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A uniform horizontal disk of radius 5.50 m turns without friction at = 2.30 rev/s on a vertical axis through its center, as in the figure below. A feedback mechanism senses the angular speed of the disk, and a drive motor at A ensures that the angular speed remain constant while a m = 1.20 kg block on top of the disk slides outward in a radial slot. The block starts at the center of the disk at time t = 0 and moves outward with constant speed v = 1.25 cm/s relative to the disk until it reaches the edge at t = 465 s. The sliding block experiences no friction. Its motion is constrained to have constant radial speed by a brake at B, producing tension in a light string tied to the block. (a) Find the torque as a function of time that the drive motor must provide while the block is sliding. Hint: The torque is given by T 2mrvw. = t N.m (b) Find the value of this torque at t = 465 s, just before the sliding block finishes its motion. 2.52 N-m (c) Find the power which the drive motor must deliver as a function of time. t W/s (d) Find the value of the power when the sliding block is just reaching the end of the slot. 18.2 x w (e) Find the string tension as a function of time. ]t N/s

A uniform horizontal disk of radius 5.50 m turns without friction at = 2.30 rev/s on a vertical axis through its center, as in the figure below. A feedback mechanism senses the angular speed of the disk, and a drive motor at A ensures that the angular speed remain constant while a m = 1.20 kg block on top of the disk slides outward in a radial slot. The block starts at the center of the disk at time t = 0 and moves outward with constant speed v = 1.25 cm/s relative to the disk until it reaches the edge at t = 465 s. The sliding block experiences no friction. Its motion is constrained to have constant radial speed by a brake at B, producing tension in a light string tied to the block. (a) Find the torque as a function of time that the drive motor must provide while the block is sliding. Hint: The torque is given by T 2mrvw. = t N.m (b) Find the value of this torque at t = 465 s, just before the sliding block finishes its motion. 2.52 N-m (c) Find the power which the drive motor must deliver as a function of time. t W/s (d) Find the value of the power when the sliding block is just reaching the end of the slot. 18.2 x w (e) Find the string tension as a function of time. ]t N/s

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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
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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