How to find the acceleration of a car ascending in an incline when a sphere makes an angle in a quarter of a circle cavity?

Question

The problem is as follows:

The figure from below shows a car going up in an incline. The car has a circular cavity on it where there is a small sphere over it. Assume the circular surface has negligible friction. Given these conditions find the acceleration in meters per second square which the wagon must have so that the ball takes the position as shown in the diagram.

The alternatives given are as follows:

$\begin{array}{ll} 1.&9.80\,\frac{m}{s^2}\\ 2.&8.33\,\frac{m}{s^2}\\ 3.&6.25\,\frac{m}{s^2}\\ 4.&5.66\,\frac{m}{s^2}\\ 5.&4.57\,\frac{m}{s^2}\\ \end{array}$

In this problem I'm not sure how to proceed. But my instinct tells me that the acceleration of ascention must be equal to the centripetal acceleration of the ball. But I'm confused exactly at how show I make FBD or something similar to see how forces are acting on the body, therefore a draw or sketch would be appreciated in order to spot exactly the justication of the following calculations.

If I were to ignore the thing that the wagon is on an incline, the bob would have:

$mg\cos 37^{\circ}=\frac{mv^2}{R}$

In this case the masses cancel, and the answer would be just $g\cos 37^{\circ}$. But this doesn't convince me much. Can someone help me here?.

Answer 1

As the car accelerates upward along the ramp with $\vec a$, the small sphere experience an effective constant gravity as $-\vec a$. Together with the downward $\vec g$ they make a net effective gravity $\overrightarrow{g_{\text{eff}}}$ that forms $16^{\circ}$ with the vertical line, as indicated in the diagrams below. This angle $16^{\circ}$ is understood as the (approximate) right triangle having the $7$-$24$-$25$ Pythagorean triple. Similarly, the angle $16^{\circ}$ is associated with the $3$-$4$-$5$ Pythagorean triple.

Denoting the magnitude of the car acceleration as $|\vec a| = a$ and the usual downward gravity as $|\vec g| = g = 10$ , the rightmost diagram reads

$$\frac{ \text{horizontal short leg} }{ \text{vertical long leg} } = \frac{\frac45 a}{ g + \frac35 a} = \frac7{24}$$ Solve for $a$ we have $$\frac{96}5 a = 7g + \frac{21}5 a \implies \frac{75}5 a = 70 \implies a = \frac{70}{15} \approx 4.6667$$

Thus the answer is the $5$th option, where the minor discrepancy is either a typo, or due to the approximated $g=10$ instead of $9.80$.

Answer 2

I think Lee David Chung Lin's answer is probably how you were intended to solve this problem, especially given the hint to assume that $\sin 16^\circ = \frac7{25}.$

I prefer the following acceleration diagram, however, decomposing the accelerations into components parallel to the ramp and perpendicular to the ramp:

Here we can make the usual assumption that $\sin 37^\circ = \frac35$. But there is no need to refer to the angle $16^\circ$ in any way.

To fill out the diagram, notice that we have two $3$-$4$-$5$ triangles, one inside the other. The only known acceleration is the hypotenuse of the smaller triangle, but you can use it to get both of the other sides of that triangle. Then you have one leg of the larger triangle and can use it to get the other sides. Finally, $a$ is the difference between the two collinear legs of the triangles.

Answer 3

As far as I understand, the ball does not move along the cavity, therefore, there's no centripetal acceleration of the ball. If so, you should write down the projected forces on the horizontal and vertical axes. Forces are following: gravitation force $mg$, supporting force $N$ and D'Alambert's force $-ma$. D'Alambert's force $-ma$ is the force opposite to acting acceleration. https://en.wikipedia.org/wiki/D%27Alembert%27s_principle

$-macos37+Nsin16=0$

$N=\frac {macos37} {sin16} $

$-mg-masin37+Ncos16=0$

$-mg-masin37+\frac {macos37} {sin16}cos16=0$

$a=\frac {g} {\frac {cos37} {sin16}cos16-sin37}=\frac {gsin16} {cos37cos16-sin37sin16}=\frac {gsin16} {cos53}=\frac {gsin16} {sin37}=\frac {10*7/25} {3/5}=\frac {14} {3}=4.67 $

$sin37=\sqrt \frac {1-cos74} {2}=\sqrt \frac {1-sin16} {2}=\sqrt \frac {1-\frac {7} {25}} {2}=\frac {3} {5}$

Seems like there's a typo in the last alternative.