Question about fuzzy dice on force and inertia

Question

I cant really show the image but I'll try to describe it best for you to solve: Imagine a rearview mirror and attached to it is your typical fuzzy dice connect with a string (that does not stretch). The fuzzy dice is at an angle swinging to the left so there will be a dashed line going straight down so inbetween would be the angle for #1.

1) The angle between the string and the vertical line is theta = 26 degrees. What is the normal component (also known as the radial component) of the acceleration of the car at this time?

[HINT: It might (and then again it might not) be useful to imagine this happening in space with a car that is accelerating upward and in a radial direction.]

2) At the time shown in the diagram, the car was going around a curve with a radius of curvature of 45 m. What was the speed of the car?

Note: ANSWER 1-2 because they correlate with each other make sure to bold answer so I will know and show work

Answer 1

Nathan, here's how to handle this.

Draw a diagram. Draw the dice, hanging on a string at a 26^o angle to the vertical, to the left of the vertical.

There is a tension, T, in the string, acting along the direction of the string.

T has vertical component T sin 26^o acting upwards, and horizontal (radial) component, acting to the right.

Since the dice is not moving, the forces in the vertical and horizontal (radial) directions must be in equilibrium, that is, balancing to zero.

Vertically, we have T sin26^o acting upward and mg, the weight of the dice, acting downward.

So, T sin26^o = mg . . . . . . . . (1)

Horizontally, we have T cos26^o acting to the right and the centrifugal force, mv²/r acting to the left

Therefore, T cos26^o = mv²/r . . . .(2) This is the required normal (radial) component. the answer to part (a)

Divide (1) by (2): tan26^o = v²/gr

: tan26^o = v²/45g

I leave the calculation of v, the car's speed, to you as an exercise.