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A gear with a diameter of 12.0 inches makes 35 revolutions every three minutes. Find the linear and angular velocities of a point on the outer edge of the gear. Give your answers in both exact and approximate forms.

User Richard A
by
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1 Answer

5 votes

Answer:

Angular velocity:
\omega = (7\pi)/(18) \,(rad)/(s) (
1.222\,(rad)/(s))

Linear velocity:
v = (14\pi)/(3)\,(m)/(s) (
14.661\,(m)/(s))

Explanation:

The gear experiments a pure rotation with axis passing through its center, the angular (
\omega), in radians per second, and linear velocities (
v), in inches per second, of a point on the outer edge of the element are, respectively:


\omega = (2\pi)/(60)\cdot \dot n (1)


v = R\cdot \omega (2)

Where:


\dot n - Rotation rate, in revolutions per minute.


R - Radius of the gear, in inches.

If we know that
\dot n = (35)/(3)\,(rev)/(min) and
R = 12\,in, then the linear and angular velocities of the gear are, respectively:


\omega = (2\pi)/(60)\cdot \dot n


\omega = (7\pi)/(18) \,(rad)/(s) (
1.222\,(rad)/(s))


v = R\cdot \omega


v = (14\pi)/(3)\,(m)/(s) (
14.661\,(m)/(s))

User Adam Merrifield
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4.3k points