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Inspired by the hard integral thread !!!
Topics of discussion : Hard questions from the A level Physics/ Mathematics (Mechanics) and the first-year undergraduate classical mechanics topics including Special Relativity.
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The second Question:

The two components of a binary star are observed to move in circles of radii r1 and r2. What is the ratio of their masses?
(Note : Binary stars orbits around the combined centre of mass.)

If you have some interesting questions please post them.
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username2452153
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This time a simpler one from the projectile motion :
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Please feel free to post any questions, interesting/confusing ideas or anything interesting related to mechanics you may have.
Feel free to tag any members who you may think might be interested in this.
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NatoHeadshot
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(Original post by tangotangopapa2)
Inspired by the hard integral thread !!!
Topics of discussion : Hard questions from the A level Physics/ Mathematics (Mechanics) and the first-year undergraduate classical mechanics topics including Special Relativity.
Name:  interesting.png
Views: 367
Size:  62.9 KB

The second Question:

The two components of a binary star are observed to move in circles of radii r1 and r2. What is the ratio of their masses?
(Note : Binary stars orbits around the combined centre of mass.)

If you have some interesting questions please post them.
will try this tonight!
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langlitz
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A balloon contains a bag of sand of initial mass m_0, and the combined mass of balloon+sand is M_0. The balloon experiences a constant upthrust R and is initially at rest and in equilibrium, with the upthrust compensating exactly for the gravitational force. At time t = 0 the sand is released at a constant rate dm/dt. It is fully disposed of after a time T.

(i) Determine the velocity of the balloon v(t) at any time between t = 0 and T. (Hint: Derive your expression in terms of  k = m_0/M_0T)
(ii) Find the height gained by the balloon as a function of time.
(iii) For small values of  \epsilon= m_0/M_0 express your results for the velocity and height gain at time T to first order in \epsilon .
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(Original post by langlitz)
A balloon contains a bag of sand of initial mass m_0, and the combined mass of balloon+sand is M_0. The balloon experiences a constant upthrust R and is initially at rest and in equilibrium, with the upthrust compensating exactly for the gravitational force. At time t = 0 the sand is released at a constant rate dm/dt. It is fully disposed of after a time T.

(i) Determine the velocity of the balloon v(t) at any time between t = 0 and T. (Hint: Derive your expression in terms of  k = m_0/M_0T)
(ii) Find the height gained by the balloon as a function of time.
(iii) For small values of  \epsilon= m_0/M_0 express your results for the velocity and height gain at time T to first order in \epsilon .
Really tough question. My previous solution for (i) was: Name:  aaa see this.png
Views: 225
Size:  2.6 KBbut now I know that I missed out something. I used F = M(x) dv/dt which is wrong. Need to do it again writing F as F = M(x) dv/dt + dM/dt times v and solve the differential equation again. I am not sure I will be able to solve the differential equation.
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langlitz
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(Original post by tangotangopapa2)
Really tough question. My previous solution for (i) was: Name:  aaa see this.png
Views: 225
Size:  2.6 KBbut now I know that I missed out something. I used F = M(x) dv/dt which is wrong. Need to do it again writing F as F = M(x) dv/dt + dM/dt times v and solve the differential equation again. I am not sure I will be able to solve the differential equation.
Yes you're right there, it's an easy mistake to make as normally the mass isn't changing in mechanics questions. The integration is not too bad actually, I have faith in you
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(Original post by tangotangopapa2)
Really tough question. My previous solution for (i) was: Name:  aaa see this.png
Views: 225
Size:  2.6 KBbut now I know that I missed out something. I used F = M(x) dv/dt which is wrong. Need to do it again writing F as F = M(x) dv/dt + dM/dt times v and solve the differential equation again. I am not sure I will be able to solve the differential equation.
(Original post by langlitz)
Yes you're right there, it's an easy mistake to make as normally the mass isn't changing in mechanics questions. The integration is not too bad actually, I have faith in you
Wait a moment, If m_0 is very small compared to M_0. and dm/dt is fairly small, M(x)dv/dt could be reasonable approximation to F. In that case, rewriting above equation using R = M_0 g, we get: Name:  aaa see this.png
Views: 207
Size:  5.8 KBand then after integrating this (had to use integration of parts to solve integration of ln(1-kt) by writing it as 1 times ln (1-kt)) we can arrive at the following equation: Name:  bbb see this.png
Views: 206
Size:  2.5 KB

I am sorry but I just looked at the solution in the following sheet, last question: http://www2.ph.ed.ac.uk/~egardi/MfP3..._Workshop3.pdf
Attached files
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One end of a thin inexstendable, but perfectly flexible, length of string 'L' with uniform mass per unit length is held at a point on a smooth table a distance 'd' (< L) away from a small vertical hole in the surface of the table. The string passes through the hole so that a length L - d of the string hangs vertically. The string is released from rest. Assuming the height of the table is greater than L, find the time taken for the end of the string to reach the top of the hole.

tangotangopapa2 this is a bit easy for you
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(Original post by Ipsooo)
One end of a thin inexstendable, but perfectly flexible, length of string 'L' with uniform mass per unit length is held at a point on a smooth table a distance 'd' (< L) away from a small vertical hole in the surface of the table. The string passes through the hole so that a length L - d of the string hangs vertically. The string is released from rest. Assuming the height of the table is greater than L, find the time taken for the end of the string to reach the top of the hole.

tangotangopapa2 this is a bit easy for you
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sqrt (L/g) arc cosh (L/ (L-D))
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Ipsooo
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(Original post by tangotangopapa2)
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sqrt (L/g) arc cosh (L/ (L-D))
As I said...a bit easy
Let me find some harder ones for you hmm
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(Original post by Ipsooo)
As I said...a bit easy
Let me find some harder ones for you hmm
Omg!!! These differential equations are nearly killing me.
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(Original post by tangotangopapa2)
Wait a moment, If m_0 is very small compared to M_0. and dm/dt is fairly small, M(x)dv/dt could be reasonable approximation to F. In that case, rewriting above equation using R = M_0 g, we get: Name:  aaa see this.png
Views: 207
Size:  5.8 KBand then after integrating this (had to use integration of parts to solve integration of ln(1-kt) by writing it as 1 times ln (1-kt)) we can arrive at the following equation: Name:  bbb see this.png
Views: 206
Size:  2.5 KB

I am sorry but I just looked at the solution in the following sheet, last question: http://www2.ph.ed.ac.uk/~egardi/MfP3..._Workshop3.pdf
Well done, that question is damned brutal (even when you cheat ). I have a similar and possibly harder one which has no answer online if you'd like to try it?
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(Original post by langlitz)
Well done, that question is damned brutal (even when you cheat ). I have a similar and possibly harder one which has no answer online if you'd like to try it?
Sure!!!
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(Original post by tangotangopapa2)
Sure!!!
Actually I just did it and it's not super hard but here

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A stick of mass density per unit length ρ rests on a circle of radius R.The stick makes an angle θ with the horizontal and is tangent to the circle at its upper end. Friction exists at all points of contact and it is large enough to keep the system at rest. Find the friction force between the ground and the circle.

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(Yes, the information given is complete. You do not need coefficient of friction or the frictional force between stick and circle/ground). Good Luck haha.

If you have something please post it.
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(Original post by langlitz)
Actually I just did it and it's not super hard but here

Name:  a1.png
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Are my answers correct?



a)  v = v_o e^{- \frac{bt}{M_o + m_o}} Where (v)s are vectors, don't know how to write vectors in LaTex.

b) Momentum is only conserved if there is no external force, as there is external drag momentum is not conserved. The change in energy accounts for the work done against the resistive force.

c) (By conservation of momentum) Mv + delta mu = 0 Differentiating with respect to time and writing, dM/dt = 0 and du/dt = 0, we get the given equation. (I have taken reference frame to be the frame stationary with respect to rocket just before the propulsion.) (This is the famous rocket equation where F = 0.)

d) Integrating the equation given in c) with limits M= M_o + m_o to M = M_o + m_o - m_1 we get the equation in v.

e) As, v_o and u are parallel, we could just write their magnitudes in the equation. Now we simply have to plug in values to obtain:
 m_1 = (M_o + m_o)(1-e^{-\frac{v_o}{10u}})
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(Original post by Ipsooo)
One end of a thin inexstendable, but perfectly flexible, length of string 'L' with uniform mass per unit length is held at a point on a smooth table a distance 'd' (< L) away from a small vertical hole in the surface of the table. The string passes through the hole so that a length L - d of the string hangs vertically. The string is released from rest. Assuming the height of the table is greater than L, find the time taken for the end of the string to reach the top of the hole.

tangotangopapa2 this is a bit easy for you
Hint pls
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mik1a
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That's a neat one. ^^

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You can treat the string as a particle falling through space, but with variable mass.
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This is because the mass of the horizontal part of the string does not cause acceleration (balanced by table reaction force), but the mass of the vertical part of the string does. So the mass of the object is effectively the mass per unit length times the length of string hanging vertically, which is a function of time. Set up the equations allowing mass to be a function of this length, and this length to be a function of the displacement of the end of the string, which (via F=ma integrated) is a function of the mass.

You should get a 2nd order differential equation in time that should be quite easy to solve.
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(Original post by NatoHeadshot)
Hint pls
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Just to add to what mik1a said, if  \frac{d^{2}y}{{dt}^2}=w^2y then  y=c_1e^{wt} + c_2e^{-wt} where c1 and c2 are constants of integration.
 cosh x = \frac{e^{x} + e^{-x}}{2} could be handy too. Good Luck!!!!
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langlitz
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(Original post by tangotangopapa2)
Are my answers correct?



a)  v = v_o e^{- \frac{bt}{M_o + m_o}} Where (v)s are vectors, don't know how to write vectors in LaTex.

b) Momentum is only conserved if there is no external force, as there is external drag momentum is not conserved. The change in energy accounts for the work done against the resistive force.

c) (By conservation of momentum) Mv + delta mu = 0 Differentiating with respect to time and writing, dM/dt = 0 and du/dt = 0, we get the given equation. (I have taken reference frame to be the frame stationary with respect to rocket just before the propulsion.) (This is the famous rocket equation where F = 0.)

d) Integrating the equation given in c) with limits M= M_o + m_o to M = M_o + m_o - m_1 we get the equation in v.

e) As, v_o and u are parallel, we could just write their magnitudes in the equation. Now we simply have to plug in values to obtain:
 m_1 = (M_o + m_o)(1-e^{-\frac{v_o}{10u}})
I think so I don't think you'd get all the marks for what you wrote in b) though.

p.s. to write vectors in latex you do \vec{b} for example for a vector \vec{b}
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