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Chapter 1: Problem 21

An insect is able to cling to the side of a car's tire when the car is going \(5 \mathrm{mph}\). How much harder is it for the insect to hold on when the speed is \(10 \mathrm{mph}\) ?

Short Answer

Expert verified
Answer: The insect has to hold on about 4 times harder when the car is going at 10 mph compared to when the car is going at 5 mph.

Step by step solution

01

Calculate the linear velocity at both speeds

Let's first convert the speeds from mph to m/s: 1 mph = 0.44704 m/s (get rid of miles and hour) At 5 mph: \(5\,\text{mph} \times 0.44704\,(\dfrac{meter}{1\,second}) = 2.2352\,\dfrac{meter}{1\,second}\) At 10 mph: \(10\,\text{mph} \times 0.44704\,(\dfrac{meter}{1\,second}) = 4.4704\,\dfrac{meter}{1\,second}\)
02

Calculate the ratio of the centrifugal forces at both speeds

Now, let's find the ratio of the forces at both speeds using the centrifugal force formula: Force at 5 mph: \(F_{5} = m(\dfrac{2.2352}{s})^2 / r\) Force at 10 mph: \(F_{10} = m(\dfrac{4.4704}{s})^2 / r\) To find the ratio, divide the force at 10 mph by the force at 5 mph: \(\dfrac{F_{10}}{F_{5}} = \dfrac{m(\dfrac{4.4704}{s})^2 / r}{m(\dfrac{2.2352}{s})^2 / r}\) As the mass (m) and the radius (r) are constant for both cases, they can be canceled out: \(\dfrac{F_{10}}{F_{5}} = \dfrac{(\dfrac{4.4704}{s})^2}{(\dfrac{2.2352}{s})^2}\) Now, square both linear velocities: \(\dfrac{F_{10}}{F_{5}} = \dfrac{(4.4704)^2}{(2.2352)^2}\)
03

Calculate the ratio

Now, let's calculate the ratio: \(\dfrac{F_{10}}{F_{5}} = \dfrac{(4.4704)^2}{(2.2352)^2} \approx 4\)
04

Conclusion

The insect has to hold on about 4 times harder when the car is going at 10 mph compared to when the car is going at 5 mph.

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