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The solid wood door of a gymnasium is 1.00 m wide and 2.00

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 92P Chapter 10

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 92P

The solid wood door of a gymnasium is 1.00 m wide and 2.00 m high, has total mass 35.0 kg, and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 N to the door for 8.00 ms. Find the angular speed of the door after the impact. [?Hint: ?Integrating Eq. (10.29) yields is called the angular impulse.]

Step-by-Step Solution:

Solution 92P Step 1 of 5: Given solid wood door has width 1m and height 2m and mass M=35 kg is hit by basketball at the center of the door with a force of 1500 N to the door for 8 ms as shown in the figure below. We need to calculate the angular speed of door after impact. Step 2 of 5: Given data, Mass, M=35 kg Distance from axis, x=0.5 m Applied force, F= 1200 N Time interval, t =0 and t = 8ms= 8× 10 s 3 1 2 We know that,for linear motion; impulse is equal to change in linear momentum. Similarly in case of rotational impulse it is equal to change in angular momentum. That is, 2 L = dt t[ ] 1 Since here the torque is due to the force applied by basketball That is = F x Step 3 of 5: Above equation becomes, t2 L = [F x dt] 1 3 Substituting t =1 and t = 82 10 s 8×103 L = [F x dt] 0 On integrating 8×10 s L = F|x t 0 Substituting F=1200 N , x =0.5 m 3 L = 1200 N (0.5 m)(8 × 10 s) L = 6 kg.m/s 2

Step 4 of 5

Chapter 10, Problem 92P is Solved
Step 5 of 5

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The answer to “The solid wood door of a gymnasium is 1.00 m wide and 2.00 m high, has total mass 35.0 kg, and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 N to the door for 8.00 ms. Find the angular speed of the door after the impact. [?Hint: ?Integrating Eq. (10.29) yields is called the angular impulse.]” is broken down into a number of easy to follow steps, and 77 words. This full solution covers the following key subjects: Door, angular, hint, average, basketball. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 92P from chapter: 10 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. Since the solution to 92P from 10 chapter was answered, more than 947 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460. This textbook survival guide was created for the textbook: University Physics, edition: 13.

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