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Rotational equilibrium, subtending, and rotational acceleration

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jakeowens
#1
Oct17-05, 07:24 PM
P: 34
The lightweight pivoted bar in Fig. MC8 will be in rotational equilibrium when a 200 N force acts
none of these
up at C only
down at E or up at C
up at B
down at D

(see attatched picture)

So the net force has to be 0, i wasnt exactly sure how to do this one because i didnt know the lengths of anything. But i just assumed the distance at point A to be 1, and point G to be 1. so B would be 2/3, C 1/3, D 0, E 1/3, and F 2/3. So after that i multiplied the forces by their respective distances, and came up with the force on the right hand side being 66.66666 N more thaen the left side, in a upward direction. So, if you had 200N force you could put it either up at C or down at E to balance that out, since 1/3 of 200N is 66.6666N, right?

The 10-point center-circle bull's eye on the official slow fire pistol target is 7/8 in. in diameter. What angle does it subtend as seen by a shooter 55 ft away?
radians?
degrees?


I have no idea how to do this problem. We've done nothing like this in class, and the book is really no help. I've tried reading the whole chapter and theres nothing that i found to help me. I just need to get pointed in the right direction. Is it just asking what the measure of the angle is if it's a big circle with a radius of 55ft, and it wants to know the measure of that 7/8 of an inch?

So on that tangent, i found the circumference of the circle to be 345.575 ft, and then divided the length of the bullseye by that to find out what percentage of the whole circle that bullseye is. 0.0729ft/345.575ft=2.11e-4%. so after that, i just multiplied that % by 360 degrees, giving me 0.0759 degrees, and also multiplied that % by 2pi, to give me 0.00132 radians.

Is that what the question is asking? did i screw up on my answer in radians? im not sure if my radians answer is correct.


36 seconds after the start button is pressed on a big electric motor, its shaft is whirling around at 480 rev/s. Determine its average angular acceleration.
Answer: rad/s2.


For this one, I used the equation alpha=change in velocity / change in time.

So, average angular acceleration = 480*2pi radians / 36 seconds right? this gives me an answer of 83.78 radians/s. Am i right here?
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whozum
#2
Oct17-05, 09:56 PM
P: 2,218
The lightweight pivoted bar in Fig. MC8 will be in rotational equilibrium when a 200 N force acts
none of these
up at C only
down at E or up at C
up at B
down at D
(see attatched picture)
I would treat each letter as one unit away from each other letter, so distance between A and B is 1, A and C is 2, etc. From there I would just list the CCW torques and their distances, and then the clockwise torques and their distances, and do the same for the CW distances.

The 10-point center-circle bull's eye on the official slow fire pistol target is 7/8 in. in diameter. What angle does it subtend as seen by a shooter 55 ft away?
It's asking you that if your eye was a point 55ft away from that, then how big of an angle would your eye sweep to see all of it? Its a really really small angle. Pretty much you are 55 ft away from a 7/8 in. object, how big is the angle made, just draw a triangle.

For this one, I used the equation alpha=change in velocity / change in time.
Thats the correct way to do it, but rev/s is not an angualr velocity but a frequency. Convert it to a velocity by converting revolutions into radians, and you should be good.


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