow v=\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{u \sqrt{m}}{\sqrt{m+2ku^2 t}}$.$

$Therefore, if $T$ is the time when the distance travelled is $d$, we have$

$$\int _0 ^d \text{ d}x =\int_0 ^T \frac{u \sqrt{m}}{\sqrt{m+2ku^2 t}} \text{ d}t$ (as $x=0$ when $t=0$)$

$\Rightarrow d=\left[\frac{u\sqrt{m}\times 2\sqrt{m+2ku^2 t}}{2ku^2} \right]_0 ^T$

$\Rightarrow d=\left[\frac{\sqrt{m}\times \sqrt{m+2ku^2 t}}{ku} \right]_0 ^T$

$$\Rightarrow d=\frac{\sqrt{m}\times \sqrt{m+2ku^2 T}}{ku} -\frac{m}{ku}$. Now we just need to solve for $T$:$

$d+\frac{m}{ku}=\frac{\sqrt{m}\times \sqrt{m+2ku^2 T}}{ku}$

$\Rightarrow kud + m=\sqrt{m}\sqrt{m+2ku^2 T}$

$$\Rightarrow \frac{kud}{\sqrt{m}} + \sqrt{m}=\sqrt{m+2ku^2 T}$ (dividing through by $\sqrt{m}$ and using $\frac{m}{\sqrt{m}}=\sqrt{m}$ on the L.H.S.)$

$\Rightarrow \left(\frac{kud}{\sqrt{m}} + \sqrt{m}\right)^2=m+2ku^2 T$

$\Rightarrow \frac{k^2u^2d^2}{m}+2\frac{kud}{\sqrt{m}}\sqrt{m}+m=m+2ku^2 T$

$\Rightarrow \frac{k^2u^2d^2}{m}+2kud=2ku^2 T$

$$\Rightarrow T=\frac{kd^2}{2m}+\frac{d}{u}$ (dividing through by $2ku^2$).$

Voldemath · Jul 9, 2015

Thanks a lot! Would've been here for a long time. Makes sense though now cheers

Voldemath · Jul 10, 2015

Another mechanics question... This time on circular motion.

10. A smooth circular disc rotates in a horizontal plane with angular velocity w rad/s about a vertical axis through its centre C. A particle P is connected by light, inextensible strings to points A and B on a diameter of the disc such that AC = BC = 5l m, AP = 8l m and BP = 6l m. If the tension in AP is P kg wt and both strings remain taut find:
i) The tension in BP.
ii) The mass of the particle.

Answers:
i) 3P/4
ii) (Pg)/(4w^2l)

Thanks in advance

InteGrand · Jul 10, 2015

Voldemath said:
Another mechanics question... This time on circular motion.

10. A smooth circular disc rotates in a horizontal plane with angular velocity w rad/s about a vertical axis through its centre C. A particle P is connected by light, inextensible strings to points A and B on a diameter of the disc such that AC = BC = 5l m, AP = 8l m and BP = 6l m. If the tension in AP is P kg wt and both strings remain taut find:
i) The tension in BP.
ii) The mass of the particle.

Answers:
i) 3P/4
ii) (Pg)/(4w^2l)

Thanks in advance

$Firstly, what does it mean by ``the tension in $AP$ is $P$ kg wt''. Units of tension should be newtons (N), which is kg m s$^{-2}$; I don't know what the ``wt'' means. Secondly, let's just relabel the tension force as having magnitude $\rho$ N, as $P$ is used already to describe a point.$

$Thirdly, I think the answer to (ii) is meant to be $m=\frac{\rho}{4\ell \omega^2}$ (no $g$ on top (gravity is not a part of this problem anyway from what I can see)). The answer with $g$ on top would not be dimensionally correct anyway (i.e. the units (or more technically, dimensions) would not match. They do in the answer $m=\frac{\rho}{4\ell \omega^2}$). (It is a useful check on your answer to see whether the dimensions match).$

$I managed the answer of $m=\frac{\rho}{4\ell \omega^2}$ for (ii) using vectors, but I'm not sure if there's a non-vector HSC-friendly way.$

$And for part (i), firstly note that the circle in question is of radius $5$, and so the diameter is 10 (using units of $\ell$ m for now). This means $P$ is situated in a triangle with sides of length $6,8$ and $10$, since it is part of triangle $ABP$, with $AB$ diameter of the circle. This is a right-angled triangle (converse of Pythagoras' Theorem), and using the converse of ``angle in a semi-circle is a right angle'', this means that $P$ actually lies on the circle.$

$This means that the net force acting on $P$ must point from $P$ to the circle centre (centripetal force), as $P$ is moving in uniform circular motion. The net force on $P$ is the sum of the two tension vectors $\vec{T}_A$ and $\vec{T}_B$ (i.e. tension in strings $AP$ and $BP$, which point from $P$ to $A$ and $B$ respectively). In order for these to sum from $P$ and end up at the origin, it is like having half the vector $\overrightarrow{PA}$ and half the vector $\overrightarrow{PB}$ and adding these, since adding the full vector $\overrightarrow{PA}$ and $\overrightarrow{PB}$ would take us from $P$ to the opposite vertex of the rectangle formed by adding these vectors.$

$Vectors $\overrightarrow{PA}$ and $\overrightarrow{PB}$ have their magnitudes in the ratio $8:6$, since $PA=8$, and $PB=6$. So their halves are also in this ratio. So our required string tension in $BP$ has magnitude six-eighths of that of the tension in $AP$. Since this latter quantity is given to be $\rho$, we must have $\left \| \vec{T_B} \right \|=\frac{6}{8}\rho=\frac{3}{4}\rho$.$

(Sorry, this makes a lot more sense if we assume knowledge of vectors, which isn't in the syllabus in too much detail; I'm not really sure if it's possible to do without using vectors. But the HSC doesn't usually ask mechanics Q's that require detailed knowledge of vectors.)

Voldemath · Jul 10, 2015

InteGrand said:
$Firstly, what does it mean by ``the tension in $AP$ is $P$ kg wt''. Units of tension should be newtons (N), which is kg m s$^{-2}$; I don't know what the ``wt'' means. Secondly, let's just relabel the tension force as having magnitude $\rho$ N, as $P$ is used already to describe a point.$

$Thirdly, I think the answer to (ii) is meant to be $m=\frac{\rho}{4\ell \omega^2}$ (no $g$ on top (gravity is not a part of this problem anyway from what I can see)). The answer with $g$ on top would not be dimensionally correct anyway (i.e. the units (or more technically, dimensions) would not match. They do in the answer $m=\frac{\rho}{4\ell \omega^2}$). (It is a useful check on your answer to see whether the dimensions match).$

$I managed the answer of $m=\frac{\rho}{4\ell \omega^2}$ for (ii) using vectors, but I'm not sure if there's a non-vector HSC-friendly way.$

$And for part (i), firstly note that the circle in question is of radius $5$, and so the diameter is 10 (using units of $\ell$ m for now). This means $P$ is situated in a triangle with sides of length $6,8$ and $10$, since it is part of triangle $ABP$, with $AB$ diameter of the circle. This is a right-angled triangle (converse of Pythagoras' Theorem), and using the converse of ``angle in a semi-circle is a right angle'', this means that $P$ actually lies on the circle.$

$This means that the net force acting on $P$ must point from $P$ to the circle centre (centripetal force), as $P$ is moving in uniform circular motion. The net force on $P$ is the sum of the two tension vectors $\vec{T}_A$ and $\vec{T}_B$ (i.e. tension in strings $AP$ and $BP$, which point from $P$ to $A$ and $B$ respectively). In order for these to sum from $P$ and end up at the origin, it is like having half the vector $\overrightarrow{PA}$ and half the vector $\overrightarrow{PB}$ and adding these, since adding the full vector $\overrightarrow{PA}$ and $\overrightarrow{PB}$ would take us from $P$ to the opposite vertex of the rectangle formed by adding these vectors.$

$Vectors $\overrightarrow{PA}$ and $\overrightarrow{PB}$ have their magnitudes in the ratio $8:6$, since $PA=8$, and $PB=6$. So their halves are also in this ratio. So our required string tension in $BP$ has magnitude six-eighths of that of the tension in $AP$. Since this latter quantity is given to be $\rho$, we must have $\left \| \vec{T_B} \right \|=\frac{6}{8}\rho=\frac{3}{4}\rho$.$

(Sorry, this makes a lot more sense if we assume knowledge of vectors, which isn't in the syllabus in too much detail; I'm not really sure if it's possible to do without using vectors. But the HSC doesn't usually ask mechanics Q's that require detailed knowledge of vectors.)

Terribly sorry about that... I thought it was a standard notation to write kg wt since my tuition uses it all the time. Basically 1kg wt is 9.8kgms^(-2) or 9.8N. Your vector method for i) makes sense to me. I do believe the answer to ii) may be correct. Would you be kind enough to post how u got to your answer for ii) as well with your vector method. I'll be willing to give my night for u to teach me a bit more about them.

InteGrand · Jul 11, 2015

Voldemath said:
Terribly sorry about that... I thought it was a standard notation to write kg wt since my tuition uses it all the time. Basically 1kg wt is 9.8kgms^(-2) or 9.8N. Your vector method for i) makes sense to me. I do believe the answer to ii) may be correct. Would you be kind enough to post how u got to your answer for ii) as well with your vector method. I'll be willing to give my night for u to teach me a bit more about them.

I searched it online and sources have said that kg wt stands for 'kilogram weight' and is equal to $g$ newtons like you said, but it's better to use N (newtons) if writing force units in the HSC (assuming they give you units (and if they do, it would generally be in newtons)). Newtons is SI units, whereas kg wt is not, and there's a greater risk probably of the marker not knowing it (there's actually an old thread about it: http://community.boredofstudies.org/14/mathematics-extension-2/14427/newtons-kg-wt.html ).

Voldemath · Jul 11, 2015

Alright thanks so much for all ur help

I'm in the middle of my 90 question mechanics marathon so I'll let u know if I have any other questions.

InteGrand · Jul 11, 2015

Voldemath said:
Alright thanks so much for all ur help I'm in the middle of my 90 question mechanics marathon so I'll let u know if I have any other questions.

I just realised why the answer had g on the numerator. It's because they used kg wt as the units, which, as you said, is essentially g newtons. So the tension force was actually $\rho g$ newtons. I was doing the Q assuming $\rho$ newtons, because I hadn't heard of kg wt before. So using $\rho g$ newtons instead, I'd get the answer you gave. (My method would still be the same.)

Voldemath · Jul 11, 2015

InteGrand said:
I just realised why the answer had g on the numerator. It's because they used kg wt as the units, which, as you said, is essentially g newtons. So the tension force was actually $\rho g$ newtons. I was doing the Q assuming $\rho$ newtons, because I hadn't heard of kg wt before. So using $\rho g$ newtons instead, I'd get the answer you gave. (My method would still be the same.)

Perfect just as I thought. Thanks for the help!

Voldemath · Jul 11, 2015

Ok my next question is on conical pendulums:
4. A smooth hollow cone of semi-vertex angle a is placed with axis vertical and vertex downward. A particle moves in a circle on its inner making nrevolutions per second. Find the distance of the particle from the axis of the cone at any time.

Answer:
(gcot(a))/(4(pi)^2 n^2)

InteGrand · Jul 11, 2015

Voldemath said:
Ok my next question is on conical pendulums:
4. A smooth hollow cone of semi-vertex angle a is placed with axis vertical and vertex downward. A particle moves in a circle on its inner making nrevolutions per second. Find the distance of the particle from the axis of the cone at any time.

Answer:
(gcot(a))/(4(pi)^2 n^2)

$Draw a diagram of an inverted cone with semi-vertical angle $\alpha$, and the particle revolving around in a circle of radius $r$ (which is what we want to find) some way up from the vertex. The forces on the particle are: a normal force $\overrightarrow{N}$ from the surface of the cone, which (as the name suggests) points normal to the surface; and a gravitational (weight) force $\overrightarrow{W}=mg$ downwards ($m$ is the particle's mass).$

$Let the vertical component of the normal force be $N_y$ (clearly pointing upwards), and the horizontal component be $N_x$ (this provides the centripetal force). From geometry, we can see that $N_x = N\cos \alpha$ and $N_y=N\sin \alpha$, where $N$ is the magnitude of the normal vector $\overrightarrow{N}$ (that is, $N=\left\|\overrightarrow{N} \right\|$).$

$Since the particle stays at the same level, there is 0 net force in the vertical direction, so we must have $N_y = mg \Rightarrow N\sin \alpha = mg \Rightarrow N = \frac{mg}{\sin \alpha}$.$

$Now, analysing horizontal forces, as mentioned earlier, $N_x$ provides centripetal force, so $N_x = mr\omega^2$, where we are given that $\omega = n \text{ rev/s} = 2\pi n \text{ rad/s}$ (since $1\text{ rev} = 2\pi \text{ rad}$).$

$\therefore N_x = mr\omega^2$

$\Rightarrow N\cos \alpha = mr\omega^2$

$\Rightarrow \frac{mg}{\sin \alpha}\times\cos \alpha = mr\omega^2$

$$\Rightarrow g\cot \alpha = r\omega^2$ (cancelling $m$)$

$$\Rightarrow r=\frac{g\cot \alpha}{\omega^2}$ (rearranging)$

$$\Rightarrow r=\frac{g\cot \alpha}{4\pi^2 n^2}$ (since $\omega = 2\pi n$).$

Voldemath · Jul 11, 2015

InteGrand said:
$Draw a diagram of an inverted cone with semi-vertical angle $\alpha$, and the particle revolving around in a circle of radius $r$ (which is what we want to find) some way up from the vertex. The forces on the particle are: a normal force $\overrightarrow{N}$ from the surface of the cone, which (as the name suggests) points normal to the surface; and a gravitational (weight) force $\overrightarrow{W}=mg$ downwards ($m$ is the particle's mass).$

$Let the magnitude of the vertical component of the normal force be $N_y$ (clearly pointing upwards), and that of the horizontal component be $N_x$ (this provides the centripetal force). From geometry, we can see that $N_x = N\cos \alpha$ and $N_y=N\sin \alpha$, where $N$ is the magnitude of the normal vector $\overrightarrow{N}$ (that is, $N=\left\|\overrightarrow{N} \right\|$).$

$Since the particle stays at the same level, there is 0 net force in the vertical direction, so we must have $N_y = mg \Rightarrow N\sin \alpha = mg \Rightarrow N = \frac{mg}{\sin \alpha}$.$

$Now, analysing horizontal forces, as mentioned earlier, $N_x$ provides centripetal force, so $N_x = mr\omega^2$, where we are given that $\omega = n \text{ rev/s} = 2\pi n \text{ rad/s}$ (since $1\text{ rev} = 2\pi \text{ rad}$).$

$\therefore N_x = mr\omega^2$

$\Rightarrow N\cos \alpha = mr\omega^2$

$\Rightarrow \frac{mg}{\sin \alpha}\times\cos \alpha = mr\omega^2$

$$\Rightarrow g\cot \alpha = r\omega^2$ (cancelling $m$)$

$$\Rightarrow r=\frac{g\cot \alpha}{\omega^2}$ (rearranging)$

$$\Rightarrow r=\frac{g\cot \alpha}{4\pi^2 n^2}$ (since $\omega = 2\pi n$).$

As usual thanks for the great in-depth reply. The solution is absolutely crystal clear. I always seem to struggle with getting the diagram in these pendulum questions, especially with multiple masses

Voldemath · Jul 11, 2015

Ok I'm back with two more questions - this time on projectile motion

8. A particle projected with speed V can just reach a certain point on the horizontal plane through the point of projection. Show that to hit a target h metres above the ground at the same horizontal distance using the same angle of projection the speed of projection must be increased to (V^2)/(V^2 -gh)^(1/2)

9. A stone is projected upwards at an angle of elevation a with an initial velocity of projection U. If at time t, the stone is moving in a direction perpendicular to the velocity of projection show that: t = U/(gsin(a)). Also, find the stone's speed at that time.

Thanks

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