Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

SAVE ME from this weird mechanics problem.

  1. Oct 14, 2004 #1
    Two persons of different mass roll down the same hill from the same starting point on identical bicycles.
    Neglect friction and air resistance and assume that the tires roll without slipping. Which person will reach the
    bottom first? Prove.

    I have mid term tomorrow.

    Best,
    Saksham
     
  2. jcsd
  3. Oct 14, 2004 #2

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    These assumptions contradict each other.

    Anyway, what ideas do you have ? What equations may be relevant here ?
     
  4. Oct 14, 2004 #3
    This is my Physics assignment. I am sure that there should be some theory behind it. As you can see that it is either true or false. We should consider rotational and translational kinetic energy.
    But who reaches the bottom first?
     
  5. Oct 14, 2004 #4
    You should start out by drawing the free-body diagrams. The person who gets down the hill the fastest accelerated faster, so you want to find the acceleration of each person.
     
  6. Oct 14, 2004 #5
    It's really the same as dropping two people from the the Empire State Building when one weighs more than the other (of course neglecting air resistance, like your question). Who hits first?
    I hope you know, cuz I don't want you to have to do the exeriment:devil:
    (I would hope you know since this is brought up all the time - I'v probablly heard a version of it fifty times, and 47 of them weren't even in physics class.)
     
  7. Oct 15, 2004 #6
    If all the conditions are obeyed then both will reach at the bottom simultaneuosly, distance coverd is same acceleration is same nofriction to oppose the motion therefor neither size nor mass willeffect it. Try equations yourself.
     
  8. Oct 15, 2004 #7
    Since they r of diff masses , normal F will be diff. But since they r rolling friction is sufficient, ( By neglect friction, they mean ROLLING FR.)
    SO, THEY WILL REACH SIMULTANEOUSLY.
     
  9. Oct 15, 2004 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Neglecting friction means that there is no kinetic friction. No slipping means that the static friction is enough to ensure roll. If a wheel rolls without slipping, the speed of its centre of mass is [tex]v=R\omega[/tex]. Static friction does not do any work, you can use conservation of energy to solve the problem. The potential energy at the top equals to the kinetic energy at the bottom. The kinetic energy is the sum of the translational and rotational ones. There is rotational kinetic energy because of the wheels.
    [tex]KE_{rot}=\frac{1}{2}I\omega^2[/tex].

    Can you continue from here?

    ehild
     
  10. Oct 15, 2004 #9
    Thanks ehild. That was what I wanted but I think we get the result that they both reach the bottom at the same time.
    Thanks to all the people who replied.
    And, I did not have that question in the midterm, had I got that then I would have got that correct.
    Thanks once again.
    Cheers,
    Saksham
     
  11. Oct 16, 2004 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You are right if the moment of inertia of the wheels can be neglected. You are wrong if it counts.

    ehild
     
  12. Oct 16, 2004 #11

    Doc Al

    User Avatar

    Staff: Mentor

    If you model the bike as having a wheel of mass m, with all its mass on its rim, then you should be able to show that the speed of the bike is independent of the mass of the rider. (Thus they reach the bottom simultaneously.)
     
    Last edited: Oct 16, 2004
  13. Oct 16, 2004 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I may do something wrong, check my derivation, please.

    m is the mass of the two wheels. M is the mass of the rider and the other parts of the bike. R is the radius of the wheels, I is the moment of inertia of the two wheels. If all the mass of the wheels is concentrated on the rims then [tex]I=mR^2[/tex]. If the wheels do not slip the angular velocity of the wheels multiplied by R is the same "v" as the velocity of the centre of mass of the whole bike and rider together.

    The bike starts from rest at the top of the slope. The angle of incline is [tex]\alpha [/tex].
    According to conservation of mechanical energy, the sum of change of KE after travelling a distance s along the incline and the change of PE equals zero.

    [tex] \frac{1}{2}[(m+M)+I/R^2]v^2 = (m+M)gs\sin(\alpha)[/tex]

    [tex] v=\sqrt{ \frac{2(m+M)gs\sin(\alpha)}{m+M+I/R^2}} = ds/dt[/tex]

    We get s(t) by integration:

    [tex]s=\frac{(m+M)g\sin(\alpha)}{2(m+M+I/R^2)}t^2[/tex]

    If all the mass m is on the rims then I/R^2=m and

    [tex] s=\frac{(m+M)g\sin(\alpha)}{2(m+M+m)}t^2=\frac{g\sin(\alpha)}{2(1+\frac{m}{m+M})}t^2[/tex]

    If m is comparable to M the time needed to rich the bottom of the slope does depend on the mass of the rider.

    ehild
     
  14. Oct 16, 2004 #13
    wouldn't heavy person reach first because since they are rolling down a hill, the "forward" component of the force of gravity on the heavier biker will be greater. Therefore, he will accelerate faster. wouldn't he?
     
  15. Oct 16, 2004 #14

    Doc Al

    User Avatar

    Staff: Mentor

    You are absolutely correct. I made an error in my derivation. (I left out the translational KE of the wheels. Oops!) Nicely done, as usual. :smile:
     
  16. Oct 16, 2004 #15

    Doc Al

    User Avatar

    Staff: Mentor

    Since the force of gravity is exactly proportional to the mass, that argument would lead to the same acceleration for each.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?