- see video for derivation
- $26.1^\circ$

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This is College Physics Answers with Shaun Dychko. Our job in this question is to figure out what is this angle that the cyclist is tilted over compared to vertical. We know that there are two forces being applied at this point here. One force is straight upwards, this is a normal force and it has to be equal in magnitude to the weight of the cyclist. There's also this friction force directed this way parallel to the ground, which is providing the centripetal force that makes the cyclists go in a circle. This resultant here has components that are the normal force, and the other component is the friction force along here, and this angle then is going to be the inverse tangent of the friction force divided by the normal force. One step at a time now. First we'll say that the sum of the vertical forces then is the normal force upwards minus the weight downwards <i>mg</i>. That equals mass times the vertical acceleration but there's no vertical acceleration. This equals zero. We can say after we add <i>mg</i> to both sides, we can say the normal force equals the weight. Then considering the x-direction, we have only the friction force acting in the horizontal direction. That's going to equal mass times its horizontal acceleration, which in this context is called centripetal acceleration. That is substituted with <i>v squared</i> over <i>r</i>. We can say that the friction force then is <i>mv</i> squared over <i></i>. We redo this triangle here and said that the tangent of this angle theta is going to be the opposite friction force divided by the adjacent, which is the normal force, and that's <i>mv</i> squared over <i>r</i> is the friction force, and then divided by the normal force. Now because this friction force is a fraction, I don't like to divide a fraction by yet another fraction because that gets confusing. So instead of multiplying by the reciprocal of the normal force, multiplying by the reciprocal of this, which is one over <i>mg</i>, and the <i>m</i>'s cancel, leaving us with <i>v squared</i> over <i>rg</i>. Then if we take the inverse tangent of both sides, we solve for theta on the left and then we're left with this expression on the right. Theta is inverse tangent of <i>v squared</i> over radius of curvature of its circle, times <i>g</i>. If a cyclist was traveling at 12 meters per second, we'd have to square that and divide by the radius of the circle he's traveling in, which is 30 meters times acceleration due to gravity, 9.8 meters per second squared, take the inverse tangent of all that and we'd get this tilt would be 26.1 degrees.