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A vertical wheel with a diameter of 39 cm starts from rest and rotates with a constant angular acceleration of 7.2 rad/s2 around a fixed axis through its center counterclockwise.

a) Through what angle (in degrees) has the point initially at the bottom of the wheel traveled when t = 14 s? (Indicate the direction with the sign of your answer.)

User Fernand
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Final answer:

The wheel's point initially at the bottom has traveled through an angle of approximately 40425.8 degrees after 14 seconds, considering its constant angular acceleration and counterclockwise direction.

Step-by-step explanation:

The student's question is about calculating the angular displacement of a wheel after a certain period of time when the wheel starts from rest and has a constant angular acceleration. Given that the wheel has a diameter of 39 cm and an angular acceleration of 7.2 rad/s², we can find the angle through which a point on the edge of the wheel has traveled after 14 seconds by using the angular kinematic equation:

\( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)

where:

  • \( \theta \) = angular displacement in radians
  • \( \omega_0 \) = initial angular velocity, which is 0 rad/s since it starts from rest
  • \( \alpha \) = angular acceleration
  • t = time in seconds

Plugging in the values, we get:

\( \theta = 0 \times 14 + \frac{1}{2} \times 7.2 \times 14^2 \)

\( \theta = \frac{1}{2} \times 7.2 \times 196 \)

\( \theta = 3.6 \times 196 \)

\( \theta = 705.6 \text{ rad} \)

To convert to degrees, we use the conversion factor where \( 1 \text{ rad} = \frac{180}{\pi} \text{ degrees} \).

\( \theta_{\text{degrees}} = 705.6 \times \frac{180}{\pi} \)

Then, the angle \

User Yiyuan Lv
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