An ant climbs onto the side of a bicycle tire a distance of 40.0 m from the hub. If the 0.010 g ant can hold onto the tire with a force of 4.34*10 to the -4, at what frequency would the tire fling the ant of assuming the wheel is spun on a horizontal plane?
so what I did for this question was!!
I used the formula for Force centripetal, and substituted everything and solved for T which is the "frequency". But that was not the answer the real answer is 1.7HZ
To solve this problem, we can start by using the formula for centripetal force:
F_centripetal = m * ω^2 * r
Where:
F_centripetal is the centripetal force
m is the mass of the ant
ω is the angular velocity of the tire (in radians per second)
r is the distance of the ant from the hub
We are given:
m = 0.010 g = 0.010 * 10^(-3) kg (convert grams to kilograms)
F_centripetal = 4.34 * 10^(-4) N
r = 40.0 m
Let's rearrange the formula to solve for ω:
ω^2 = F_centripetal / (m * r)
ω = √(F_centripetal / (m * r))
Substituting the given values:
ω = √(4.34 * 10^(-4) N / (0.010 * 10^(-3) kg * 40.0 m))
ω = √(4.34 * 10^(-4) / (0.010 * 40.0)) rad/s
Now, we can calculate the frequency using the relation between ω and frequency:
frequency = ω / (2π)
Substituting the value of ω:
frequency = √(4.34 * 10^(-4) / (0.010 * 40.0)) / (2π)
frequency ≈ 1.67 Hz
So, the calculated frequency is approximately 1.67 Hz. The given answer of 1.7 Hz may be an approximation or rounded value.