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Chapter 12: Q. 17 (page 330)

A 25 kg solid door is 220 cm tall, 91 cm wide. What is the door's moment of inertia for (a) rotation on its hinges and (b) rotation about a vertical axis inside the door, 15 cm from one edge?

Short Answer

Expert verified

The moment of inertia about the edge of slab is 6.9kgm2

The moment of inertia about 15 cm from the edge of slab is4.1kgm2

Step by step solution

01

Step 1. Given information

A mass of solid door m=25kg, height of door a=220cm, , and 91cm wide.

02

Part (a)

The door is a slab of uniform density.

The hinges are at the edge of the door, So moment of inertia about the edge of slab is

I=13Ma2I=1325kg0.91m2=6.9kgm2

The moment of inertia about the edge of slab is6.9kgm2

03

Part (b)

The distance from the axis through the center of mass along the height of the door is

d=0.91m2鈭�0.15m=0.305m

. Using the parallel-axis theorem,

The moment of inertia about 15 cm from the edge of slab is

I=Icm+Md2=11225kg0.91m2+25kg0.305cm2=4.1kgm2

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