Question

At what angle shall a curve be raised in order for a 1500.0-kg vehicle to make a turn around a 100.0-m circular track at a velocity of 30.0 m/s? 

Answer 1

Solution.

$m=1500.0 kg;$

$r=100.0 m;$

$v=30 m/s;$

$\theta - ?;$

$Nsin\theta=\dfrac{mv^2}{r};$

$N=\dfrac{mg}{cos\theta};$

$mg\dfrac{sin\theta}{cos \theta}=\dfrac{mv^2}{r};$

$mgtan\theta=\dfrac{mv^2}{r};$

$tan \theta=\dfrac{v^2}{rg};$

$tan \theta=\dfrac{900}{100.0\sdot9.81}=0.9174;$

$\theta=42^o;$

Answer: $\theta=42^o.$