A rotating wheel requires 6.00 seconds to complete 29.0 revolutions. Its angular velocity at the end of the 6.00-second interval is 97.0 rad/s. What is the constant angular acceleration (in rad/s²) of the wheel?

Question

12-02-2025
Physics

High School

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A rotating wheel requires 6.00 seconds to complete 29.0 revolutions. Its angular velocity at the end of the 6.00-second interval is 97.0 rad/s. What is the constant angular acceleration (in rad/s²) of the wheel?

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damonsalvator7878 damonsalvator7878 · Answer 1 · 2023-10-12T18:31:39-04:00

The constant angular acceleration of the wheel is approximately [tex]\( -5.17 \, \text{rad/s}^2 \).[/tex]

We can use the equations of rotational motion to solve this problem. The angular acceleration [tex]\( \alpha \)[/tex] can be calculated using the formula:

[tex]\[ \alpha = \frac{\omega_f - \omega_i}{t} \][/tex]

where [tex]\( \omega_f \)[/tex] is the final angular velocity, [tex]\( \omega_i \)[/tex] is the initial angular velocity, and t is the time interval.

Given that the wheel rotates 29.0 revolutions in 6.00 seconds, we first find the initial angular velocity [tex]\( \omega_i \)[/tex] in radians per second:

[tex]\[ \omega_i = \frac{2 \pi \times \text{number of revolutions}}{\text{time}} \][/tex]

[tex]\[ \omega_i = \frac{2 \pi \times 29.0}{6.00} \][/tex]

[tex]\[ \omega_i \approx 30.49 \, \text{rad/s} \][/tex]

Substituting the given values into the formula for angular acceleration:

[tex]\[ \alpha = \frac{97.0 \, \text{rad/s} - 30.49 \, \text{rad/s}}{6.00 \, \text{s}} \][/tex]

[tex]\[ \alpha = \frac{66.51 \, \text{rad/s}}{6.00 \, \text{s}} \][/tex]

[tex]\[ \alpha \approx -5.17 \, \text{rad/s}^2 \][/tex]

So, the constant angular acceleration of the wheel is approximately [tex]\( -5.17 \, \text{rad/s}^2 \)[/tex]. The negative sign indicates that the wheel is slowing down.