College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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### Description of the Disk and Block System

A uniform horizontal disk with a radius of 5.50 m rotates without friction at an angular velocity of \( \omega = 2.30 \) revolutions per second. The disk has a feedback mechanism that senses its angular speed. A drive motor at point A maintains the system's constant angular speed. A block with a mass of \( m = 1.20 \) kg sits on the disk's center and slides outward in a radial slot. Starting at \( t = 0 \), the block moves at a constant speed \( v = 1.25 \) cm/s relative to the disk until reaching the edge at \( t = 465 \) s. There's no friction during this sliding, and the block's motion is further constrained by a brake at B, which induces tension in a string tied to the block.

#### Key Tasks and Calculations

- **(a) Calculating Torque:** Determine the function of time that describes the torque the drive motor provides while the block is sliding. The formula to use is \( \tau = 2mr\omega v \).

- **(b) Torque at a Specific Time:** Calculate the torque at \( t = 465 \) s, right before the block finishes sliding.
  - **Answer:** 2.52 N·m

- **(c) Calculating Power:** Find the function for power delivered by the drive motor over time.

- **(d) Power at the End of the Slot:** Identify the power when the block reaches the end of the slot.
  - **Answer:** 18.2 W (incorrect in the original transcript)

- **(e) String Tension:** Compute the string tension as a function of time.

- **(f) Motor Work Done:** Calculate the work done by the drive motor during the 465 s motion. The formula given is \( W_{\text{motor}} = mv^2\omega^2t^2 \).
  - **Answer:** 8.47 kJ

- **(g) Work Done by String Brake:** Find the work done by the string brake on the sliding block, which is negative due to the opposing direction.
  - **Formula:** \( W_{\text{block}} = -\frac{1}{2}mv^2\omega^2t^2 \).
  - **Answer:** -4.23 kJ

- **(h
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Transcribed Image Text:### Description of the Disk and Block System A uniform horizontal disk with a radius of 5.50 m rotates without friction at an angular velocity of \( \omega = 2.30 \) revolutions per second. The disk has a feedback mechanism that senses its angular speed. A drive motor at point A maintains the system's constant angular speed. A block with a mass of \( m = 1.20 \) kg sits on the disk's center and slides outward in a radial slot. Starting at \( t = 0 \), the block moves at a constant speed \( v = 1.25 \) cm/s relative to the disk until reaching the edge at \( t = 465 \) s. There's no friction during this sliding, and the block's motion is further constrained by a brake at B, which induces tension in a string tied to the block. #### Key Tasks and Calculations - **(a) Calculating Torque:** Determine the function of time that describes the torque the drive motor provides while the block is sliding. The formula to use is \( \tau = 2mr\omega v \). - **(b) Torque at a Specific Time:** Calculate the torque at \( t = 465 \) s, right before the block finishes sliding. - **Answer:** 2.52 N·m - **(c) Calculating Power:** Find the function for power delivered by the drive motor over time. - **(d) Power at the End of the Slot:** Identify the power when the block reaches the end of the slot. - **Answer:** 18.2 W (incorrect in the original transcript) - **(e) String Tension:** Compute the string tension as a function of time. - **(f) Motor Work Done:** Calculate the work done by the drive motor during the 465 s motion. The formula given is \( W_{\text{motor}} = mv^2\omega^2t^2 \). - **Answer:** 8.47 kJ - **(g) Work Done by String Brake:** Find the work done by the string brake on the sliding block, which is negative due to the opposing direction. - **Formula:** \( W_{\text{block}} = -\frac{1}{2}mv^2\omega^2t^2 \). - **Answer:** -4.23 kJ - **(h
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