How to find the volume of this pulley?

[SagarPatil]

Homework Statement

The problem is asking to find the volume of this pulley

Formulas that is in the chapter.
Area = r^2*θ/(2)
Where θ is in radians

Homework Equations

I drew the drawing in solid works and got the volume to come out to 6.27 cubic inch. But I am not sure if this is the correct answer.

The Attempt at a Solution

I have no idea where to start. There is question in the chapter that is asking for the volume except this.

 

[berkeman]

Homework Statement

The problem is asking to find the volume of this pulley
View attachment 84150
Formulas that is in the chapter.
Area = r^2*θ/(2)
Where θ is in radians

Homework Equations

I drew the drawing in solid works and got the volume to come out to 6.27 cubic inch. But I am not sure if this is the correct answer.

The Attempt at a Solution

I have no idea where to start. There is question in the chapter that is asking for the volume except this.

Volume is just Area * Thickness, correct? So start by finding the top area of the figure. It is composed of the pie-shaped arc and the cylinder, with part of each cut away.

Make a sketch of the top of the figure, and start adding and subtracting areas. Show us what you get...

 

[SagarPatil]

Ok, makes sense

So,

Area of the sector = 5.270^2 * 0.97(rad) / 2 = 13.5
Area of the big circle = 1.23
Area of small circle = 0.2

I subtract 0.2 so it comes out to 14.53

14.53 * 0.250 = 3.63 cubic inch.

 

[berkeman]

Ok, makes sense

So,

Area of the sector = 5.270^2 * 0.97(rad) / 2 = 13.5
Area of the big circle = 1.23
Area of small circle = 0.2

I subtract 0.2 so it comes out to 14.53

14.53 * 0.250 = 3.63 cubic inch.

Careful -- it looks like you may be double-counting a small piece of area. Do you see where it is?

Also, it's a good idea to carry units along in your calculations. A number without units can be a problem...

 

[SagarPatil]

I don't see where I did wrong. I did it a couple of times and got the same answer.

 

[Mark44]

Ok, makes sense

So,

Area of the sector = 5.270^2 * 0.97(rad) / 2 = 13.5

Your area for the sector is wrong, and as berkeman said, you're counting part of the area twice. You're calling the radius of the sector 5.270". You need to subtract off the part of the sector that is in the circular portion.

Area of the big circle = 1.23
Area of small circle = 0.2

I subtract 0.2 so it comes out to 14.53

14.53 * 0.250 = 3.63 cubic inch.

 

[Mark44]

I don't see where I did wrong. I did it a couple of times and got the same answer.

If you do the same sequence of steps, but they aren't the right steps, you won't get the right answer. If you're careful, you'll get the same wrong answer twice.

 

[SagarPatil]

Ok I understand that i am counting it twice

If I subtract 13.5-1.23-0.2 = 12.07 this gives me the area of the sector

So now

12.07 + 1.23 - 0.2 = 13.1 is the total area

13.1*0.250 = 3.275 cubic inch

 

[SagarPatil]

is the answer correct ?

 

[SteamKing]

Ok I understand that i am counting it twice
View attachment 84155
If I subtract 13.5-1.23-0.2 = 12.07 this gives me the area of the sector

So now

12.07 + 1.23 - 0.2 = 13.1 is the total area

13.1*0.250 = 3.275 cubic inch

Remember, subtracting the area of the hub, 1.23 in², from the area of the sector, 13.5 in², automatically includes the area of the hole, 0.20 in². Subtracting 0.20 in² for the hole again is incorrect.

 

[SagarPatil]

Hello, I got it.

You so here is what I did

Area of the Big sector = 5.270^2 * 0.97(rad) / 2 = 13.5
Area of Small Sector = 0.625^2 * 0.97(rad) / 2 = 0.189
Area of the big circle = 1.23
Area of small circle = 0.2

So,

13.5+1.23-0.2-0.189 = 14.341

14.341 * 0.250 = 3.58 in^3

 

[theodoros.mihos]

Fix your formula for area.

 

[Mark44]

Fix your formula for area.

Do you want to elaborate on which formula Sagar wrote? He is showing four area calculations.

 

[theodoros.mihos]

There is no division by 2.

 

[Mark44]

There is no division by 2.

Are you saying that the OP didn't divide by 2 or are you saying that the formula for the area of a sector shouldn't have a divisor of 2? If you are saying the latter, then you're mistaken. The area of a sector of a circle of radius r, where the sector subtends an arc of $\theta$ radians, is $A = \frac 1 2 r^2 \theta$.