Question

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.

Answer 1

Answer:

Angular velocity: $\omega = \frac{7\pi}{18} \,\frac{rad}{s}$ ($1.222\,\frac{rad}{s}$)

Linear velocity: $v = \frac{14\pi}{3}\,\frac{m}{s}$ ($14.661\,\frac{m}{s}$)

Step-by-step 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:

$\omega = \frac{2\pi}{60}\cdot \dot n$ (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 $\dot n = \frac{35}{3}\,\frac{rev}{min}$ and $R = 12\,in$, then the linear and angular velocities of the gear are, respectively:

$\omega = \frac{2\pi}{60}\cdot \dot n$

$\omega = \frac{7\pi}{18} \,\frac{rad}{s}$ ($1.222\,\frac{rad}{s}$)

$v = R\cdot \omega$

$v = \frac{14\pi}{3}\,\frac{m}{s}$ ($14.661\,\frac{m}{s}$)