Mathematical Pendulum Revisited
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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Exercise:
We consider a mathematical pulum as a special case of a physical pulum. abcliste abc Derive the formal expression for the period of a mathematical pulum from the more general expression for a physical pulum. abc Calculate the length of a mathematical pulum with a period of TO. abc Derive an expression for the period of a mathematical pulum where the mass m_s of the string is a fraction of the pulum mass m: m_sf m. abcliste
Solution:
abcliste abc The pulum mass of a mathematical pulum is idealised to be a po mass. We can think of it as a spherical mass m with radius r at a distance sell from the pivot po. Using the expression for the period of a physical pulum we find T & approx pisqrtfracJ_s+ms^mgs pisqrtfracfracmr^+mell^mgell pisqrtfrac+mell^mgell pisqrtfracellg labelperiod This is the well-known expression for the period of a mathematical pulum. abc Solving refperiod for the pulum length we find ell LF fracncgtimesT^pi^ L approx resultLP abc The moment of inertia is the of the two partial moments: J sscJstring+sscJmass We idealise the string as a thin rod. It oscillates about one of the string i.e. we have to shift the pivot po from the center by ell/ Steiner's theorem: sscJstring fracm_s ell^+m_s ell/^ fracm_s ell^ The total moment of inertia is J fracm_sell^+mell^ leftm_s/+mrightell^ The torque on the pulum can also be expressed as the of the torques acting on the string and on the mass: tau ssctaustring+ssctaumass -m_s gfracellsinphi -m gellsinphi & approx -m_s/+mgellphi The approximation in the last step is valid for small angles phill . Using the basic for rotational motion we find Jddotphit & taut m_s/+mell^ddotphit &approx -m_s/+mgellphit Solving for the angular acceleration leads to the characteristic of the simple harmonic motion: ddotphit &approx -fracm_s/+mm_s/+mfracgellphit The angular frequency is omega sqrtfracm_s/+mm_s/+mfracgell and the period is T &approx fracpiomega pisqrtfracellgsqrtfracf/+f/+ T_sqrtfracf/+f/+ where T_ is the period of a normal mathematical pulum. As an example we can consider the case where the mass of the string is fO of the pulum mass. In this case the correction factor is corrF sqrtfracf/+f/+ corrP i.e. the effect is quite small and can usually be neglected. abcliste
We consider a mathematical pulum as a special case of a physical pulum. abcliste abc Derive the formal expression for the period of a mathematical pulum from the more general expression for a physical pulum. abc Calculate the length of a mathematical pulum with a period of TO. abc Derive an expression for the period of a mathematical pulum where the mass m_s of the string is a fraction of the pulum mass m: m_sf m. abcliste
Solution:
abcliste abc The pulum mass of a mathematical pulum is idealised to be a po mass. We can think of it as a spherical mass m with radius r at a distance sell from the pivot po. Using the expression for the period of a physical pulum we find T & approx pisqrtfracJ_s+ms^mgs pisqrtfracfracmr^+mell^mgell pisqrtfrac+mell^mgell pisqrtfracellg labelperiod This is the well-known expression for the period of a mathematical pulum. abc Solving refperiod for the pulum length we find ell LF fracncgtimesT^pi^ L approx resultLP abc The moment of inertia is the of the two partial moments: J sscJstring+sscJmass We idealise the string as a thin rod. It oscillates about one of the string i.e. we have to shift the pivot po from the center by ell/ Steiner's theorem: sscJstring fracm_s ell^+m_s ell/^ fracm_s ell^ The total moment of inertia is J fracm_sell^+mell^ leftm_s/+mrightell^ The torque on the pulum can also be expressed as the of the torques acting on the string and on the mass: tau ssctaustring+ssctaumass -m_s gfracellsinphi -m gellsinphi & approx -m_s/+mgellphi The approximation in the last step is valid for small angles phill . Using the basic for rotational motion we find Jddotphit & taut m_s/+mell^ddotphit &approx -m_s/+mgellphit Solving for the angular acceleration leads to the characteristic of the simple harmonic motion: ddotphit &approx -fracm_s/+mm_s/+mfracgellphit The angular frequency is omega sqrtfracm_s/+mm_s/+mfracgell and the period is T &approx fracpiomega pisqrtfracellgsqrtfracf/+f/+ T_sqrtfracf/+f/+ where T_ is the period of a normal mathematical pulum. As an example we can consider the case where the mass of the string is fO of the pulum mass. In this case the correction factor is corrF sqrtfracf/+f/+ corrP i.e. the effect is quite small and can usually be neglected. abcliste
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Exercise:
We consider a mathematical pulum as a special case of a physical pulum. abcliste abc Derive the formal expression for the period of a mathematical pulum from the more general expression for a physical pulum. abc Calculate the length of a mathematical pulum with a period of TO. abc Derive an expression for the period of a mathematical pulum where the mass m_s of the string is a fraction of the pulum mass m: m_sf m. abcliste
Solution:
abcliste abc The pulum mass of a mathematical pulum is idealised to be a po mass. We can think of it as a spherical mass m with radius r at a distance sell from the pivot po. Using the expression for the period of a physical pulum we find T & approx pisqrtfracJ_s+ms^mgs pisqrtfracfracmr^+mell^mgell pisqrtfrac+mell^mgell pisqrtfracellg labelperiod This is the well-known expression for the period of a mathematical pulum. abc Solving refperiod for the pulum length we find ell LF fracncgtimesT^pi^ L approx resultLP abc The moment of inertia is the of the two partial moments: J sscJstring+sscJmass We idealise the string as a thin rod. It oscillates about one of the string i.e. we have to shift the pivot po from the center by ell/ Steiner's theorem: sscJstring fracm_s ell^+m_s ell/^ fracm_s ell^ The total moment of inertia is J fracm_sell^+mell^ leftm_s/+mrightell^ The torque on the pulum can also be expressed as the of the torques acting on the string and on the mass: tau ssctaustring+ssctaumass -m_s gfracellsinphi -m gellsinphi & approx -m_s/+mgellphi The approximation in the last step is valid for small angles phill . Using the basic for rotational motion we find Jddotphit & taut m_s/+mell^ddotphit &approx -m_s/+mgellphit Solving for the angular acceleration leads to the characteristic of the simple harmonic motion: ddotphit &approx -fracm_s/+mm_s/+mfracgellphit The angular frequency is omega sqrtfracm_s/+mm_s/+mfracgell and the period is T &approx fracpiomega pisqrtfracellgsqrtfracf/+f/+ T_sqrtfracf/+f/+ where T_ is the period of a normal mathematical pulum. As an example we can consider the case where the mass of the string is fO of the pulum mass. In this case the correction factor is corrF sqrtfracf/+f/+ corrP i.e. the effect is quite small and can usually be neglected. abcliste
We consider a mathematical pulum as a special case of a physical pulum. abcliste abc Derive the formal expression for the period of a mathematical pulum from the more general expression for a physical pulum. abc Calculate the length of a mathematical pulum with a period of TO. abc Derive an expression for the period of a mathematical pulum where the mass m_s of the string is a fraction of the pulum mass m: m_sf m. abcliste
Solution:
abcliste abc The pulum mass of a mathematical pulum is idealised to be a po mass. We can think of it as a spherical mass m with radius r at a distance sell from the pivot po. Using the expression for the period of a physical pulum we find T & approx pisqrtfracJ_s+ms^mgs pisqrtfracfracmr^+mell^mgell pisqrtfrac+mell^mgell pisqrtfracellg labelperiod This is the well-known expression for the period of a mathematical pulum. abc Solving refperiod for the pulum length we find ell LF fracncgtimesT^pi^ L approx resultLP abc The moment of inertia is the of the two partial moments: J sscJstring+sscJmass We idealise the string as a thin rod. It oscillates about one of the string i.e. we have to shift the pivot po from the center by ell/ Steiner's theorem: sscJstring fracm_s ell^+m_s ell/^ fracm_s ell^ The total moment of inertia is J fracm_sell^+mell^ leftm_s/+mrightell^ The torque on the pulum can also be expressed as the of the torques acting on the string and on the mass: tau ssctaustring+ssctaumass -m_s gfracellsinphi -m gellsinphi & approx -m_s/+mgellphi The approximation in the last step is valid for small angles phill . Using the basic for rotational motion we find Jddotphit & taut m_s/+mell^ddotphit &approx -m_s/+mgellphit Solving for the angular acceleration leads to the characteristic of the simple harmonic motion: ddotphit &approx -fracm_s/+mm_s/+mfracgellphit The angular frequency is omega sqrtfracm_s/+mm_s/+mfracgell and the period is T &approx fracpiomega pisqrtfracellgsqrtfracf/+f/+ T_sqrtfracf/+f/+ where T_ is the period of a normal mathematical pulum. As an example we can consider the case where the mass of the string is fO of the pulum mass. In this case the correction factor is corrF sqrtfracf/+f/+ corrP i.e. the effect is quite small and can usually be neglected. abcliste
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