Question

You are riding in a school bus. as the bus rounds a flat curve at constant speed, a lunch box with a mass of 0.470 kg suspended from the ceiling of the bus by a string of length 1.75 m is found to hang at rest relative to the bus when the string makes an angle of 30.0 ∘ with the vertical. in this position the lunch box is a distance 49.0 m from the center of curvature of the curve.

Answer 1

Here as we know that

m = mass of the box = 0.470 kg

L = Length of the string = 1.75 m

r = radius of the path = 49 m

$\theta$ = angle of the thread

Now from the condition of equilibrium of lunch box with respect to bus we can say

$Tsin30 = \frac{mv^2}{r}$

$Tcos30 = mg$

now divide the above two equations

$tan30 = \frac{v^2}{rg}$

solving the above equation for velocity

$v^2 = rg tan30$

$v = \sqrt{rgtan30}$

$v = \sqrt{49\times 9.8\times tan30}$

$v = 16.65 m/s$

Answer 2

Missing question: " What is the speed v of the bus?"

Solution:
Let's solve the problem by writing the equilibrium conditions on both x- and y- axis:
$mg=T cos \alpha$
$m \frac{v^2}{r}=T sin \alpha$
where $mg$ is the weight of the lunch box, $m \frac{v^2}{r}$ is the centripetal force, T is the tension of the string and $\alpha=30^{\circ}$.

The radius r is the one with respect to the vertical position of the string, therefore
$r=L sin30^{\circ}=0.875 m$.

From the first equation we find
$T= \frac{mg}{cos \alpha}$
and if we replace this into the second one, we find
$m \frac{v^2}{r}=mg \tan \alpha$
from which we can find the velocity:
$v= \sqrt{rg \tan \aplha}= \sqrt{(0.875m)(9.81m/s^2)(\tan 35^{\circ})}=2.45 m/s$