Tangential Speed and Time in a Horizontal Circle Formula | Physics Problem

[Xerxes1986]

ok this stupid program is starting to piss me off again (sry for the french)

ok the question is:

What tangential speed, , must the bob have so that it moves in a horizontal circle with the string always making an angle theta from the vertical?
Express your answer in terms of some or all of the variables m, L, and theta, as well as the acceleration due to gravity .

and my answer...after several right answers marked wrong, was sqrt(L*sin(theta)*g*tan(theta)) and i got it right

the next question is:

How long does it take the bob to make one full revolution (one complete trip around the circle)?
Express your answer in terms of some or all of the variables m, L, and theta, as well 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...piece of cake to get time right? wrong. i entered this for time:

(6.28 *L*sin(theta))/(sqrt(L*sin(theta)*g*tan(theta)))

which basically equates to distance/velocity = time

BUT I GET IT WRONG! WTF?! i am not entering it wrong or anything...any input would be appreciated

attached is the picture for the problem if that helps

 

[Doc Al]

i entered this for time:

(6.28 *L*sin(theta))/(sqrt(L*sin(theta)*g*tan(theta)))

which basically equates to distance/velocity = time

BUT I GET IT WRONG! WTF?! i am not entering it wrong or anything...any input would be appreciated

Perfectly correct. But it can be greatly simplified. Also, use $\pi$ instead of a number. (I suspect the system is looking for the simplified expression.)

 

[Xerxes1986]

figured it out

the right answer wasn't a simplified version of my answer...it was completely different

it wanted us to convert to angular velocity and then use that velocity equation and distance of 2pi to find time...both work but it only accepts the one version...i am going to rip my professor a new one tommorow during class. i freaking hate this class now. i know I am right, you agree that I am right, there isn't any error...gah! i give up shoot me now

oh and i did try entering 2 pi instead of 6.28 and it still marked it wrong

 

[Doc Al]

the right answer wasn't a simplified version of my answer...it was completely different

I don't know what you mean. What was the "right answer"? (If it's not equivalent to your answer, then one of them is wrong.)

it wanted us to convert to angular velocity and then use that velocity equation and distance of 2pi to find time...both work but it only accepts the one version...

This doesn't make sense. Either way, the answer is the same. (Unless they want you to show your work.)

oh and i did try entering 2 pi instead of 6.28 and it still marked it wrong

Did you trying simplifying the answer before submitting it? (I would not be surprised if your answer was rejected; the simplified answer may be the one the system is looking for.)

 

[Heart]

I have encoutered another problem with MasteringPhysics just now. The display/format looks weird and different. I'm using the same computer, same settings; everything's the same. Now it's confusing and hard to read the text. Some weird fonts and images showed up. I don't know why profs are so fond of this masteringphysics thing.

My other online assignments never give me as much problems. Some times I got the answers right but just because the way the program/site perceive/interpret the answers, I got them wrong - some points deducted.

At the beginning of last semester, the Knight Physics textbooks weren't even in the bookstore. It took like a month for the books to come in and when they arrived, they didn't arrive in a sufficient number that some students had to live without for another 2 - 3 weeks.

I'm wondering how much the profs get paid to use certain company textbooks/materials as the requirements.

 

[Xerxes1986]

well this is the correct answer:

2*pi*sqrt(L*cos(theta)/g)

...how they got that i have no ****ing clue so don't ask

edit

this is what my prof just sent me:

Your error is that you did not simplify the algebra by reducing the
dependence on L.

uhhhhh...god i hate this prof

 

[Gamma]

6.28 *L*sin(theta))/(sqrt(L*sin(theta)*g*tan(theta)))

It is obvious that this could be simplified further.

uhhhhh...god i hate this prof

Don't blame your prof.

 

[learningphysics]

well this is the correct answer:

2*pi*sqrt(L*cos(theta)/g)

...how they got that i have no ****ing clue so don't ask

edit

this is what my prof just 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(L*sin(theta)*g*tan(theta)))

Simplify the above expression fully. Bring everything except 2Pi under the square root. Change tan(theta) to sin(theta)/cos(theta)

You should get:
2*Pi*sqrt(L*cos(theta)/g)

 

[Doc Al]

Just as I suspected.