Drive Phonograph Turntable
February 25th, 2010
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Drive Phonograph Turntable
A 35-cm-diameter phonograph record is dropped onto a turntable being driven at 33 1/3...?
A 35-cm-diameter phonograph record is dropped onto a turntable being driven at 33 1/3.
If the coefficient of friction between the record and turntable is 0.20, how far will the turntable rotate between the time when the record first contacts it and when the record is rotating at the full 33 1/3 Assume that the record is a homogeneous disk. (do an integral to calculate the torque.)
couldnt figure it out
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Kenwood KD-600 Turntable TT Phonograph Direct Drive |  |
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Pyle PNGTT1B Black Horn PC-Recordable Phonograph/ Turntable $220.89 This turntable echoes the classic styling of Edison's original phonograph with today's silicon-based technology. Play your favorite 45 and 33 RPM records and listen to your music through the horn or use the free Audacity software to transfer your music to your computer.Classic turntable with PC recording functionUSB connection to enable recording to pc or macFiles can be saved in MP3, WAV, OGG, WMA, and more formatsAuto returnAuto stopRCA stereo audio outRCA stereo audio inCeramic cartridgeBelt driveAC power: 120 v 60 hzTwo (2) mid-range/bass speakers, one (1) treble speaker for full soundColor: black |
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Shure M44G Phonograph Product $90 Shure M44G Phonograph Product Re-created to the original M44 specs, this companion DJ cartridge to the Shure M44-7 provides an ideal balance between skip resistance and accuracy of sound reproduction for the DJ. The turntable needle is perfect for both intensive scratching and mixing. Features: Skip resistance: very high Output: high Record wear: very low Sound emphasis: detailed bass and lows Get your Shure M44G Phonograph Product Today! |
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Christmas 78's - Yingle Bells - Yogi Yorgesson
Challenging Mechanic Physics problem HELP!? please?
A 35-cm-diameter phonograph record is dropped onto a turntable being driven at 33⅓ rpm.
If the coefficient of friction between the record and turntable is 0.20, how far will the turntable rotate between the time when the record first contacts it and when the record is rotating at the full 33⅓ rpm Assume that the record is a homogeneous disk.
calculate the torque?
Good problem. I might have a solution for you
Alright, a few concepts to start with.
Sum of the torques = I alpha
where I is the moment of intertia, and alpha is the angular acceleration. ( I = 1/2 m R^2 for a disc)
Sum of the forces = m a
and Torque is r cross F
Alright, so the tricky part about this problem is that our object is drastically NOT a point mass. This throws out a lot of the beautiful simplifications we love to make, but it's still doable.
If you imagine the disc as a combination of rings rather than a whole, the problem begins to reveal itself. If we sum up the forces on one of these differential rings, we can get the differential forces.
dFx= dfrictionk
dFy= dN- dW
(dN is the differential normal force, dW is the differential weight force.)
So, now we've got some place to start. Since it's kinetic friction, we can rewrite it as uk*N, or in this case, uk*dN. Here, uk is the coefficient of friction. Also, we know that the record is not accelerating in the y direction, so dN= dW. But, let's rewrite dW as g*dm.
dm is the differential mass and g is gravity.
Bringing this all together,
dFriction= uk *g*dm
K, so we've got the forces taken care of so let's turn back to the torques.
Right now, we've still got a differential force, so we've also got a differential torque. But, after we integrate it, it'll still be equal to I alpha.
dT= r cross dF
We can gloss over the cross product because the two vectors should be perpendicular, so it simplifies to
dT= r*dF
dT= uk*g*r*dm
Now it's time to address the dm.
Since I assumed that we could break up the record into a bunch of differential rings, we can represent the mass as a density multiplied by an area.
dm= sigma* dA
where sigma is m/(pi *R^2)
and the differential area of the ring dA is 2 pi r dr (Circumference multiplied by dr)
Watch out for the r's. The big R is what I'm calling the total radius of the record. Little r is a variable.
K. Bring it together. I'm leaving sigma as sigma for now.
dT= uk*g*sigma*2 pi *r^2 dr
This is a pretty easy integral. Everything is constant but r, so integrate over the ring. (Bounds are from zero to R). I'm going to gloss over the math here.
T= 2 pi R^3 uk sigma g /3
If we return sigma as the mass area density, (Check the algebra)
T= 2 uk g m R/3
The end is in sight! Set this equal to I alpha. We can even sub in the inertia to simplify it. (I = 1/2 m R^2) Solve for alpha. Again, going to gloss over the algebra.
alpha = 4 uk g /(3 R)
Now we can integrate and get the angular velocity! (I'm calling omega angular velocity)
omega= Integral of alpha with respect to time
Since all that stuff is constant in time, it's a simple integral.
omega = 4 uk g/(3 R) *t
Sweet. Now, to find the time needed, set it equal to the final velocity, and solve for time. Be sure to change that 33 1/2 rpm into radians per second. I think it works out to 3.5 rad/s
tfinal= 3 omegafinal*R/(4 uk g)
Plugging in the numbers I get something about half a second.
I know this was pretty lengthy write up, but it's an involved problem. Hopefully having it laid out step by step gives some sense to the problem.
Goodluck!
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