Solution - Turning Scooter | Collection of Physics Problems

Solution - Turning Scooter

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If the scooter does not slip to the plane, path of both wheels must be parallel with its direction.

From the picture above, we can get

$R_1=\frac{l}{\tan\theta}\ldots(1)$

Hence, the radius of CM's path is

$R_{cm}=\sqrt{\left(\frac{l}{2}\right)^2+R_1^2}$

$R_{cm}=\frac{l}{2\tan\theta}\sqrt{4+\tan^2\theta}\ldots(2)$

So the total friction force must equal to

$f=\frac{mv^2}{R_{cm}}$

$f=\frac{2mv^2\tan\theta}{l\sqrt{4+\tan^2\theta}}\ldots(3)$

Hence, the x and y component of the friction is

$f_x=f\frac{R_1}{R_{cm}}$

$f_x=\frac{4mv^2\tan\theta}{l(4+\tan^2\theta)}\ldots(4)$

$f_y=f\frac{l}{2R_{cm}}$

$f_y=\frac{2mv^2\tan^2\theta}{l(4+\tan^2\theta)}\ldots(5)$

Now, draw the friction forces work on both wheels. Each friction has perpendicular direction to each wheel. Sum of these two frictions must equal to the total friction (both in x and y direction).

First, for y-direction,

$f_y=f_2\sin\theta$

From equation (5), we can get

$f_2=\frac{2mv^2\tan^2\theta}{l\sin\theta(4+\tan^2\theta)}\ldots(6)$

Then for x-direction,

$f_x=f_1+f_2\cos\theta$

Substitute fx from equation (4) and f2 from equation (6),

$f_1=\frac{2mv^2\tan\theta}{l(4+\tan^2\theta)}\ldots(7)$

From equation (6) and (7) above, we can see that f2 is greater than f1, so the coefficient of friction must equal to

$\mu=\frac{f_2}{N_2}\ldots(8)$

For the normal forces, because the CM of the scooter is at the middle, we can say that

And the total normal forces must equal to its weight, so

Hence,

$N_1=N_2=\frac{mg}{2}\ldots(9)$

Back to coefficient of friction. From equation (6) and (8), we can get the coefficient of friction,

$\mu=\frac{4v^2\tan^2\theta}{gl\sin\theta(4+\tan^2\theta)}$

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

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