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cdingdong
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Homework Statement
A rectangular loop of sides a = 0.3 cm and b = 0.8 cm pivots without friction about a fixed axis (z-axis) that coincides with its left end (see figure). The net current in the loop is I = 3.2 A. A spatially uniform magnetic field with B = 0.005 T points in the +y-direction. The loop initially makes an angle q = 35° with respect to the x-z plane.
https://tycho-s.phys.washington.edu/cgi/courses/shell/common/showme.pl?courses/phys122/autumn08/homework/09/rectangular_loop_MFR/p8.gif
The moment of inertia of the loop about its axis of rotation (left end) is J = 2.9 × 10-6 kg m2. If the loop is released from rest at q = 35°, calculate its angular velocity [tex]\omega[/tex] at q = 0°.
Homework Equations
this is kinetic energy:
KE = 1/2I[tex]\omega^{2}[/tex] but moment of inertia is specified to be J, so
KE = 1/2J[tex]\omega^{2}[/tex]
this is potential energy for a loop with current and magnetic field acting on it:
PE = [tex]\mu[/tex]Bcos[tex]\theta[/tex]
moment vector [tex]\mu[/tex] = NIA where N = number of loops, I = current, A = area, B = magnetic field
so, PE = NIABcos[tex]\theta[/tex]
The Attempt at a Solution
I recognize that the angular kinetic energy equation can be used to find the angular velocity. KE = 1/2J[tex]\omega^{2}[/tex].
Final kinetic energy = initial potential energy. Initial potential energy = NIABcos[tex]\theta[/tex]. so, we could say
1/2J[tex]\omega^{2}[/tex] = NIABcos[tex]\theta[/tex]
the right side is potential energy. that equals
N = 1 turn; I = 3.2 Amps; A = 0.003*0.008 = 2.5e-5 meters; B = 0.005 Teslas; [tex]\theta[/tex] = 35 degrees
(1)(3.2)(2.5e-5)(0.005)cos(35) = 3.276608177 * 10^-7 Joules. we'll call this 3.2766e-7
so, now we haveN = 1 turn; I = 3.2 Amps; A = 0.003*0.008 = 2.5e-5 meters; B = 0.005 Teslas; [tex]\theta[/tex] = 35 degrees
(1)(3.2)(2.5e-5)(0.005)cos(35) = 3.276608177 * 10^-7 Joules. we'll call this 3.2766e-7
1/2J[tex]\omega^{2}[/tex] = 3.2766e-7
J = 2.9 × 10-6 kilogram meters squared. we'll call this 2.9e-6
1/2(2.9e-6)[tex]\omega^{2}[/tex] = 3.2766e-7
[tex]\omega[/tex] = [tex]\sqrt{(2*3.2766e-7)/2.9e-6}[/tex] = 4.753655581 * 10^-1
BUT, the answer happens to be what I did except that they did not take the square root at the end.
so, they got 2.259724138 * 10^-1.
did i do something wrong? what i did makes sense in my mind. is their answer wrong? why did they not not take a square root at the end? is kinetic energy not = 1/2J[tex]\omega^{2}[/tex], but instead 1/2J[tex]\omega[/tex] without the square?
i thank everybody in advance!
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