Mechanics Question (1 Viewer)

1729 · Apr 16, 2018

Screen Shot 2018-04-16 at 9.32.04 pm.png

Thanks in advance

InteGrand · Apr 16, 2018

1729 said:
View attachment 34525

Thanks in advance

$\noindent Here are some hints.$

$\noindent 1. Show that $\ddot{\theta} = \frac{g}{L}\sin \theta$ (where dots refer to time derivatives).$

$\noindent 2. Recall that $\ddot{\theta} = \frac{d}{d\theta}\left(\frac{1}{2}\omega^{2}\right)$ (same as the formula $\ddot{x} = \frac{d}{dx}\left(\frac{1}{2}v^{2}\right)$ that you would have seen in 3U probably).$

$\noindent 3. Observe that it follows from the previous two hints that $\frac{d}{d\theta}\left( \frac{1}{2}\omega^{2}\right) = \frac{g}{L}\sin \theta$.$

$\noindent 4. Integrate both sides from $\theta_{0}$ to $\theta$. (Given that the bob is released from rest, what is the value of $\omega_{0}$?)$

$\noindent 5. What is the formula for $\cos \theta$ and $\cos \theta_{0}$ in terms of tension, $m$ and $g$ (it follows from resolving of forces)? Use this to finish off.$

1729 · Apr 16, 2018

InteGrand said:
$\noindent Here are some hints.$

$\noindent 1. Show that $\ddot{\theta} = \frac{g}{L}\sin \theta$ (where dots refer to time derivatives).$

$\noindent 2. Recall that $\ddot{\theta} = \frac{d}{d\theta}\left(\frac{1}{2}\omega^{2}\right)$ (same as the formula $\ddot{x} = \frac{d}{dx}\left(\frac{1}{2}v^{2}\right)$ that you would have seen in 3U probably).$

$\noindent 3. Observe that it follows from the previous two hints that $\frac{d}{d\theta}\left( \frac{1}{2}\omega^{2}\right) = \frac{g}{L}\sin \theta$.$

$\noindent 4. Integrate both sides from $\theta_{0}$ to $\theta$. (Given that the bob is released from rest, what is the value of $\omega_{0}$?)$

$\noindent 5. What is the formula for $\cos \theta$ and $\cos \theta_{0}$ in terms of tension, $m$ and $g$ (it follows from resolving of forces)? Use this to finish off.$

Do you use Newton's Second Law tangentially to show $\ddot{\theta} = \frac{g}{L}\sin \theta$$

With $\sum F = -mg\sin \theta$ and $a = L\frac{d^2\theta}{dt^2}$

Wouldn't it then be $\ddot{\theta} = -\frac{g}{L}\sin \theta$$ and how come $a = L\frac{d^2\theta}{dt^2}$ instead of $a = \frac{d^2\theta}{dt^2}$

Thanks in advance

InteGrand · Apr 16, 2018

1729 said:
Do you use Newton's Second Law tangentially to show $\ddot{\theta} = \frac{g}{L}\sin \theta$$

With $\sum F = -mg\sin \theta$ and $a = L\frac{d^2\theta}{dt^2}$

Wouldn't it then be $\ddot{\theta} = -\frac{g}{L}\sin \theta$$ and how come $a = L\frac{d^2\theta}{dt^2}$ instead of $a = \frac{d^2\theta}{dt^2}$

Thanks in advance

$$\noindent Ah yes, I meant $-g/L$, not $g/L$, sorry.$

InteGrand · Apr 16, 2018

1729 said:
how come $a = L\frac{d^2\theta}{dt^2}$ instead of $a = \frac{d^2\theta}{dt^2}$

Thanks in advance

$\noindent It's because the displacement along the arc is given by $s = L\theta$, so $\ddot{s} = a = L\ddot{\theta}$. (Acceleration being the second derivative wrt time of displacement.)$

1729 · Apr 17, 2018

InteGrand said:
$\noindent It's because the displacement along the arc is given by $s = L\theta$, so $\ddot{s} = a = L\ddot{\theta}$. (Acceleration being the second derivative wrt time of displacement.)$

Why do we only consider half of the entire arc? Is it because the net force would then be 0?

Also what is the physical significance of integrating from $\theta _0$ to $\theta$

InteGrand · Apr 17, 2018

1729 said:
Why do we only consider half of the entire arc? Is it because the net force would then be 0?

Also what is the physical significance of integrating from $\theta _0$ to $\theta$

$\noindent Well basically the acceleration of a particle at any moment in time is independent of its ``initial location''. So it suffices to only consider displacement from the ``middle'' of the arc when deducing the acceleration expression.$

$\noindent The reason I said to integrate from $\theta_{0}$ to $\theta$ is because we have given data at $\theta_{0}$ (i.e. we know $T_{0}$ then and we know the value $\omega_{0}$ at that point in time (since it was released \underline{from rest})).$

$\noindent Integrating from $\theta_{0}$ to $\theta$ is like integrating over the ``path taken so far''; i.e. view a start of the pendular cycle as being at angle $\theta_{0}$ (the right ``extreme'' of the motion) and integrating up to its current location, $\theta$. If you don't like this, you can just integrate using indefinite integrals over $\theta$, and then you'll have to introduce a $+C$ to one side. Then use the initial condition (at $\theta_{0}$) to find the $C$. But it is quicker to just integrate from $\theta_{0}$ to $\theta$ straightaway.$

1729 · Apr 17, 2018

InteGrand said:
$\noindent Well basically the acceleration of a particle at any moment in time is independent of its ``initial location''. So it suffices to only consider displacement from the ``middle'' of the arc when deducing the acceleration expression.$

$\noindent The reason I said to integrate from $\theta_{0}$ to $\theta$ is because we have given data at $\theta_{0}$ (i.e. we know $T_{0}$ then and we know the value $\omega_{0}$ at that point in time (since it was released \underline{from rest})).$

$\noindent Integrating from $\theta_{0}$ to $\theta$ is like integrating over the ``path taken so far''; i.e. view a start of the pendular cycle as being at angle $\theta_{0}$ (the right ``extreme'' of the motion) and integrating up to its current location, $\theta$. If you don't like this, you can just integrate using indefinite integrals over $\theta$, and then you'll have to introduce a $+C$ to one side. Then use the initial condition (at $\theta_{0}$) to find the $C$. But it is quicker to just integrate from $\theta_{0}$ to $\theta$ straightaway.$

Thank you

1729 · Apr 17, 2018

Another mechanics question (sry trying to self-learn)

I let x = 0 be the wharf and t = 0 be at 1:10 am which would mean the endpoints of the tidal motion are x = -1 and x = 0.4. Since we are told it is SHM, we can deduce $x = a\cos{(nt+\alpha)} + b$

Can we immediately deduce b = -0.3 since it is the centre of motion and a = 0.7? And if t is in hours, the period would be 12 hours and 25 minutes or T = 149/12 which would make n = 24pi/149.

And using $\frac{dx}{dt} = -na\sin{(nt+\alpha)}$ and dx/dt = 0 when t = 0 I got alpha = 0.

Meaning I got $x = 0.7\cos{\left(\frac{24\pi}{149}t\right)} - 0.3$ which is clearly wrong since at t = 0, x = 0.4 and not x = -1. What have I done wrong?

Thanks in advance

Screen Shot 2018-04-17 at 2.30.37 am.png

InteGrand · Apr 17, 2018

1729 said:
Another mechanics question (sry trying to self-learn)

I let x = 0 be the wharf and t = 0 be at 1:10 am which would mean the endpoints of the tidal motion are x = -1 and x = 0.4. Since we are told it is SHM, we can deduce $x = a\cos{(nt+\alpha)} + b$

Can we immediately deduce b = -0.3 since it is the centre of motion and a = 0.7? And if t is in hours, the period would be 12 hours and 25 minutes or T = 149/12 which would make n = 24pi/149.

And using $\frac{dx}{dt} = -na\sin{(nt+\alpha)}$ and dx/dt = 0 when t = 0 I got alpha = 0.

Meaning I got $x = 0.7\cos{\left(\frac{24\pi}{149}t\right)} - 0.3$ which is clearly wrong since at t = 0, x = 0.4 and not x = -1. What have I done wrong?

Thanks in advance

View attachment 34526

$\noindent The mistake is in this line:$

1729 said:
And using $\frac{dx}{dt} = -na\sin{(nt+\alpha)}$ and dx/dt = 0 when t = 0 I got alpha = 0.

$\noindent Putting $t = 0$ there, we have $\sin \alpha = 0$. This does not imply that $\alpha = 0$, but only that $\alpha = 0$ or $\alpha = \pi$ (note it suffices to take $\alpha$ between $0$ and $2\pi$ due to periodicity). The correct one to choose turns out to be $\alpha = \pi$.$

$\noindent Think of an easier question: fit a cosine wave of the form $y = \cos(x + \alpha)$ with period $2\pi$ that goes through $(0,-1)$ with slope $0$ at $x = 0$. (By periodicity of $\cos$, it suffices to only consider $\alpha \in [0,2\pi)$.) By doing your ``slope $=$ 0 at $x = 0$'' method, we are basically not narrowing down the possibilities enough, since we admit both $y = \cos x$ and $y= -\cos x$ (these both have slope $0$ at the origin). So the slope isn't really the right thing to look at, the right thing to look at first is the \textbf{value} at $ x= 0$. Doing this, we have $y = -1$ at $x = 0$, so $\cos(0 + \alpha) = -1\Rightarrow \cos \alpha = -1$. Hence $\alpha = \pi$.$

$\noindent So in your question, try looking at the value at $t = 0$ first rather than the slope. Also you can do these sorts of questions more intuitively quickly. Since we're starting at low tide and moving towards high tide, this is like using a cosine curve that is starting at its minimum point, and moving up. This looks like a $-\cos t$ curve. So the form of our function will be $y = -A\cos (nt) + b$, where $A$ is the amplitude (note that $A > 0$ because we took care of sign by putting the minus sign out front). Now you can just inspect the values of $A, n, b$ as you did before to get the answer. Upshot: \color{blue}{We don't to worry about $\alpha$ by using this method}\color{black}; it's accounted for already since we know the curve must look like a negative cosine, due to it being a sinusoid that starts off at its minimum.$

1729 · Apr 18, 2018

InteGrand said:
$\noindent The mistake is in this line:$

$\noindent Putting $t = 0$ there, we have $\sin \alpha = 0$. This does not imply that $\alpha = 0$, but only that $\alpha = 0$ or $\alpha = \pi$ (note it suffices to take $\alpha$ between $0$ and $2\pi/n$ due to periodicity). The correct one to choose turns out to be $\alpha = \pi$.$

$\noindent Think of an easier question: fit a cosine wave of the form $y = \cos(x + \alpha)$ with period $2\pi$ that goes through $(0,-1)$ with slope $0$ at $x = 0$. (By periodicity of $\cos$, it suffices to only consider $\alpha \in [0,2\pi)$.) By doing your ``slope $=$ 0 at $x = 0$'' method, we are basically not narrowing down the possibilities enough, since we admit both $y = \cos x$ and $y= -\cos x$ (these both have slope $0$ at the origin). So the slope isn't really the right thing to look at, the right thing to look at first is the \textbf{value} at $ x= 0$. Doing this, we have $y = -1$ at $x = 0$, so $\cos(0 + \alpha) = -1\Rightarrow \cos \alpha = -1$. Hence $\alpha = \pi$.$

$\noindent So in your question, try looking at the value at $t = 0$ first rather than the slope. Also you can do these sorts of questions more intuitively quickly. Since we're starting at low tide and moving towards high tide, this is like using a cosine curve that is starting at its minimum point, and moving up. This looks like a $-\cos t$ curve. So the form of our function will be $y = -A\cos (nt) + b$, where $A$ is the amplitude (note that $A > 0$ because we took care of sign by putting the minus sign out front). Now you can just inspect the values of $A, n, b$ as you did before to get the answer. Upshot: \color{blue}{We don't to worry about $\alpha$ by using this method}\color{black}; it's accounted for already since we know the curve must look like a negative cosine, due to it being a sinusoid that starts off at its minimum.$

Thank you

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Mechanics Question (1 Viewer)

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