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The particle in the peripheral of the wheel.
Help me with this probem...?
2007-03-18 3:28 pm
Find the angular acceleration ε of the wheel, if 2 seconds after the start of the move, the vector of the acceleration of the particle, creates the angle α =60^o with the vector of the linear speed. (Instruction: We know that the centripetal acceleration is v^2/R and the tangential acceleration is εR.)
回答 (2)
2007-03-18 4:37 pm
✔ 最佳答案
I'm sorry, but I need more information to solve the problem. Are we looking at a wheel, or a particle on the wheel, or something else? Is the wheel moving in a circle or in a straight line?Please clarify (if you can) so I can assist you.
2007-03-18 4:54 pm
I had this question answered a moment ago! Where is it? did anyone deleted it? Or I did not submit it properly???
Here's the key pts:
1. find angular vel at t=2 s:
omg^2 = 1/2 * e * t^2
= 2e
2. find v at t=2s:
v = omg * r = 2er
3. draw a vector diagram consisting of the centripetal and tangential accel. The resultant accel is the vector sum of these two. These three sides form a right angle triangle with the longest side being the resultant accel making 60 deg with the tangential accel.
4. apply trigon, tan(60 deg) = centri acc / tang acc
where cetri acc = v^2/r and tang accel = er, and subt v from (3)... solve for e.
Hope this help!
Here's the key pts:
1. find angular vel at t=2 s:
omg^2 = 1/2 * e * t^2
= 2e
2. find v at t=2s:
v = omg * r = 2er
3. draw a vector diagram consisting of the centripetal and tangential accel. The resultant accel is the vector sum of these two. These three sides form a right angle triangle with the longest side being the resultant accel making 60 deg with the tangential accel.
4. apply trigon, tan(60 deg) = centri acc / tang acc
where cetri acc = v^2/r and tang accel = er, and subt v from (3)... solve for e.
Hope this help!
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