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Find the minimum value of the initial velocity $u$ of the particle such that the particle crosses the wheel of radius $R$.
enter image description here

Details and assumptions
$R=2m$
$g=9.8m/s^2$
Neglect air resistance.
All surfaces are frictionless.
The value of $\theta$ (angle the projectile makes either with vertical or horizontal), range and $u$ is not known.
Consider the motion in 2-D space only.

I tried setting the maximum height equal to $2R$ and then finding the corresponding minimum value of $u$, but my answer was incorrect.
Then I tried to set the latus rectum of the parabola (equation of trajectory) equal to $2R$ but that too didn't work.
Can anybody suggest a way to do this question?
Thanks in advance!

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  • $\begingroup$ The constraints on the trajectory are that it touches the wheel, i.e., at some x0 the value and derivative of the trajectory y(x) match those of f(x)=sqrt(R^2-x^2)+R; and the peak of the trajectory is above the axis of the wheel, x=0. $\endgroup$
    – Maxim Umansky
    Commented Mar 7, 2014 at 15:46
  • $\begingroup$ From where are you throwing it? Do we know that? $\endgroup$
    – evil999man
    Commented Mar 7, 2014 at 17:58
  • $\begingroup$ The range is said to be unknown. $\endgroup$
    – jazzwhiz
    Commented Mar 7, 2014 at 18:01
  • $\begingroup$ Nice problem. Is this HS level? $\endgroup$
    – John Alexiou
    Commented Mar 7, 2014 at 20:04
  • $\begingroup$ I find that the minimum velocity is $u=\sqrt{(1+\sqrt{2})2gR}$. $\endgroup$
    – G. Smith
    Commented Jul 12, 2019 at 20:10

2 Answers 2

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The value of initial velocity will be different for different angles θ with the horizontal.. So I got this result.

$$ u=(gR/(sinθcosθ-cos^{2}θ))^{1/2} $$ or $$ u=(2gR/(sin2θ-2cos^{2}θ))^{1/2} $$ or $$ u=(39.2/(sin2θ-2cos^{2}θ))^{1/2} $$

This is my attempt for the solution(i have attached image):

From A to B displacement is FB

From C to B displacement is EB

my solution

and θ should be greater than π/4 so that the particle will touch at two points

is the answer correct?

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    $\begingroup$ I don't know my solution is correct or not. Is is the answer given in your book? $\endgroup$
    – Tea is life
    Commented Mar 8, 2014 at 18:03
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Well basically what you said is true, the maximum height, in this case, depends on the initial velocity and the angle $\theta$. So if you consider that the maximum height to be 2R, and by using the trajectory equation, while replacing $x_{y_{max}}=\frac{Total \;Distance}{2}=f(u)$, you'll get $u=f(R,\theta)$.

Now I did the calculation and I got $$u=\frac{4Rg}{\sin^2(\theta)}$$ which means for a maximum value of $\sin^2(\theta)$ you should have a minimum value of $u$ and you'll get something like: $$u_{min}=4Rg$$ which the value you've been searching for.

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  • $\begingroup$ As I have stated in the problem, the answer that we get from here doesn't match with the answer given in the book. $\endgroup$
    – MathGod
    Commented Mar 8, 2014 at 16:55

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