Question

1. in order for a wheel of radius 1.5ft to have a max linear velocity of 80ft/s, what does the max rpm need to be? round to nearest whole number.
2. in order for a wheel to have a max linear velocity of 100ft/s at a max rpm of 200, what does the radius of the wheel need to be? round to nearest 10th.

Explanation:

Step1: Recall linear-angular velocity relation

The formula relating linear velocity \(v\), angular velocity \(\omega\) (in radians per second), and radius \(r\) is \(v = r\omega\). Also, \(\omega = \frac{2\pi \times \text{RPM}}{60}\), so substituting gives \(v = r \times \frac{2\pi \times \text{RPM}}{60}\).

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For Question 3:

Step2: Rearrange for RPM

Isolate RPM: \(\text{RPM} = \frac{v \times 60}{2\pi r}\)

Step3: Plug in values (\(v=80\), \(r=1.5\))

\[
\text{RPM} = \frac{80 \times 60}{2\pi \times 1.5} = \frac{4800}{3\pi} \approx \frac{4800}{9.4248} \approx 509.3
\]

Step4: Round to whole number

Round 509.3 to the nearest integer.

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For Question 4:

Step5: Rearrange for radius \(r\)

From \(v = r \times \frac{2\pi \times \text{RPM}}{60}\), solve for \(r\): \(r = \frac{v \times 60}{2\pi \times \text{RPM}}\)

Step6: Plug in values (\(v=100\), \(\text{RPM}=200\))

\[
r = \frac{100 \times 60}{2\pi \times 200} = \frac{6000}{400\pi} = \frac{15}{\pi} \approx 4.7746
\]

Step7: Round to nearest 10th

Round 4.7746 to one decimal place.

Answer:

1. 509
2. 4.8