Hi I have a problem here and I have no idea how to tackle it:
A uniform disc of radius R and mass M is attached at its center to one end of a massless,rigid rod of length L.The other end of the rod is attached at the ceiling and pivots freely about that point.
The system makes a physical pendulum. What is the period of small oscillations of the pendulum?
Can someone help me?
B
QUOTE
A uniform disc of radius R and mass M is attached at its center to one end of a massless,rigid rod of length L.The other end of the rod is attached at the ceiling and pivots freely about that point.
The system makes a physical pendulum. What is the period of small oscillations of the pendulum?
The system makes a physical pendulum. What is the period of small oscillations of the pendulum?
Radius = R
Mass = M
L = ridig rod
This is a very simple question if you understand the oscillation:
Oscillation is any motion that has a repetative sinosidal fucntion.
we know that in oscilation the T = 2л √ m / k
k is a constant and should be derived for each system independently.
I am not sure whe he wants in terms of oscillation but this looks like a special form of a simple pendulum so
we have
T = T = 2л √ m / mg/L = T = 2л √ L / g
so we have:
T = 2л √ L / g
Maybe he wants to test you to see if you understand whether or not mass and shape of the pendulum is irrelevant ..i dont konw. this seems to be a pretty straight forward case.
Thank you for your help.
Frankly ,I was searching for something more complicated!! but as you said it is very simple.
B
Frankly ,I was searching for something more complicated!! but as you said it is very simple.
B
The asnwer supplied was for a pendulum. Not a physical pendulum. (a pendulum is where all the mass is considered to be a point mass). A physical pendulum is quite different. T= 2pi * root( I / MgR)
QUOTE
The asnwer supplied was for a pendulum. Not a physical pendulum. (a pendulum is where all the mass is considered to be a point mass). A physical pendulum is quite different. T= 2pi * root( I / MgR)
yes but every mass can be thought of as a point mass in the CM so if the rod for instance is uniform in density then you can assume that the CM is d/2 down the length and we call d/2=L.
T = 2pi * root ( I /MgR)
so if we have:
I = 1/3ML^2 (moment of inertia for a uniform rod rotating around a pivot like a door)
and that gives:
I/MgR = 1/3 ML^2 / MgL = 1/3 L / g = 1/3 (L/g)
so then we have:
T(physical) = 2pi * root ( I /MgR) = 2 pi * root ( 1/3L/g) = .58 *2pi * root(L/g) = .58* T(Normal)
so T(Physical pendulum) = .6 T(Normal)
since this is a "Small oscilation" we can for sake of simplicity ignore the .58, unless of course if the oscilation is large or if the gravity is large.
QUOTE (Drude+Nov 30 2005, 07:23 AM)
QUOTE
A uniform disc of radius R and mass M is attached at its center to one end of a massless,rigid rod of length L.The other end of the rod is attached at the ceiling and pivots freely about that point.
The system makes a physical pendulum. What is the period of small oscillations of the pendulum?
The system makes a physical pendulum. What is the period of small oscillations of the pendulum?
Radius = R
Mass = M
L = ridig rod
This is a very simple question if you understand the oscillation:
Oscillation is any motion that has a repetative sinosidal fucntion.
we know that in oscilation the T = 2ë √ m / k
k is a constant and should be derived for each system independently.
I am not sure whe he wants in terms of oscillation but this looks like a special form of a simple pendulum so
we have
T = T = 2ë √ m / mg/L = T = 2ë √ L / g
so we have:
T = 2ë √ L / g
Maybe he wants to test you to see if you understand whether or not mass and shape of the pendulum is irrelevant ..i dont konw. this seems to be a pretty straight forward case.
very interesting, i agree with your help.
thanks, your hint saves some our of my weekend
seekXL
I had the same problem, but for longer look in the InterNet I landed and the problem here is solved, seek believed I you also, many thanks to drude
Thanks for this informations, very interstings.
you are very welcomed.
i just wanted to thank you for this topic. i just found the information i needed for an essay i had to write.
Great Site. Good topics.
Further so.
With friendly regards
SMSER
*****************************************************************
SMSChat Kontaktanzeigen SMS Chat Partnersuche
Further so.
With friendly regards
SMSER
*****************************************************************
SMSChat Kontaktanzeigen SMS Chat Partnersuche
QUOTE (Drude+Nov 30 2005, 07:23 AM)
This is a very simple question if you understand the oscillation:
Oscillation is any motion that has a repetative sinosidal fucntion.
we know that in oscilation the T = 2л √ m / k
k is a constant and should be derived for each system independently.
I am not sure whe he wants in terms of oscillation but this looks like a special form of a simple pendulum so
we have
T = T = 2л √ m / mg/L = T = 2л √ L / g
so we have:
T = 2л √ L / g
Maybe he wants to test you to see if you understand whether or not mass and shape of the pendulum is irrelevant ..i dont konw. this seems to be a pretty straight forward case.
sometimes we just think to difficult and don't see the easy way that is so close to us.
Oscillation is any motion that has a repetative sinosidal fucntion.
we know that in oscilation the T = 2л √ m / k
k is a constant and should be derived for each system independently.
I am not sure whe he wants in terms of oscillation but this looks like a special form of a simple pendulum so
we have
T = T = 2л √ m / mg/L = T = 2л √ L / g
so we have:
T = 2л √ L / g
Maybe he wants to test you to see if you understand whether or not mass and shape of the pendulum is irrelevant ..i dont konw. this seems to be a pretty straight forward case.
sometimes we just think to difficult and don't see the easy way that is so close to us.
Since people keep finding and resurrecting this thread, I've got to post here to straighten out the information. The ONLY correct statement in this thread is the one from Guest_Mike, who observed that this is a PHYSICAL pendulum, not a SIMPLE pendulum.
In a simple pendulum, all the mass is concentrated at a point at the bottom of a massless rod or string. The only motion is the motion of this mass point along a small circular arc. For a simple pendulum, the period is 2*pi*sqrt(L/g), where L is the length of the string.
In case you want to know, this formula is derived by looking at the force pulling the pendulum back toward the low point, and comparing it with the formula for a mass on a spring. For a simple pendulum doing small oscillations, this force turns out to be approximately -mgx/L (you can find this from the free body diagram if you want), so the pendulum acts like a spring with a -kx = -mgx/L, so k = mg/L. Since the period of a mass and spring is 2*pi*sqrt(m/k), you just plug in the new formula for k, and you get the simple pendulum period.
In a physical pendulum, the mass is not concentrated at a point, and the mass is forced to rotate as the pendulum swings. In the present problem, the disk does not remain horizontal, but tips as the pendulum swings. This is an additional motion on top of the swinging of the center of mass point. Since you have rotational motion, you CANNOT just use force to analyze the situation, you must use torque. In this situation, the formula for the period is 2*pi*sqrt(I/mgR).
In case you want to know, this formula is derived by looking at the torque twisting the pendulum back to the vertical orientation, and once again comparing it with the formula for a mass on a spring. For a physical pendulum doing small oscillations, this torque turns out to be mgL*theta, where L is the distance form the point where the pendulum hangs to the center of mass. For rotation, you use torque = I*(angular acceleration) in place of F=ma, so you replace F with torque, m with I, and x with theta in the formula, to get period = 2*pi*sqrt(m/k) = 2*pi*sqrt(m/(F/x)) ---> 2*pi*sqrt(I/(torque/theta)) = 2*pi*sqrt(I/(mgL*theta/theta)) = 2*pi*sqrt(I/(mgL)).
For this problem, the rod is massless, and the mass is a disk of radius R. A disk has I = 1/4*mR^2 when it is rotated around an axis across its middle. (Most people might remember the formula I = 1/2*mR^2 for a disk, but that is only for an axis perpendicular to the disk and through its center, like the axle of a wheel. This axis is across the diameter of the disk, so the rotation motion is like flipping a pancake.) The axis the whole pendulum is turning around is at the point where the massless rod attaches to the ceiling, so the moment of inertial must be calculated around this axis. The parallel axis theorem says that if you know I on an axis through the center of mass, you can find I around a different axis by adding md^2, where d is the distance you moved the axis. In this case, you need to move the axis from the center of the disk up to the point the pendulum hangs from, which is a distance L. Therefore, the true moment of inertia is I = 1/4mR^2 + mL^2.
Now you can plug this into the formula for period. Period = 2*pi*sqrt((1/4mR^2 + mL^2)/mgL) = 2*pi*sqrt(1/4*(R/L)^2*L/g + L/g) = 2*pi*sqrt(L/g)*sqrt(1+(R/4L)^2). So this pendulum period is longer than for a simple pendulum by a factor of sqrt(1+(R/4L)^2).
Drude has said in this thread that the factor was 0.56, but that was for a different pendulum, with the mass in the rod, not in the disk. The factor for the actual pendulum in the given problem is > 1. Also, Drude mentioned that the 0.56 was negligible for small amplitudes. This is not true because the small amplitude was already taken into account earlier. The 0.56 does not go away. Also, it cannot be considered "small" because it is a factor, not a term being added to the period. It means that the pendulum in Drude's example swings almost twice as fast as a simple pendulum. Ours swings slower. Finally, Drude mentioned that mass can all be considered as if it were collected together at the center of mass point. This is true as far as forces are concerned, but it is NOT true for torques, and we must use torques to analyze the physical pendulum. That is why it is not safe to put the mass at a point. That would give the simple pendulum answer, which is wrong for a physical pendulum.
Hope this clears things up!
--Stuart Anderson
In a simple pendulum, all the mass is concentrated at a point at the bottom of a massless rod or string. The only motion is the motion of this mass point along a small circular arc. For a simple pendulum, the period is 2*pi*sqrt(L/g), where L is the length of the string.
In case you want to know, this formula is derived by looking at the force pulling the pendulum back toward the low point, and comparing it with the formula for a mass on a spring. For a simple pendulum doing small oscillations, this force turns out to be approximately -mgx/L (you can find this from the free body diagram if you want), so the pendulum acts like a spring with a -kx = -mgx/L, so k = mg/L. Since the period of a mass and spring is 2*pi*sqrt(m/k), you just plug in the new formula for k, and you get the simple pendulum period.
In a physical pendulum, the mass is not concentrated at a point, and the mass is forced to rotate as the pendulum swings. In the present problem, the disk does not remain horizontal, but tips as the pendulum swings. This is an additional motion on top of the swinging of the center of mass point. Since you have rotational motion, you CANNOT just use force to analyze the situation, you must use torque. In this situation, the formula for the period is 2*pi*sqrt(I/mgR).
In case you want to know, this formula is derived by looking at the torque twisting the pendulum back to the vertical orientation, and once again comparing it with the formula for a mass on a spring. For a physical pendulum doing small oscillations, this torque turns out to be mgL*theta, where L is the distance form the point where the pendulum hangs to the center of mass. For rotation, you use torque = I*(angular acceleration) in place of F=ma, so you replace F with torque, m with I, and x with theta in the formula, to get period = 2*pi*sqrt(m/k) = 2*pi*sqrt(m/(F/x)) ---> 2*pi*sqrt(I/(torque/theta)) = 2*pi*sqrt(I/(mgL*theta/theta)) = 2*pi*sqrt(I/(mgL)).
For this problem, the rod is massless, and the mass is a disk of radius R. A disk has I = 1/4*mR^2 when it is rotated around an axis across its middle. (Most people might remember the formula I = 1/2*mR^2 for a disk, but that is only for an axis perpendicular to the disk and through its center, like the axle of a wheel. This axis is across the diameter of the disk, so the rotation motion is like flipping a pancake.) The axis the whole pendulum is turning around is at the point where the massless rod attaches to the ceiling, so the moment of inertial must be calculated around this axis. The parallel axis theorem says that if you know I on an axis through the center of mass, you can find I around a different axis by adding md^2, where d is the distance you moved the axis. In this case, you need to move the axis from the center of the disk up to the point the pendulum hangs from, which is a distance L. Therefore, the true moment of inertia is I = 1/4mR^2 + mL^2.
Now you can plug this into the formula for period. Period = 2*pi*sqrt((1/4mR^2 + mL^2)/mgL) = 2*pi*sqrt(1/4*(R/L)^2*L/g + L/g) = 2*pi*sqrt(L/g)*sqrt(1+(R/4L)^2). So this pendulum period is longer than for a simple pendulum by a factor of sqrt(1+(R/4L)^2).
Drude has said in this thread that the factor was 0.56, but that was for a different pendulum, with the mass in the rod, not in the disk. The factor for the actual pendulum in the given problem is > 1. Also, Drude mentioned that the 0.56 was negligible for small amplitudes. This is not true because the small amplitude was already taken into account earlier. The 0.56 does not go away. Also, it cannot be considered "small" because it is a factor, not a term being added to the period. It means that the pendulum in Drude's example swings almost twice as fast as a simple pendulum. Ours swings slower. Finally, Drude mentioned that mass can all be considered as if it were collected together at the center of mass point. This is true as far as forces are concerned, but it is NOT true for torques, and we must use torques to analyze the physical pendulum. That is why it is not safe to put the mass at a point. That would give the simple pendulum answer, which is wrong for a physical pendulum.
Hope this clears things up!
--Stuart Anderson
QUOTE (911turbo+Oct 7 2006, 04:19 PM)
Thanks for this informations, very interstings
i bookmark it
i bookmark it
Hello together,
it´s a really difficult question - there i can do nothing because this is to high for me.
Can i use this for my cooking recipes?
Well now i don´t think so.
best regards
tom the chef
p.s. thank you for this great site and info's
it´s a really difficult question - there i can do nothing because this is to high for me.
Can i use this for my cooking recipes?
Well now i don´t think so.
best regards
tom the chef
p.s. thank you for this great site and info's
Hi I have a problem here and I have no idea how to tackle it:
A uniform disc of radius R and mass M is attached at its center to one end of a massless,rigid rod of length L.The other end of the rod is attached at the ceiling and pivots freely about that point.
The system makes a physical pendulum. What is the period of small oscillations of the pendulum?
Can someone help me?
A uniform disc of radius R and mass M is attached at its center to one end of a massless,rigid rod of length L.The other end of the rod is attached at the ceiling and pivots freely about that point.
The system makes a physical pendulum. What is the period of small oscillations of the pendulum?
Can someone help me?
Why are you posting a question that is word-for-work identical to the question that started this thread? This question has already been answered in the thread half a year ago, so all you have to do is read the answer:
If you have some numerical data, just plug your data into the very last formula on the right. If you just want the formula, that last one is your answer.
Hope this helps!
--Stuart Anderson
QUOTE
Now you can plug this into the formula for period. Period = 2*pi*sqrt((1/4mR^2 + mL^2)/mgL) = 2*pi*sqrt(1/4*(R/L)^2*L/g + L/g) = 2*pi*sqrt(L/g)*sqrt(1+(R/4L)^2).
If you have some numerical data, just plug your data into the very last formula on the right. If you just want the formula, that last one is your answer.
Hope this helps!
--Stuart Anderson
oscilation was always tricky for me 
PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.
To quit out of "lo-fi" mode and return to the regular forums, please click here.