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In the equation: at = alpha x r , you know the value of at and also r. I am confused on the algebra when there is a cross product involved. How do you get alpha by itself? And how do you determine that the answer is in the k direction?
8 comments:
Anonymous
said...
For that problem, you don't have to use a cross product since both a and r just have one component; you can just divide them. Then use the right hand rule to figure out the direction of alpha, or you can always remember i cross j is k. But since you're dividing a by r to get alpha rather than multiplying to alpha and r to get a, alpha will be in the opposite direction of what you get when you divide. I hope this helps some.
You also could have taken the cross product of a and r, and you would get the opposite of alpha, so divide it by -1 to get alpha. That should work for other problems where you have to use a cross product.
To find alpha in that problem just divide the i component of the acceleration by -r. This is because k x j = -i. Don't forget that omega and alpha will always be in the k direction.
I was able to find the angular acceleration, but am having trouble with the angular velocity. I have the equation v_p = v_0+omega *dot* r_p/o I know that v_0 = 0, but I don't know what to do from there.
You do not need to use that equation to find omega. You can find omega and alpha from the rigid body acceleration equation. Omega in the acceleration equation is a magnitude, not a vector.
All of these responses are good, and right to the point. Let me attempt a response that pulls together all that was said.
Write down the rigid body acceleration equation that relates the motion of points O and P:
aP = aO + alpha x rP/O - omega^2*rP/O
The acceleration of P is given by: aP = -3*i - 4*j. Since O is a stationary point, aO = 0. The angular acceleration vector is given by alpha = alpha*k. The position vector rP/O is given by rP/O = 0.5*j. Substituting these into the kinematics equation and performing the cross product gives:
-3*i -4*j = -r*alpha*i -r*omega^2*j
where r = 0.5.
From this vector equation you get two scalar equations by balancing the coefficients of the x- and y-components:
i: -3 = -r*alpha j: -4 = -r*omega^2
As ebcox and nablock say, you find alpha from the first equation: alpha = 3/r . You find omega from the second equation: omega=sqrt(4/r). This is scott's point in his comment.
The alpha vector is ALWAYS perpendicular to the plane of the motion of the rigid bodies (k-direction). Note that the direction of the angular acceleration vector represents the axis of rotation of the body.
Your question might be more directed whether the alpha vector points in the positive or negative k-direction. You can determine this by the right-hand-rule: when you cross alpha with the position vector your thumb points in the direction of alpha x r. From this you can figure out the sign on alpha. (Note that the r vector referenced above goes from point A to point B, where B is point whose acceleration is desired.
8 comments:
For that problem, you don't have to use a cross product since both a and r just have one component; you can just divide them. Then use the right hand rule to figure out the direction of alpha, or you can always remember i cross j is k. But since you're dividing a by r to get alpha rather than multiplying to alpha and r to get a, alpha will be in the opposite direction of what you get when you divide. I hope this helps some.
You also could have taken the cross product of a and r, and you would get the opposite of alpha, so divide it by -1 to get alpha. That should work for other problems where you have to use a cross product.
To find alpha in that problem just divide the i component of the acceleration by -r. This is because k x j = -i. Don't forget that omega and alpha will always be in the k direction.
I was able to find the angular acceleration, but am having trouble with the angular velocity.
I have the equation v_p = v_0+omega *dot* r_p/o
I know that v_0 = 0, but I don't know what to do from there.
You do not need to use that equation to find omega. You can find omega and alpha from the rigid body acceleration equation. Omega in the acceleration equation is a magnitude, not a vector.
All of these responses are good, and right to the point. Let me attempt a response that pulls together all that was said.
Write down the rigid body acceleration equation that relates the motion of points O and P:
aP = aO + alpha x rP/O - omega^2*rP/O
The acceleration of P is given by: aP = -3*i - 4*j. Since O is a stationary point, aO = 0. The angular acceleration vector is given by alpha = alpha*k. The position vector rP/O is given by rP/O = 0.5*j. Substituting these into the kinematics equation and performing the cross product gives:
-3*i -4*j = -r*alpha*i -r*omega^2*j
where r = 0.5.
From this vector equation you get two scalar equations by balancing the coefficients of the x- and y-components:
i: -3 = -r*alpha
j: -4 = -r*omega^2
As ebcox and nablock say, you find alpha from the first equation: alpha = 3/r . You find omega from the second equation: omega=sqrt(4/r). This is scott's point in his comment.
Good job by all in this discussion! Thanks.
How do we know that alpha points in the k direction? I can't seem to figure this out.
The alpha vector is ALWAYS perpendicular to the plane of the motion of the rigid bodies (k-direction). Note that the direction of the angular acceleration vector represents the axis of rotation of the body.
Your question might be more directed whether the alpha vector points in the positive or negative k-direction. You can determine this by the right-hand-rule: when you cross alpha with the position vector your thumb points in the direction of alpha x r. From this you can figure out the sign on alpha. (Note that the r vector referenced above goes from point A to point B, where B is point whose acceleration is desired.
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