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I have been working on this problem for three days! Someone please help me. I am using a physics book, but it seems so unrepresentative of the question... In the book, either the velocity or the angle is always provided.
A particle P travels with constant speed on a circle of radius r = 4.70 m (Figure 4-56) and completes one revolution in 20.0 s. The particle passes through O at time t = 0.
At t = 5.00 s, what is the particle's position vector? Give (a) magnitude and (b) direction (as an angle relative to the positive direction of x.)
At t = 7.50 s, what is the particle's position vector? Give (c) magnitude and (d) direction (as an angle relative to the positive direction of x.)
At t = 10.00 s, what is the particle's position vector? Give (e) magnitude and (f) direction (as an angle relative to the positive direction of x.)
Link to picture: http://gold3nlily.shutterfly.com/pictures/9
If you cannot see the picture, the bottom-center of the circle is on the origin (O). And the particle is moving counter clockwise.
Okay, i have tried this problem a couple of different ways... But this is what I am doing now:
1) V= 1rev/20sec (2pi x r/rev) = 1.4766m/s
2) a=v^2/r
a=1.46766^2/4.7 -> 0.463871
(I'm confused about the units here: if it is m/s then divided be m, wouldn't the answer be in seconds b/c the meters cancels? But this cannot be becasue this is the equation in my book)
This brings me to my next question:
If the particle travels with "constant speed", the velocity found above would not change, and the radius doesn't change, then the acceleration doesn't change... so does this mean that the magnitude is the same all over the circle and (a) (c) and (e) are all the same answer (0.464)?
Finally, the angles.
I thought that since one rev = 20s, then in 5s the particle would be 1/4 the way round and the angle would be 0 degrees (not so...). At 10s I thought it would be 1/2 way round the circle so 90 degree angle (this one was actually right). At 7.5 seconds i thought the particle would be 1/2 way btwn 90 degrees and 0 degrees so 45 degrees (also wrong). How should I approach this differently?
A particle P travels with constant speed on a circle of radius r = 4.70 m (Figure 4-56) and completes one revolution in 20.0 s. The particle passes through O at time t = 0.
At t = 5.00 s, what is the particle's position vector? Give (a) magnitude and (b) direction (as an angle relative to the positive direction of x.)
At t = 7.50 s, what is the particle's position vector? Give (c) magnitude and (d) direction (as an angle relative to the positive direction of x.)
At t = 10.00 s, what is the particle's position vector? Give (e) magnitude and (f) direction (as an angle relative to the positive direction of x.)
Link to picture: http://gold3nlily.shutterfly.com/pictures/9
If you cannot see the picture, the bottom-center of the circle is on the origin (O). And the particle is moving counter clockwise.
Okay, i have tried this problem a couple of different ways... But this is what I am doing now:
1) V= 1rev/20sec (2pi x r/rev) = 1.4766m/s
2) a=v^2/r
a=1.46766^2/4.7 -> 0.463871
(I'm confused about the units here: if it is m/s then divided be m, wouldn't the answer be in seconds b/c the meters cancels? But this cannot be becasue this is the equation in my book)
This brings me to my next question:
If the particle travels with "constant speed", the velocity found above would not change, and the radius doesn't change, then the acceleration doesn't change... so does this mean that the magnitude is the same all over the circle and (a) (c) and (e) are all the same answer (0.464)?
Finally, the angles.
I thought that since one rev = 20s, then in 5s the particle would be 1/4 the way round and the angle would be 0 degrees (not so...). At 10s I thought it would be 1/2 way round the circle so 90 degree angle (this one was actually right). At 7.5 seconds i thought the particle would be 1/2 way btwn 90 degrees and 0 degrees so 45 degrees (also wrong). How should I approach this differently?
