MHB Angular Speed of Smaller Wheel

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To determine the angular speed of the smaller wheel, the relationship between the linear speeds of both wheels must be used, given by the equations v = rω1 and v = Rω2. With r at 6 cm and R at 10 cm, and the larger wheel's angular speed (ω2) at 100 rpm, the formula ω1 = (Rω2)/r leads to ω1 = (10 * 100)/6. This calculation results in ω1 = 500/3 rpm. Converting this to radians per minute gives the final answer as 1000π/3 radians/min.
mathdad
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Let r = radius

See picture.

If r = 6 cm, R = 10 cm, and the angular speed of the larger wheel is 100 rpm, determine the angular speed of the smaller wheel in radians per minute.

Again, 1 revolution = 2 pi radians.

I need to use w = θ/t.

So, θ = 100 rpm • 2 pi radians.

θ = 200 pi radians

w = 200 pi radians/min, which is my answer.

At this point, I doubted that my answer is right because it did not take long to find w or omega representing the angular speed.

The book's answer is 1000/3 radians/min.

What did I forget to do?

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f r = 6 cm, R = 10 cm, and the angular speed of the larger wheel is 100 rpm, determine the angular speed of the smaller wheel in radians per minute.

$v = r \omega_1$, where $\omega_1$ is the angular speed of the smaller wheel

$v = R \omega_2$, where $\omega_2$ is the angular speed of the larger wheel.

the belt connecting the two wheels has a fixed linear speed, $v$ ...

$\implies r \omega_1 = R \omega_2$

$\omega_1 = \dfrac{R \omega_2}{r} = \dfrac{10 \cdot 100}{6} = \dfrac{500}{3} \, rpm$

convert $\omega_1$ to radians per minute ...

$\dfrac{500 \, rev}{3 \, min} \cdot \dfrac{2\pi \, rad}{rev} = \dfrac{1000\pi}{3} \, rad/min$
 
Nicely-done. I will try your solution steps for this question to answer the textbook question where r = 6 cm and R = 10 cm.
 
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