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A math professor cycles around a corner which is banked (inclined downwards towards the center of the radius of curvature). The centripetal force that causes the bike to go round a bend in the road is usually provided by friction. This road is very icy, however. Could the force required to turn the corner be provided if the friction coefficient between the bikes tires and the road goes to 0.01? Find the range of velocities required by the bike if the radius of curvature of the curve is 120 m and the road is banked at 10o to the horizontal.

Answer :

Olajidey

Answer:

[tex]V_{max} = 14.8m/s[/tex]

[tex]V_{min} = 13.9m/s[/tex]

Explanation:

Given that,

[tex]\mu_s = 0.0[/tex]

[tex]U = 120m[/tex]

[tex]\theta = 10^\circ[/tex]

max speed for cycle for safe turning on the corner

[tex]V_{max}=\sqrt{\frac{Ug(\mu_s+ \tan \theta}{1 - \mu_s \tan \theta} }[/tex]

[tex]V_{max} = \sqrt{\frac{120\times 9.8 (0.01\+ \tan 10^0)}{1-0.01 \tan 10^0} } \\\\V_{max} = 14.8m/s[/tex]

min speed for cycle for safe turning on the corner

[tex]V_{min} = \sqrt{\frac{Ug( \tan \theta - \mu_s)}{1+ \mu_s \tan \theta} }[/tex]

[tex]V_{min} = \sqrt{\frac{120 \times 9.8(\tan 10^0 - 0.01)}{1+0.01 \tan 10^0} } \\\\V_{min} = 13.9m/s[/tex]

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