Answer :
Final answer:
The angular acceleration of the potter's wheel is -3.84 rad/s², indicating a deceleration as the wheel comes to a stop.
Explanation:
The question asks for the angular acceleration of a potter's wheel. The wheel is initially rotating at 66 revolutions per minute (rev/min) and is brought to a stop with a duration of 1.8 seconds by applying a wet rag to the rim. To find the angular acceleration, the initial angular velocity needs to be converted to radians per second (rad/s), the final angular velocity is 0 rad/s (since the wheel stops), and then the angular acceleration can be calculated using the formula α = (ωf - ωi) / t, where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time in seconds.
Conversion from rev/min to rad/s is done by multiplying by (2π rad/1 rev) × (1 min/60 s). The initial angular velocity in rad/s is (66 rev/min) × (2π rad/1 rev) × (1 min/60 s) = 6.91 rad/s (rounded to two decimal places). Plugging into the angular acceleration formula:
Angular acceleration (α) = (0 rad/s - 6.91 rad/s) / 1.8 s = -3.84 rad/s² (negative sign indicates deceleration).
