How Long Does It Take The Bob To Make One Full Revolution (One Complete Trip Around The Circle)?
Stupid masteringphysics once again
- Thread starter Xerxes1986
- Kickoff date
ok the question is:
What tangential speed, , must the bob have so that it moves in a horizontal circle with the cord always making an angle theta from the vertical?
Limited your answer in terms of some or all of the variables thou, L, and theta, as well as the acceleration due to gravity .
and my reply...afterwards several right answers marked wrong, was sqrt(L*sin(theta)*g*tan(theta)) and i got it correct
the next question is:
How long does information technology take the bob to make one full revolution (ane complete trip effectually the circumvolve)?
Express your reply in terms of some or all of the variables yard, 50, and theta, also as the acceleration due to gravity .
easy i thought...i already knew that the radius was going to be L*sin(theta) from the previous problem and so i had the distance and the velocity...slice of cake to get time right? wrong. i entered this for time:
(six.28 *Fifty*sin(theta))/(sqrt(L*sin(theta)*g*tan(theta)))
which basically equates to distance/velocity = time
BUT I Go It WRONG! WTF?! i am not inbound it wrong or anything...whatever input would exist appreciated
attached is the flick for the problem if that helps
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Answers and Replies
Perfectly correct. But information technology tin be greatly simplified. Also, use [itex]\pi[/itex] instead of a number. (I suspect the system is looking for the simplified expression.)i entered this for time:(six.28 *L*sin(theta))/(sqrt(L*sin(theta)*g*tan(theta)))
which basically equates to distance/velocity = time
BUT I Become Information technology Incorrect! WTF?! i am not entering it wrong or anything...any input would be appreciated
the right respond wasn't a simplified version of my answer...information technology was completely different
it wanted usa to convert to angular velocity and and then employ that velocity equation and distance of 2pi to find time...both work merely it only accepts the ane version...i am going to rip my professor a new one tommorow during class. i freaking hate this course now. i know im correct, you agree that im right, there isn't any mistake...gah! i surrender shoot me now
oh and i did try entering 2 pi instead of half dozen.28 and it however marked information technology wrong
I don't know what you mean. What was the "right answer"? (If it's not equivalent to your answer, and then i of them is incorrect.)the correct answer wasn't a simplified version of my answer...information technology was completely unlike
This doesn't brand sense. Either manner, the answer is the aforementioned. (Unless they want you to testify your work.)it wanted u.s.a. to catechumen to angular velocity and then use that velocity equation and distance of 2pi to find time...both work only information technology only accepts the 1 version...
Did you trying simplifying the answer before submitting information technology? (I would not be surprised if your answer was rejected; the simplified answer may be the one the system is looking for.)oh and i did try entering 2 pi instead of half-dozen.28 and it still marked it wrong
My other online assignments never give me as much problems. Some times I got the answers correct simply just considering the way the program/site perceive/translate the answers, I got them wrong - some points deducted.
At the commencement of concluding semester, the Knight Physics textbooks weren't even in the bookstore. Information technology took like a month for the books to come in and when they arrived, they didn't make it in a sufficient number that some students had to live without for another 2 - 3 weeks.
I'm wondering how much the profs become paid to use certain company textbooks/materials every bit the requirements.
ii*pi*sqrt(Fifty*cos(theta)/g)
...how they got that i take no ****ing clue then don't enquire
edit
this is what my prof simply sent me:
Your error is that you did non simplify the algebra by reducing the
dependence on Fifty.
uhhhhh...god i hate this prof
6.28 *L*sin(theta))/(sqrt(L*sin(theta)*g*tan(theta)))
It is obvious that this could exist simplified further.
uhhhhh...god i hate this prof
Don't blame your prof.
well this is the correct answer:2*pi*sqrt(L*cos(theta)/g)
...how they got that i have no ****ing inkling so don't ask
edit
this is what my prof simply sent me:
Your error is that you did not simplify the algebra by reducing the
dependence on L.uhhhhh...god i hate this prof
(2Pi *L*sin(theta))/(sqrt(50*sin(theta)*g*tan(theta)))
Simplify the above expression fully. Bring everything except 2Pi nether the square root. Change tan(theta) to sin(theta)/cos(theta)
You should get:
two*Pi*sqrt(L*cos(theta)/g)
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