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Rider in a loop problem

  1. Oct 24, 2012 #1
    3. Consider the motorcyclist riding on the inside of a vertical circle of radius b. His initial speed at the bottom of the circle is u.

    a) Explain why the three domains (i) u^2 < 2gb, (ii) 5gb > u^2 > 2gb and (iii) u^2 > 5gb are qualitatively different.

    b) Describe what happens in domain (ii). Give a sketch of the trajectory, indicating the height h where the rider is released from the hoop. Calculate h(u).

    So for (a) I said that the domains are the speed of the motorcyclist at different points along the circle, domain i being the top of the track since speed would be smallest, domain being the bottom where speed is greatest, and domain ii being everywhere else. Would I be correct in saying this?

    For (b), if my explanation above is correct then I know what happens in domain ii, where gravity affects the speed of the motorcyclist in such a way that it decreases his speed as he travels from the bottom to the top and increases it as he goes back from the top to the bottom. I'm just confused on the wording of the rest of (b) and not sure how to approach solving this since I thought the rider is in a closed hoop so how can he be released from it. Am I not reading the problem correctly?
    Last edited: Oct 24, 2012
  2. jcsd
  3. Oct 24, 2012 #2
    There are three types of motion depending on the initial velocity. For the first type of motion let the initial velocity be very very small,

    (mv^2/2 << 2mgb).

    What happens?

    Now keep adding energy and imagine or draw what happens. Then do the math.
  4. Oct 24, 2012 #3
    Be careful not to oversimplify energy by only talking about speed(velocity). Energy (as you know) includes potential as well as kinetic energy. I'd try recasting the domains as explicit energy expressions.
  5. Oct 24, 2012 #4
    The total energy is essentially the potential energy right? Since kinetic energy is negligible when it is much smaller than potential.
  6. Oct 24, 2012 #5
    Ok so domain i is where energy is essentially potential energy, which is at the top of the hoop, domain iii is where energy is essentially just kinetic energy which is at the bottom and domain ii is everywhere else.
  7. Oct 24, 2012 #6
    Keep in mind that the rider starts at the bottom of the circle.
  8. Oct 24, 2012 #7
    So if initial velocity is really small he won't have enough kinetic energy to counteract gravity, and just stay at the bottom?
  9. Oct 24, 2012 #8
    The questions aren't asking, "what point is he at, when x > y?" They are asking for you to explain what happens if the given relationships are true *when he enters the circle*.
  10. Oct 24, 2012 #9
    Yup, now we're getting somewhere. He won't necessarily stay right at the bottom, but he will not reach the top.
  11. Oct 24, 2012 #10
    Oh ok, so I need to say that in domain i he won't be able to move up the side of the hoop, in domain iii he can move in a complete circle and in domain ii he will stop before he gets past the top of the hoop?
  12. Oct 24, 2012 #11
    Almost, except for domain (ii). Here we need to consider the velocity he must have at the top of the circle in order to maintain contact with the track. Hint: at this point, the centripetal force is in the opposite direction of the gravitational force. In order to maintain contact, he must have enough centripetal force to cancel out the gravitational force.

    Formula for centripetal force is:
    [itex] F_{c} = \frac{mv^{2}}{r} [/itex]

    Edit: rereading the question, I suppose you don't have to prove this as it only asks for a qualitative answer.
  13. Oct 24, 2012 #12
    Ok so for part (a) what I said is fine? Would I need to show this relation where centripetal force equals potential at the top to find h in part (b) of the question?
  14. Oct 24, 2012 #13
    Ahh yes, except remember that it's not [itex] F_{c} = U_{g} = mgh [/itex]

    but is instead [itex] F_{c} = F_{g} = mg [/itex]
  15. Oct 24, 2012 #14
    So m(v^2)/r = mg which means v = sqrt(gr) for the rider to make it to the top, so for domain ii v < sqrt(gr). In order to calculate h(u) like the problem says, do I merely need to find some relationship like a > h(u) > b where a is 2 times the radius and b is some value determined by conservation of energy using v = sqrt(gr)?
  16. Oct 24, 2012 #15
    I thought about my last post some more and thought of something that confuses me. If the speed required to make it to the top of the loop is sqrt(gr) then wouldn't the rider make the loop in domain ii and iii since u^2 > 2gr in both cases? Also since domain i isn't u^2 < gr, doesn't that mean that the rider could make it past the top of the loop for this domain as well?
  17. Oct 24, 2012 #16


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    Wrt (i), I think you need to be a little more precise about "won't make it up the loop".

    Wrt (ii), how much KE will the rider still have at the top? What speed does that imply? What is the min speed there to remain in contact with the track?
  18. Oct 24, 2012 #17
    I did some calculations and found that the min speed required is sqrt(4gr). So for (a) of the problem I need to say that in domain i the rider will not reach the top of the hoop since u^2 < 2gb which is less than 4gb (where b = r). Domain iii the rider will definitely make it to the top of the hoop and will continue since his speed is greater than the min required to stay on the track, and in domain ii it depends on the speed, if u^2 > 4gb then he will make it past the top of the hoop, if u^2 < 4gb then he will not.
  19. Oct 24, 2012 #18


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    For domain (i), there's more to it than merely "won't make it to the top". (Otherwise it would be the same as domain (ii).) There are two ways in which the rider might not make it to the top, with a considerable difference in the health & safety consequences.
    For domain (ii), you should get a 5 in there, not 4. Please show your working.
  20. Oct 24, 2012 #19
    Careful, I think you have missed a step for v. The v we're talking about for centripetal force is what he must *have left* at the top. This means we need [itex] KE_{o} \geq KE_{t} + PE_{t} [/itex]

    I'll leave the expansion to you, but we can simplify this down to find the relationship [itex] v_{o} \geq \sqrt{5gr} [/itex]

    I'm working on the h(u) function, I'll get back to you in a few.
  21. Oct 24, 2012 #20
    For the min speed velocity at the top will be zero, and at the bottom of the track potential energy is zero. So (1/2)m(u^2) = 2mgb, the m's cancel, and I multiplied both sides by 2 so
    u^2 = 4gb.
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