Question

The friction force acting on the bike is

• A. $\frac{2m{v}^{2}tan\theta }{\ell \sqrt{4+ta{n}^2\theta }}$
• B. $\frac{2m{v}^{2}ta{n}^2\theta }{\ell \sqrt{2+ta{n}^2\theta }}$
• C. $\frac{m{v}^{2}ta{n}^2\theta }{\ell \sqrt{4+ta{n}^2\theta }}$
• D. $\frac{m{v}^{2}ta{n}^2\theta }{\ell \sqrt{2+ta{n}^2\theta }}$

Answer 1

The correct option is A $\frac{2m{v}^{2}tan\theta }{\ell \sqrt{4+ta{n}^2\theta }}$

If the bike does not slip, path of both the wheel must be parallel with its direction as shown in the figure.
$R_1=\ell /tan\theta$

$R_{cm}=\sqrt{\left(\begin{array}{c}\frac{\ell }{2}\end{array}\right)+R_1^2}=\frac{\ell }{2\tan \theta }\sqrt{4+ta{n}^2\theta }$

Net friction force $f=\frac{m{v}^2}{R_{cm}}=\frac{2m{v}^{2}tan\theta }{\ell \sqrt{4+ta{n}^2\theta }}$