Iam looking for the entire formula on how to solve this.
If a person on a 35.0 m diameter Ferris wheel feels 30% heavier at the bottom of the wheel, how fast is it going around?
Is N/Mg = 1 + v^2/Rg actual weight
mv^2/R * m + mg=1.30mg
.3gR=v^2
To solve this problem, we can use the concept of apparent weight and the formula for centripetal acceleration.
The formula you mentioned, N/ Mg = 1 + v^2/Rg, represents the relationship between the apparent weight (N) of a person on a rotating object, the actual weight (Mg) of the person, the speed of the object (v), the radius of the object (R), and the acceleration due to gravity (g).
In this case, the person feels 30% heavier at the bottom of the Ferris wheel, so their apparent weight (N) is 30% more than their actual weight (Mg). Mathematically, this can be expressed as N = (1 + 0.30) Mg = 1.30 Mg.
To determine the speed at which the Ferris wheel is rotating (v), we need to rearrange the formula:
N/ Mg = 1 + v^2/Rg
Substituting the value of N, we get:
1.30 Mg/ Mg = 1 + v^2/Rg
Simplifying:
1.30 = 1 + v^2/Rg
Rearranging the equation further, we can isolate v^2:
v^2/Rg = 0.30
Now, multiply both sides of the equation by Rg:
v^2 = 0.30 × Rg
Finally, to solve for v, take the square root of both sides of the equation:
v = √(0.30 × Rg)
Thus, the formula to find the speed at which the Ferris wheel is rotating is v = √(0.30 × Rg). You can plug in the values for R and g to calculate the speed.