# Thread: Mechanics Question

2. ## Re: Mechanics Question

Originally Posted by 1729
$\noindent Here are some hints.$

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

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

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

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

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

3. ## Re: Mechanics Question

Originally Posted by InteGrand
$\noindent Here are some hints.$

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

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

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

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

 Spoiler (rollover to view): $\noindent 5. What is the formula for \cos \theta and \cos \theta_{0} in terms of tension, m and g (it follows from resolving of forces)? Use this to finish off.$
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 $a = \frac{d^2\theta}{dt^2}$

4. ## Re: Mechanics Question

Originally Posted by 1729
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 $a = \frac{d^2\theta}{dt^2}$

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

5. ## Re: Mechanics Question

Originally Posted by 1729
how come $a = L\frac{d^2\theta}{dt^2}$ instead of $a = \frac{d^2\theta}{dt^2}$

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

6. ## Re: Mechanics Question

Originally Posted by InteGrand
$\noindent It's because the displacement along the arc is given by s = L\theta, so \ddot{s} = a = L\ddot{\theta}. (Acceleration being the second derivative wrt time of displacement.)$
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 $\theta _0 to \theta$

7. ## Re: Mechanics Question

Originally Posted by 1729
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 $\theta _0 to \theta$
$\noindent Well basically the acceleration of a particle at any moment in time is independent of its initial location''. So it suffices to only consider displacement from the middle'' of the arc when deducing the acceleration expression.$

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

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

8. ## Re: Mechanics Question

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

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

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

9. ## Re: Mechanics Question

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?

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

10. ## Re: Mechanics Question

Originally Posted by 1729
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?

Screen Shot 2018-04-17 at 2.30.37 am.png
$\noindent The mistake is in this line:$

Originally Posted by 1729
And using $\frac{dx}{dt} = -na\sin{(nt+\alpha)}$ and dx/dt = 0 when t = 0 I got alpha = 0.
$\noindent Putting t = 0 there, we have \sin \alpha = 0. This does not imply that \alpha = 0, but only that \alpha = 0 or \alpha = \pi (note it suffices to take \alpha between 0 and 2\pi due to periodicity). The correct one to choose turns out to be \alpha = \pi.$

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

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

11. ## Re: Mechanics Question

Originally Posted by InteGrand
$\noindent The mistake is in this line:$

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

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

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

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