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This is just an additoin to the hint given by CMK. make sure you are consistent with the units in the problem, that might be one place to go wrong.
** There is also a different way to think why Vc = 0. If you see the cable, thats round around the pulley, its fixed at a point to a the wall, that is it hangs from a wall hook, since the wall hook is fixed its velocity is zero, and also since Vc is along the same cable Vc=0. Hope this way of thinking helps. Kindly comment if anything is wtong in what I said.
I found the correct answer for the velocity of point O, but I can't seem to get the correct answer for the velocity of B. It is very close but not the same. I am doing the velocity equation with point B and point O because we know the distance between the two and the velocity at C is zero. Any suggestions as to where I am going wrong?
deepak makes two good points. Be careful on units since speed is in m/sec and the dimensions are in mm. Also his explanation as to why vC = 0 is right on the money -- since all points on the cable above C have zero velocity, then so does C since the cable is not stretch.
Your procedure looks correct -- you are writing the velocity of B in terms of the velocity of C. Since vC = 0, then the velocity vector vB = omega x rB/C. Check to make sure that you have the r-vector correct. If that does not produce a correct answer, please send a scan of your solution to the ME274 email address; I would be happy to look over your solution and advise.
Use the rigid body velocity equation to go between points C and D (D is the point on the left-side cable immediately below A ... D has the same velocity as A).
Just to make sure that we are talking about the same thing, it is my understanding that you are wanting to determine the angular velocity of the pulley using the following kinematics equation:
v_D = v_C + omega x r_D/C
where C is the contact point of the cable on the right side of the pulley on the right side and D is the contact point of the cable on the left side of the pulley (in the figure attached to the original post, I had labeled this point D as A').
Is this correct?
If so, you will use v_D = v_A and v_C = 0. For this equation you also need to use:
r_D/C = vector FROM C TO D = -0.270*i
The "-" sign results from going FROM C TO D in the negative x-direction. The distance of 0.270 represents the distance from C to O (0.180 m) plus the distance from O to D (0.090 m).
I might be missing the point of your question. If so, please let me know.
The answer should be the same whether you use point O or point C. For C, the velocity is zero (as you have used) and for O the velocity is 0.6*j (as you have used).
Email me a scan (or post the scan) and I will look over it to see if I can find your error.
12 comments:
This is just an additoin to the hint given by CMK.
make sure you are consistent with the units in the problem, that might be one place to go wrong.
** There is also a different way to think why Vc = 0. If you see the cable, thats round around the pulley, its fixed at a point to a the wall, that is it hangs from a wall hook, since the wall hook is fixed its velocity is zero, and also since Vc is along the same cable Vc=0. Hope this way of thinking helps.
Kindly comment if anything is wtong in what I said.
I found the correct answer for the velocity of point O, but I can't seem to get the correct answer for the velocity of B. It is very close but not the same. I am doing the velocity equation with point B and point O because we know the distance between the two and the velocity at C is zero. Any suggestions as to where I am going wrong?
deepak makes two good points. Be careful on units since speed is in m/sec and the dimensions are in mm. Also his explanation as to why vC = 0 is right on the money -- since all points on the cable above C have zero velocity, then so does C since the cable is not stretch.
courtney,
Your procedure looks correct -- you are writing the velocity of B in terms of the velocity of C. Since vC = 0, then the velocity vector vB = omega x rB/C. Check to make sure that you have the r-vector correct. If that does not produce a correct answer, please send a scan of your solution to the ME274 email address; I would be happy to look over your solution and advise.
I'm unsure of how to find omega. Are there two different omegas - one for the smaller pulley and one for the other?
Consider the pulley to be a single rigid body.
Use the rigid body velocity equation to go between points C and D (D is the point on the left-side cable immediately below A ... D has the same velocity as A).
Is the r_(d/c)= ( 180j) need to find omega?
-- to deborah --
Yes, you will need the position vector of r_D/C in order to find the angular velocity of the pulley. For this problem you will have:
r_D/C = -0.270*i mm
where D is the point on the pulley immediately below A.
i am having trouble finding the r vaules for the problem, how did you determine to use r_d/c=-.270.
Just to make sure that we are talking about the same thing, it is my understanding that you are wanting to determine the angular velocity of the pulley using the following kinematics equation:
v_D = v_C + omega x r_D/C
where C is the contact point of the cable on the right side of the pulley on the right side and D is the contact point of the cable on the left side of the pulley (in the figure attached to the original post, I had labeled this point D as A').
Is this correct?
If so, you will use v_D = v_A and v_C = 0. For this equation you also need to use:
r_D/C = vector FROM C TO D = -0.270*i
The "-" sign results from going FROM C TO D in the negative x-direction. The distance of 0.270 represents the distance from C to O (0.180 m) plus the distance from O to D (0.090 m).
I might be missing the point of your question. If so, please let me know.
i found the answer for V_B by using
w = -3.33k rad/s
V_B = V_C+ (w * r_B/C)
where V_C = 0
why can't i get the right answer using:
V_B = V_0 + (w * r_B/0)
where V_0 = 0.6j m/s
Can anyone kindly help me?
--to ali azhar--
The answer should be the same whether you use point O or point C. For C, the velocity is zero (as you have used) and for O the velocity is 0.6*j (as you have used).
Email me a scan (or post the scan) and I will look over it to see if I can find your error.
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