A pair of fuzzy dice m = 0.10 kg hang from the end of a string connected to the rearview mirror of a car.?

Question

If the car accelerates from a traffick light at 4.9 m/s2, what angle will the string make with the vertical?

Answer 1

Draw a force diagram, it will help you visualize what is occurring and make the problem much easier to solve.

Consider all of the forces acting on the pair of fuzzy dice….
There is, of course, the force of gravity (weight) which always acts on the dice in the straight downward direction. But the pair of dice are not accelerating downward, so this force must be getting canceled out somehow. Also, there is some accelerating force which is required for the dice to accelerate with the car in the horizontal direction.
Since the fuzzy dice are hanging on a string, there is also the force if tension in the string applied to the dice.

The force of tension is divided up into a vertical and horizontal component since the dice are hanging at an angle with respect to the vertical…that is some of the tension force is pulling the dice “up” and some of it is pulling it “sideways”. The upward component of the tension force is canceling out the force of gravity so the dice experience no net acceleration downward. The horizontal tension force is providing the force require for the dice to accelerate horizontally along with the car. The vector combination of these two horizontal and vertical forces yields the overall force of tension.

The angle between the horizontal and vertical forces will be the same as the angle of the string with respect to the vertical and can be found as,
Angle = arctan (ma / mg) = tan^-1 (ma / mg) = arctan (a / g)
Where m is the mass of the dice (which is actually not necessary to know in this problem), a is the horizontal acceleration of the car and/or dice (4.9 m/s^2), and g is the gravitational acceleration (9.81 m/s^2).
This angle becomes more apparent if you draw and label the force diagram.

Plugging into the equation to find the angle we get,
Angle = arctan (4.9 / 9.81)
Angle = 26.5° from the vertical.

Answer 2

0.10kg. = 0.22 lbs.

F = k/r²

0.22lbs = k/4.9m/s²

y=1/3x²

dy/dt = (1/3)(2x) dx/dt

Vy = 1/3 xVx

Vy = 1/3 (2) (1.078 lbs. m/s²) = 0.718 lbs. m/s²

V = √(4.9²) + (0.718²) = 4.95 lbs.m/s² ......magnitude

tan α = 4.9/0.718

tan α = 6.824512535

α = 81.66° ...............ans.