Need help with motion question
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http://i44.tinypic.com/xle6a8.jpg any ideas? cant seem to see pics in this window*?
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wo ai mei mei, mei mei ai wo! ha ha ha
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Wo ai tai mei, tai mei ai wo.
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See 2 pervs above me :X
Want to chio gals go sex forum la XD
(a) It is given that x=200cos(0.3t) and y=x(squared)/200, whereas the time is 6s. Hence to find the position, u just sub t=6 in and find x, and sub the x-value to find y.
For velocity, V=dx/dt= -(200)(0.3)sin(0.3t), and acceleration,
A=dv/dt= -(200)(0.3)(0.3)cos(0.3t), so you just need to sub in t=6 to get the answer.
For (b) and (c).....I haven't touch physics in quite a while wor, so very rusty liao, think need to get someone else to help you.
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Me in office, all photo sharing sites are blocked :(
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Originally posted by eagle:
Me in office, all photo sharing sites are blocked :(
The Question is:
A guide with a vertical slot is given a horizontal motion of x=200cos(0.3t), where x is in mm, t is in seconds and (0.3t) in radians. The guide controls the motion of pin P, which is also constrained to move along a fixed parabolic slot as shown below. The parabolic slot has a shape given by the equation y=x(squared)/200, where y is in mm.
For the instant where t=6.0seconds,
(a) Determine the position, velocity and acceleration of P in X-Y components.
(b) Determine the rate at which the speed of P is changing.
(c) Determine the instantaneous radius of curvature of the path.
Forget the diagram, with your level of 'awesomeness', these should be enough. : D
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Originally posted by Forbiddensinner:
The Question is:
A guide with a vertical slot is given a horizontal motion of x=200cos(0.3t), where x is in mm, t is in seconds and (0.3t) in radians. The guide controls the motion of pin P, which is also constrained to move along a fixed parabolic slot as shown below. The parabolic slot has a shape given by the equation y=x(squared)/200, where y is in mm.
For the instant where t=6.0seconds,
(a) Determine the position, velocity and acceleration of P in X-Y components.
(b) Determine the rate at which the speed of P is changing.
(c) Determine the instantaneous radius of curvature of the path.
Forget the diagram, with your level of 'awesomeness', these should be enough. : D
a) y=(x^2)/200 ---> Sub in x = 200cos(0.3t) ---> y= 200[cos(0.3t) ]^2
Therefore position of P, (x,y) given by x=200 cos(0.3t) , y=200[cos(0.3)]^2
Velocity of P, (vx, vy), given by vx = dx/dt , vy = dy/dt (sub in x and y from above)
acceleration of P, (ax, ay) given by ax= dvx/dt, ay = dvy/dt (sub in vx and vy from above)
b) rate at which speed of P is changing = rate of change of velocity. Simply use pythagaros theorem with answer = sqrt( ax^2 + ay^2)
c) acceleration = v^2 /r
acceleration = answer from part b
v = sqrt(vx^2 + vy^2)
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Originally posted by crimson soldier:
a) y=(x^2)/200 ---> Sub in x = 200cos(0.3t) ---> y= 200[cos(0.3t) ]^2
Therefore position of P, (x,y) given by x=200 cos(0.3t) , y=200[cos(0.3)]^2
Velocity of P, (vx, vy), given by vx = dx/dt , vy = dy/dt (sub in x and y from above)
acceleration of P, (ax, ay) given by ax= dvx/dt, ay = dvy/dt (sub in vx and vy from above)
b) rate at which speed of P is changing = rate of change of velocity. Simply use pythagaros theorem with answer = sqrt( ax^2 + ay^2)
c) acceleration = v^2 /r
acceleration = answer from part b
v = sqrt(vx^2 + vy^2)
Pro wor... I even forgot that there is BOTH a X and Y to velocity and accleration too...haha
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I no need do liao
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I plotting to steal eagle's place as moderator
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okok
I can retire liao
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KNS, why NUS exams haven't start yet
my productivity divingNTU almost all finished le.
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lol u want exams to start so fast?
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Productivity diving le mah...might as well get it over with
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thanks for the help (: Im further asked to draw a scale diagram of the path for -100mm < x < 100 mm. "show the position of the pin at t = 6 secs and the velocity and acceleration vectors. Also show the position of the instantaneous centre of curvature." What steps do i go about drawing this? iirc, velocity is tangential to the path and acceleration is the 90 degree angle to velocity. right?
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A levels physics?
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Originally posted by crimson soldier:
a) y=(x^2)/200 ---> Sub in x = 200cos(0.3t) ---> y= 200[cos(0.3t) ]^2
Therefore position of P, (x,y) given by x=200 cos(0.3t) , y=200[cos(0.3)]^2
Velocity of P, (vx, vy), given by vx = dx/dt , vy = dy/dt (sub in x and y from above)
acceleration of P, (ax, ay) given by ax= dvx/dt, ay = dvy/dt (sub in vx and vy from above)
b) rate at which speed of P is changing = rate of change of velocity. Simply use pythagaros theorem with answer = sqrt( ax^2 + ay^2)
c) acceleration = v^2 /r
acceleration = answer from part b
v = sqrt(vx^2 + vy^2)
yourTherefore position of P, (x,y) given by x=200 cos(0.3t) , y=200[cos(0.3)]^2
... y=200[cos(0.3)]^2 is missing a "t" right? -
Originally posted by WoAiMeiMei:
yourTherefore position of P, (x,y) given by x=200 cos(0.3t) , y=200[cos(0.3)]^2
... y=200[cos(0.3)]^2 is missing a "t" right?
Yes, he is missing a t.
But everyone makes mistakes right? : D
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o.O
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Originally posted by Forbiddensinner:
Yes, he is missing a t.
But everyone makes mistakes right? : D
hahaha im looking at your post for 1a) and his post for 1a) and now im confused which is correct :(
now im lost if u actually have to sub in t=6secs or leave it in its component form! :s
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oh, i think u use the 6secs in part b :D
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ps: the scale diagram would be figured out by the data calculated in part b and c
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Question :
for Part C: i used the v (speed) from part a . apparently that would be wrong because you need to use velocity. huh?
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Well TS, it seems that I made a mistake for part C. I used the formula for circular motion directly without realising that only the tangential acceleration = v^2/r. And to find tangential acceleration is pretty tedious in this case. But it can be done.
Alternatively
a) You want to find the radius of curvature.
b) You know that the radius vector is perpendicular to the vector of the velocity.
so
1) Find the velocity vector in terms of magnitude and direction. Magnitude given by sqrt( vx^2 + vy^2). Angle(direction) given by arctan(vy/vx).
2) Find the direction perpendicular to the velocity vector. It actually gives you the direction of the radius vector.
3) So you need to find the magnitude of the radius vector. Simple. You now have the angle and the horizontal distance of P from the origin (your x value basically).
4) Construct the basic right angled triangle and use cosine to find the magnitude of the radius.
A bit busy with my exams preparation or I would try to upload a peekture. It's probably university first year question.. either that or I'm trying too hard.