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Acceleration of flywheel?

  1. Dec 6, 2011 #1
    A flywheel is driven by a pulley system. friction in the system exerts a braking force of 2Nm. The moment of inertia is 400kgm2.
    The forces in the pulley belt are 120N on the tight side and 80N on the slack side.
    Pulley diameter is 0.2m
    Calculate the acceleration of the wheel.

    the pulley has the greater diameter but doesn't say the diameter of the motor but doesn't give a diameter for the motor. also doesn't say which way the flywheel is moving.


    I have made several attempts using equations like the square root of I/m but i dont know the mass. i did 400/120 which i think gives me the radius of gyration. but as i already know the inertia i dont then thought i dont need this.

    so i then did T= F x R
    But as i'm not sure which is the side with the pulley i just guessed that it was the bigger force of 120N
    120/0.1 = 1200Nm

    T=I x a
    a= T/I
    1200/400 = 3rad/s2

    or is the using the other 80N
    or is it using the 80N and then subtracting the 1st from the second?
    how does the 2N of braking force get accounted for?
     
  2. jcsd
  3. Dec 6, 2011 #2
    You wrote,

    "so i then did T= F x R
    But as i'm not sure which is the side with the pulley i just guessed that it was the bigger force of 120N
    120/0.1 = 1200Nm"

    Why did you divide by the radius? Don't you want to multiply by r? Also there are three torques in this problem, the torque from friction and the two torques from the belt?
     
  4. Dec 7, 2011 #3
    Well Torque is equal to inertia multiplied by alpha and as alpha is the acceleration which i need to find out.

    I have pages and pages of equations and workings out ranging from using omega to mechanical advantage but it would take a long time to write it all down.. obviously as i don't know the answer and me doing 120/0.1 is a mistake.

    as i only have one radius how can i work out the torque running through the other side of the system? I am given forces along the belt but it doesnt state which are for what.

    I am on here for help else i wouldnt have posted and your clearly not helping
     
  5. Dec 7, 2011 #4
    You wrote,

    "I am on here for help else i wouldnt have posted and your clearly not helping"

    Not my intention.

    I may be mistaken but,

    You wrote,


    "120/0.1 = 1200Nm"

    I thought the 120 above was 120N and the 0.1 above was .1m ? If so when you divide above you don't get Nm?
     
  6. Dec 7, 2011 #5
    don't get huffy - you divided where you should have multiplied to solve the equation torque = F cross R

    It's the sum of the torques that is equal to I alpha and the torque in this case will be the sum of the forces from the pulleys minus force due to friction if they are all acting on the same radius. My problem with questions of this sort is that not every physics student is a gear head and knows what a flywheel is and how it works, what the slack side does and what the tight side does so you best option here is to google or go to wikipedia flywheel and look at how one works. From there, sum your forces, if they all act on the same radius, to get your net torques and then use your torque = I alpha from there just as you are doing. So your problem is not being able to visualize the system. Lucky you has google or wikipedia or a gear head friend.
     
  7. Dec 8, 2011 #6
    i think i'v got it now.. i took the 80N and 120N and subtracted them together to give me the net force which i then multiplied by the radius of 0.1m to give me 4Nm i then added the friction to give me 6Nm.

    From there i used the equation T=Ialpha and rearranged so alpha = T/I
    6/400 = alpha (angular acceleration)

    which gives 0.015 rad/s

    This to me sounds small but it does seem like the right way of going around it and getting to the answer
     
  8. Dec 8, 2011 #7
    You do have one problem with the original equation though - it say the braking force is 2Nm. Forces don't come in Nm. IF it's supposed to read a braking force produces a negative torque on the system of 2Nm you'd be OK but it doesn't say that. Also why would you ADD the braking force/torque to the torque moving the system? also check your units on alpha - acceleration comes in rad/sec squared
     
  9. Dec 11, 2011 #8
    well a breaking force is a negative force.. and i added it as it needs to have xNm to accelerate the wheel but also needs an extra 2Nm force to overcome the friction
     
  10. Dec 11, 2011 #9
    2 Nm is NOT a force. It is a torque. You need to check with your teacher on this. If it is a braking torque then yes you can subtract it but the problem says it is a FORCE which it is NOT because of the units Nm. This is an error in the problem.
     
  11. Dec 12, 2011 #10
    yeah i did check with the tutor who said what i had done was right
     
  12. Dec 12, 2011 #11
    The problem is incorrect, not you. It should have state that a torque caused by friction is 2 NM, not that there is a force of friction of 2 NM. Forces don't come in Newton meters, they come in Newtons.
     
  13. Dec 13, 2011 #12
    yes i'm not disagreeing with you but as the solution i have got is correct why are you going any further with it? i know it should be Nm but its a typo
     
  14. Dec 13, 2011 #13
    Because whoever wrote the problem needs to fix it. Even though you got the "right" answer according to whatever source gave you the answer, you really didn't didn't solve this problem because it is not solvable as written. Their proof reader didn't catch this mistake and it's the better student who will get penalized because they would know it wasn't solvable. Just a good thing to know. You probably know that on the SAT and the GRE, I'm thinking, you can appeal a question and it will get fixed.
     
  15. Dec 13, 2011 #14
    well the tutor of the class gave me the answer and how to solve it.. i'm not even studying mechanics. its all just additional knowledge that i asked for to broaden my knowledge. i'm studying electrical engineering. What is SAT and GRE? i'm from the uk not the states
     
  16. Dec 13, 2011 #15
    Scholastic aptitude test to get into college and Graduate Record Exam to get into graduate school in the US.

    Regardless, the problem itself has a "fatal flaw" and at least you know about it.
     
  17. Dec 14, 2011 #16
    well those exams are invalid to me as i live in the uk lol.. i'm taking foundation to electrical engineering degree so like i say this has nothing to do with the course i'm studying just for own personal knowledge.. thank you for pointing out the mistake though
     
  18. Dec 14, 2011 #17
    True, but a lot of people from the UK do study in the US at some time in their educational careers so it's good to know about the tests. You can buy books with practice tests in them just to see how well you would do, "for personal knowledge". The GRE also has specific course tests like physics, chemistry, and so on. They are pretty wicked. One of the universities I attended actually required a major in EE to get a physics major. As I had no interest in that area, I did transfer to a different university. :-)

    Good luck!
     
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