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# A solid wood door 1.00 m wide and 2.00 m high is hinged ISBN: 9780321675460 31

## Solution for problem 46E Chapter 10

University Physics | 13th Edition

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Problem 46E

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 kg, traveling perpendicular to the door at 12.0 m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

Step-by-Step Solution:

Solution 46E Step 1: Length of the door l = 1.00 m Mass of the door M = 40.00 Kg Mass of the mud m = 0.5 Kg Speed of the door v = 12.0 m/s Step 2: To find the angular speed = Angular momentum of the mud (L-----(1) Moment of Inertia (I) Step 3: Since the mud is stuck at the centre of the door, radius of the circular path made by the mud r = l/2. Hence the Angular momentum of the mud would be equal to L = mvr = mvl------(2) 2 Step 4: Moment of Inertia of door and mud are combined to calculate total moment of inertia. I = Moment of inertia of door + moment of inertia of mud 2 Moment of inertia of door = Ml 1 3 1 1 2 Moment of inertia of the mud = m( ) 2 2 I = Ml + m( ) ------(3) 3 2 2

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##### ISBN: 9780321675460

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A solid wood door 1.00 m wide and 2.00 m high is hinged

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