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发表于 31-7-2010 11:19 PM
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isnt that tan teta = 0.5/2 then teta =14 + sth??? then why in 4th quadrant?? |
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发表于 31-7-2010 11:25 PM
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回复 361# vick5821
Okay, it may not necessarily be in the fourth quadrant; it depends on how are you going to draw the diagram. I didn't draw the diagram, but I processed it directly in my brain.
Ya, thanks for correction: tan^-1(0.5/2) = 14.04  |
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发表于 31-7-2010 11:28 PM
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then what ans is correct for the angle?? 14.04? or 180 + 14.04?? 04 360 - 14.04? Welcome |
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发表于 31-7-2010 11:29 PM
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Give a diagram, and follow your diagram. |
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发表于 31-7-2010 11:53 PM
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回复 Lov瑜瑜4ever
不用客气。其实你会做那题，不用道谢的
今年upper six?
数学系 发表于 31-7-2010 10:18 PM 
没。。。我lower6
我刚刚只是找出magnitude罢了
还没有找出direction |
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发表于 1-8-2010 09:40 PM
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I still dono which quadrant leh..suppose third quadrant?? |
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发表于 1-8-2010 10:41 PM
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Urgent!!!
A simple pendulum is suspended freely on the ceiling of a bas moving in a horizontal circular track of radius 80 m. The angle between the string of the pendulum and the vertical line is 36 degree. Calculate the centripetal acceleration of the pendulum |
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发表于 1-8-2010 10:44 PM
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回复 367# vick5821
Before posting my solution, could I know how are you analyzing this question? |
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发表于 1-8-2010 10:49 PM
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i cant get the ques..the radius of the circular motion is wat?? |
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发表于 1-8-2010 10:53 PM
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本帖最后由 whyyie 于 1-8-2010 10:56 PM 编辑
回复 367# vick5821
tan* = v^2 / rg .. (1)
a = v^2/r... (2)
a = g tan*
= 9.81 tan 36
= 7.13ms-1
刚才打错了 |
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发表于 1-8-2010 10:54 PM
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回复 369# vick5821
I don't understand the sentence "the radius of the circular motion is what". Are you asking "what is the meaning of the radius of the circular motion" or "what is the value of the radius of the circular motion"? You could write your reply in Chinese, I can read Chinese... |
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发表于 1-8-2010 11:00 PM
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radius of circular motion of the pendulum is what??? given de is for radius of the road ma |
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发表于 1-8-2010 11:06 PM
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回复 370# whyyie
why tan* = v^2/rg?? |
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发表于 1-8-2010 11:09 PM
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回复 373# vick5821
T sin* = mv^2 / r
T cos* = mg
________________
tan * = v^2/rg
where * is angle from the vertical
应该有学过的. |
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发表于 1-8-2010 11:14 PM
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so, the given radius of the circular track is useless??? |
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发表于 1-8-2010 11:15 PM
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回复 373# vick5821
whyyie 做对了。
不过我的想法是简化的。
T sin54' = g ... (1)
T cos54' = a ... (2)
Thus, a = g tan 36' |
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发表于 1-8-2010 11:16 PM
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typo, should be ma and mg |
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发表于 1-8-2010 11:43 PM
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is it no use de the given radius??? |
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发表于 2-8-2010 12:14 AM
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发表于 2-8-2010 12:40 AM
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cheh..make me confused nia..I thought need use de..thanks all ^^ |
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