A jewel smith wishing to buff a finished piece of jewelry attaches a buffing disk to his drill. The radius of the disk is 3.90 mm, and he operates it at 2.05  104 rad/s. The jeweler increases the operating speed so that the tangential speed of the rim of the disk is now 275 m/s. What is the period of rotation, in seconds, of the disk now?

To find the period of rotation (\( T \)) of the disk, we need to understand that the jeweler increases the speed such that the tangential velocity (\( v \)) is now 275 m/s. 

We are given:
- Radius of the disk: \( r = 3.90 \, \text{mm} = 3.90 \times 10^{-3} \, \text{m} \) (since 1 mm = \( 10^{-3} \) m)
- Tangential speed: \( v = 275 \, \text{m/s} \)

### Step 1: Determine the Angular Velocity (\( \omega \))

We relate tangential speed and angular velocity (\( \omega \)) by the equation:

\[
v = r \cdot \omega
\]

Rearranging to find \( \omega \):

\[
\omega = \frac{v}{r}
\]

Substituting the given values:

\[
\omega = \frac{275 \, \text{m/s}}{3.90 \times 10^{-3} \, \text{m}}
\]

\[
\omega = 70,512.82 \, \text{rad/s}
\]

### Step 2: Determine the Period of Rotation

The relationship between the period of rotation (\( T \)) and the angular velocity (\( \omega \)) is given by:

\[
\omega = \frac{2 \pi}{T}
\]

Rearranging to solve for \( T \):

\[
T = \frac{2 \pi}{\omega}
\]

Substituting the calculated \( \omega \):

\[
T = \frac{2 \pi}{70,512.82}
\]

\[
T \approx 8.91 \times 10^{-5} \, \text{s}
\]

### Final Answer
The period of rotation of the disk is approximately \( 8.91 \times 10^{-5} \) seconds.

Would you like further details on any step or have any questions?

Here are 5 related questions to deepen your understanding:
1. How does increasing the radius affect the period of rotation if the tangential speed is kept constant?
2. What would be the period if the tangential speed was doubled, but the radius remained the same?
3. How can you calculate the number of rotations per minute from the period?
4. If the radius was given in centimeters, how would the conversion affect your calculations?
5. How does the angular velocity relate to linear velocity in other rotating objects like wheels or gears?

**Tip:** When solving rotational motion problems, always check the units to avoid confusion, especially with radius measurements, since it's common to be given in mm or cm.

A jewel smith wishing to buff a finished piece of jewelry attaches a buffing disk to his drill. The radius of the disk is 3.90 mm, and he operates it at 2.05 x 10^4 rad/s. The jeweler increases the operating speed so that the tangential speed of the rim of the disk is now 275 m/s. What is the period of rotation, in seconds, of the disk now?

Discover how to calculate the period of rotation of a buffing disk used in jewelry making, with an increased tangential speed of 275 m/s and a radius of 3.90 mm. This problem covers essential concepts in rotational motion, including angular velocity and period calculation, suitable for Grades 11-12.

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