Question

question 2
5 pts
the sun gear has 20 teeth and the outer ring gear has 54 teeth. the ring gear is going at 2000 rpm how fast is the sun gear going?
rpm

Explanation:

Step1: Recall Gear Ratio Formula

For gears, the speed (RPM) is inversely proportional to the number of teeth. The formula is \( \frac{N_1}{N_2}=\frac{\text{RPM}_2}{\text{RPM}_1} \), where \( N \) is the number of teeth and \( \text{RPM} \) is the rotational speed. Here, sun gear (1) has \( N_1 = 20 \), ring gear (2) has \( N_2 = 54 \), and ring gear RPM \( \text{RPM}_2 = 2000 \). We need to find \( \text{RPM}_1 \).

Step2: Rearrange Formula

From \( \frac{N_1}{N_2}=\frac{\text{RPM}_2}{\text{RPM}_1} \), rearrange to \( \text{RPM}_1=\frac{N_2\times\text{RPM}_2}{N_1} \).

Step3: Substitute Values

Substitute \( N_1 = 20 \), \( N_2 = 54 \), \( \text{RPM}_2 = 2000 \) into the formula: \( \text{RPM}_1=\frac{54\times2000}{20} \).

Step4: Calculate

Simplify \( \frac{54\times2000}{20}=54\times100 = 5400 \)? Wait, no, wait. Wait, the ring gear is going at 2000 RPM? Wait, the question says "The ring gear is going at 2000 RPM how fast is the sun gear going?" Wait, maybe I mixed up. Wait, sun gear (S) and ring gear (R). In a planetary gear system, the relationship is \( \text{RPM}_S\times N_S=\text{RPM}_R\times N_R \) (since speed is inversely proportional to teeth for meshing gears). So \( \text{RPM}_S=\frac{\text{RPM}_R\times N_R}{N_S} \). Wait, if ring gear (R) has 54 teeth, sun (S) has 20, and \( \text{RPM}_R = 2000 \), then \( \text{RPM}_S=\frac{2000\times54}{20}=2000\times2.7 = 5400 \)? Wait, no, maybe the direction? Wait, no, the formula is based on inverse proportionality. Wait, maybe I had the ratio reversed. Let's check again. For two gears meshing, the number of teeth times RPM is equal (since each tooth must mesh, so total teeth moved per minute are equal). So \( N_1\times\text{RPM}_1 = N_2\times\text{RPM}_2 \). So if sun gear is 1, ring gear is 2: \( 20\times\text{RPM}_S = 54\times2000 \). Then \( \text{RPM}_S=\frac{54\times2000}{20}=54\times100 = 5400 \). Wait, but that seems fast. Wait, maybe the ring gear is the driven or driving? Wait, the problem says "The ring gear is going at 2000 RPM how fast is the sun gear going?" So ring gear RPM is 2000, sun gear teeth 20, ring gear teeth 54. So using \( N_S\times\text{RPM}_S = N_R\times\text{RPM}_R \), so \( \text{RPM}_S=\frac{N_R\times\text{RPM}_R}{N_S}=\frac{54\times2000}{20}=5400 \). Wait, but maybe I made a mistake. Wait, no, the formula is correct for meshing gears: the product of teeth and RPM is constant (since each tooth must pass a point, so teeth per minute = N*RPM, which is equal for meshing gears). So yes, \( 20\times\text{RPM}_S = 54\times2000 \), so \( \text{RPM}_S = (54\times2000)/20 = 5400 \) RPM. Wait, but that seems high. Wait, maybe the ring gear is the driven gear? Wait, no, the question is as stated. So the calculation is \( \frac{54\times2000}{20}=5400 \).

Answer:

The sun gear is going at \(\boxed{5400}\) RPM. Wait, but wait, maybe I mixed up the gears. Wait, in a planetary gear system, the ring gear and sun gear rotate in the same direction? No, the formula is about the meshing, so the teeth and RPM are inversely proportional. So if ring gear has more teeth, it should rotate slower. Wait, wait, the ring gear is going at 2000 RPM, sun gear has fewer teeth, so sun gear should rotate faster. So 5400 RPM is faster than 2000, which makes sense because fewer teeth mean higher RPM for the same teeth per minute. So yes, the calculation is correct.