1 the wheel consists of a 5 kg slender ring with radius of...

Name: 1 the wheel consists of a 5 kg slender ring with radius of 05 m and four 2 kg rods each 05 m long what is the mass moment of inertia around the wheels center of gravity a 192 kg m2 b 079 kg 31003
Uploaded: 2025-02-12T00:18:44-08:00
Duration: 5 min 42 s
Channel: Eduard Sanchez
Description: 1 the wheel consists of a 5 kg slender ring with radius of 05 m and four 2 kg rods each 05 m long what is the mass moment of inertia around the wheels center of gravity a 192 kg m2 b 079 kg 31003

1. The wheel consists of a 5 kg slender ring with radius of 0.5 m and four, 2 kg rods (each 0.5 m long). What is the mass moment of inertia around the wheel's center of gravity? a. 1.92 kg-m² b. 0.79 kg-m² c. 0.625 kg-m² d. 2.96 kg-m² KNOWN: 1 ring = 5 kg r_s = 0.5 m 4 rods = 2 kg each r_r = 0.5 (0.5)² + 4 [1/12 (2/9) (0.5)²] = 0.288 .288 + 0.04 (0.5) = need || axis then too. 500 mm

Transcript

00:01 Hi there, so for this problem, the first problem in here is about moment of inertia of the wagon wheel.

00:13 So the first thing that we have for this is the radius for this that is 0 .3 meters.

00:21 The mass of the rim, it is 1 .40 kilograms.

00:30 Now, each spoke is a slender rod lying along the diameter.

00:35 The length for all for them will be 0 .3 meters.

00:41 The mass of one spoke, same about the last ms, it is 0 .280, and the number of spouts, it is 8.

00:55 Now, let's calculate the moment of inertia of the rim.

00:59 So for the moment of the dimension of the ring, that will be the mass of the rim times its radius to the square.

01:05 So now we substitute the values, 1 .0.

01:08 40 times 0 .3 to the square.

01:12 So this will give us about 0 .126 in units of kilograms times meters square.

01:21 Now the moment of inertia of this boats, each spoke is a thing rot rotating about its center.

01:31 So the moment of inertia of one spoke about the center is one divided by 3 times the moment of one spoke, this mass of one spoke times its length to the square.

01:43 So that will be 1 divided by 3 times 0 .28 times 0 .3 to the square.

01:51 So this will give us a value of 0 .0084 kilograms times give it meters, a meters square.

02:02 Sorry.

02:04 Now, since for the total is boats in here, we have 8, so we need to multiply that value by 8.

02:12 So that will be 0 .0672 kilograms times meters square.

02:19 So the total for this will be the sum of the 0 .126 for the ring plus 0 .6272.

02:30 So this will give us 0 .1932 in units of kilograms times meters square.

02:38 And that will be the solution for problem one.

02:42 Now for problem two, we are talking about rotational kinetic energy of the airplane propeller.

02:51 The length of the propeller for this, it is given, that is 2 .08 meters.

02:57 So then the radius is half of this value, that is 1 .04 meters.

03:01 The mass of the propeller also given, that is 117 kilograms, the rotational speed for this will be 2 ,400 revolutions per minute...

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