You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with a mass of 0.500 kg suspended from the ceiling of the bus by a string of length 1.88 m is found to hang at rest relative to the bus when the string makes an angle of 31.0 ∘ with the vertical. In this position the lunch box is a distance 52.0 m from the center of curvature of the curve.

What is the speed v of the bus?

yo this question is sick.

i'm missing something fundamental that links centripetal force experienced by the bus and the force experienced by the lunch box :/ curious to know how to get to the answer tho

ok i thought about it for another min or so. here's a starting idea

the centripetal force is pointed from the bus towards the center. Thus, the force experienced by the lunch box towards the center should be m*a = m*g*sin(31). If this is the centripetal force, and centripetal force = m*v^2/r, then these two expressions should be equal.

combining this information together, we have m*g*sin(31) = m*v^2/r

this simplifies to v^2 = g*sin(31)*r, where r = 52 m.

one caveat here is that we never use the string length of 1.88m :/ so this may not be right

I figured it out, the answer ended being 17.5m. You actually never use the string.

Masterying physics is a troll....

This post was edited by desertwolf on Oct 4 2015 11:40pm