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a_lawson_2k
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I'm brushing up on the physics, and am teaching myself from University Physics w/ Modern Physics 10th Edition (I got 11th off eBay with the solution manual, it should come in about a week). I haven't had too much trouble with the problems up until the section about the dynamics of rotational motion, either way, I got through the exercises (easy problems), but when I get to the regular problems, things start getting tricky. What may be happening here is that I didn't quite pick up on how to set up the problem over the two days I've worked on the chapter. I looked back over it, and it didn't appear I had missed anything, hopefully seeing what I'm doing wrong with these problems may shed light on the situation.
First problem:
I drew the free body diagram, and applied Newton's Second to the (adjusted) x and y axes (offset 36,9 degrees).
[tex]\Sigma F_y = n-mg cos(\theta) = 0[/tex]
[tex]\Sigma F_x = mg sin(\theta)-F_f-T = m*a_y[/tex]
The natural force became mg cos (36.9*) which worked out to be 39,2N
mg sin (36.9*) came out to be 29.4N, so the sum of the forces in the Y direction came out to:
[tex]\Sigma F_x = 19.6-T = m*a_y[/tex]
Here is what really started puzzling me, how do I get the flywheel to come into the equation? It doesn't specify a radius from which I can evaluate the torque, and from there the angular acceleration, so I'm stuck.
Second problem:
I don't know where to start here, sadly. I have no clue how to even approach it, the book didn't provide an example even remotely like this one. I could probably set it up in energy terms and using the sum of the forces in the radial direction as being (mv^2)/r, but I'd have no idea what to do from there. Forgive me if it looks like I'm not trying, it's just I'm not sure how to reconcile the stuff in the problem.
This one I have more of an idea of how to solve:
for the block: [tex]\Sigma F_y = M*a = M*g-T_1[/tex]
for the flywheel: [tex]\Sigma \tau = I*\alpha = (1/2*M*R^2)(a/R)=1/2*M*R*a = (T_2-T_1)*R[/tex]
For the cylinder with the yoke: [tex]\Sigma F_x = T_2 = M*a, a=(2R)\alpha[/tex]
Here is where I'm stuck:
Again for the cylinder with the yoke: [tex]\Sigma \tau = (1/2*M*(2R)^2)(a/(2R)) = a*M*R = ? [/tex]
I didn't think there would be torque in the system since the axle acted on the center of mass, thus torque would be zero, but otherwise, I can't solve this since there are three unknowns and only two equations.
Thanks in advance for any and all help.
First problem:
10-55 said:A block with mass m=5 kg slides down a surface inclined at 36,9 degrees to the horizontal. The coefficient of kinetic friction is 0.25. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 25 kg and moment of inertia of 0.5 kg-m^2 w/ respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0,2 m from the axis.
a) what is the acceleration of of the block down the plane?
b) what is the tension in the string?
I drew the free body diagram, and applied Newton's Second to the (adjusted) x and y axes (offset 36,9 degrees).
[tex]\Sigma F_y = n-mg cos(\theta) = 0[/tex]
[tex]\Sigma F_x = mg sin(\theta)-F_f-T = m*a_y[/tex]
The natural force became mg cos (36.9*) which worked out to be 39,2N
mg sin (36.9*) came out to be 29.4N, so the sum of the forces in the Y direction came out to:
[tex]\Sigma F_x = 19.6-T = m*a_y[/tex]
Here is what really started puzzling me, how do I get the flywheel to come into the equation? It doesn't specify a radius from which I can evaluate the torque, and from there the angular acceleration, so I'm stuck.
Second problem:
10-61 said:A uniform marble, radius r, starts from rest with its center of mass at heigh h above the lowest point of a loop-the-loop track of radius R, a track shaped like the one in figure 7-26 (http://img144.imageshack.us/img144/7875/untitledjj5.gif" ). It rolls without slipping. Rolling friction and air resistance are negligible.
a) what is the minimum value of h such that the marble will not leave the track at the top of the loop?
--NOTE: radius r is NOT negligible compared to radius R
b) what is the answer to part (a) if the track if lubricated making friction negligible?
I don't know where to start here, sadly. I have no clue how to even approach it, the book didn't provide an example even remotely like this one. I could probably set it up in energy terms and using the sum of the forces in the radial direction as being (mv^2)/r, but I'd have no idea what to do from there. Forgive me if it looks like I'm not trying, it's just I'm not sure how to reconcile the stuff in the problem.
This one I have more of an idea of how to solve:
10-69 said:Uniform solid cylinder, mass M and radius 2R rests on horizontal table top. A string is attached by a yoke to a frictionless axle through the center so the cylinder can rotate about the axle. The string runs over a disk-shaped pulley, mass M, radius R, mounted on a frictionless axle through its center. A block of mass M is suspended from the free end of the string. The string doesn't slip over the pulley surface and the cylinder rolls without slipping on the table top.
a) find the acceleration of the block after the system is released from rest
for the block: [tex]\Sigma F_y = M*a = M*g-T_1[/tex]
for the flywheel: [tex]\Sigma \tau = I*\alpha = (1/2*M*R^2)(a/R)=1/2*M*R*a = (T_2-T_1)*R[/tex]
For the cylinder with the yoke: [tex]\Sigma F_x = T_2 = M*a, a=(2R)\alpha[/tex]
Here is where I'm stuck:
Again for the cylinder with the yoke: [tex]\Sigma \tau = (1/2*M*(2R)^2)(a/(2R)) = a*M*R = ? [/tex]
I didn't think there would be torque in the system since the axle acted on the center of mass, thus torque would be zero, but otherwise, I can't solve this since there are three unknowns and only two equations.
Thanks in advance for any and all help.
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