Physics I should be able to do

[JonF]

For some reason I can’t solve this:

10. [PSE6 1.P.039.] One cubic meter (1.00 m3) of aluminum has a mass of 2.70 103 kg, and 1.00 m3 of iron has a mass of 7.86 103 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 3.10 cm on an equal-arm balance.

I keep getting $$r =(\frac{2.7*3.1^3}{7.86} )^{1/3}$$

 

[jamesrc]

Since it's an equal arm balance, the mass of the iron ball will equal the mass of the aluminum ball:

$$m_{\rm Fe} = m_{\rm Al}$$
$$\rho_{\rm Fe}V_{\rm Fe} = \rho_{\rm Al}V_{\rm Al}$$
$$\rho_{\rm Fe}\frac{4\pi r_{\rm Fe}^3}{3} = \rho_{\rm Al}\frac{4\pi r_{\rm Al}^3}{3}$$

Cancel out the common terms and solve for r_Fe in terms of the given parameters. If you follow that, you should find your mistakes.

 

[JonF]

i thought that's what i did

 

[JonF]

here are some other ones we arn't getting:

10.An auditorium measures 40.0 m 15.0 m 11.0 m. The density of air is 1.20 kg/m3.
(b) What is the weight of air in the room in pounds?
i found the volume to be: 233077ft^3 not sure what do form there.

11(b)Consider a solid disk of mass M=12.3kg and areal density =Dx3. Determine the value D if the radius of the disk is 0.25m.
I don’t even know where to start.

2. What level of precision is implied in an order-of-magnitude calculation?
i thought it was 3

 

[Leong]

Rearrange the equation by jamesrc :

$$r_{Al} =\sqrt[3]{\frac{\rho_{Fe}}{\rho_{Al}}r^3_{Fe}}$$

$$r_{Al}=\sqrt[3]{\frac{7.86103}{2.70103}(0.0310)^3}$$
= 0.0443 m

$$\rho=\frac{m}{v}$$
NOT
$$\rho=\frac{v}{m}$$

* Keep track of the units. Here, I used SI units.

 

[Leong]

here are some other ones we arn't getting:

10.An auditorium measures 40.0 m 15.0 m 11.0 m. The density of air is 1.20 kg/m3.
(b) What is the weight of air in the room in pounds?
i found the volume to be: 233077ft^3 not sure what do form there.

11(b)Consider a solid disk of mass M=12.3kg and areal density =Dx3. Determine the value D if the radius of the disk is 0.25m.
I don’t even know where to start.

10.(a) It is better to use the units given and then convert it lastly.
$$233077ft^3=6600m^3$$
Using the formula for density:
$$\rho_{air}=\frac{m_{air}}{v_{air}}$$
will help you to find the mass of the air.
Then convert the mass to pound unit.

(b) I think it asks us to modify the formula for density a little bit so that it can be applied in the question. Since the disk is flat, i think we should use this :

$$\rho_{disk} = \frac{m_{disk}}{A_{disk}} = 3D$$

 

[JonF]

sorry for 11(b) it is dx^3.

so i get 12.3/pi(.25)^2 = dx^3 not sure where to go from there


but thank you guys i got all the rest

 

[Leong]

JonF,
Omit the hint given for 11(b). it looks useless now since the question has changed. what is x ? i am stuck as well.

 

[JonF]

Well the first question was
11. (a)Consider a two meterstick of mass M=18.2kg and linear density =Cx^4. Determine the value C.

So we did this: $$c*\int_0^2 x^4 dx= 18.2$$ and it was right.

This is the other question.

b) Consider a solid disk of mass M=12.3kg and areal density =Dx^3. Determine the value D if the radius of the disk is 0.25m.

 

[Leong]

$$m_{disk}=\rho_{disk}A_{disk}$$
See the attached image :
$$dm=Dx^3[\frac{1}{2}(d\theta)(r^2)]$$ ...(1)
$$x=rcos\theta$$...(2)
(2)$$\rightarrow$$(1)
$$dm=\frac{D}{2}r^5cos^3\theta d\theta$$
$$\int_0^m dm = \int_0^{2\pi} \frac{D}{2}r^5 cos^3\theta d\theta$$
Hint :

1. $$cos^3\theta = cos\theta cos^2\theta$$
2.$$cos2\theta = 2cos^2\theta -1$$
3.$$cosAcosB = \frac{1}{2}[cos(A+B) + cos (A-B)]$$

 

[JonF]

uh you so lost me, but thanks for you help

 

[HallsofIvy]

The very complicated formula Leong used was from saying that x, in polar coordinates, is given by x= r cos(θ). If the problem has said the density was
Dr³, that would be much simpler.

 

[Leong]

JonF,
I tried to integrate the equation and found that it equaled to zero!

I want to say sorry because you might think that i was such a *** person trying to teach others to do some problems when he himself didn't even try it first!

Since the disk areal density is given by $$Dx^3$$; then $$x \geq 0$$ because the density can never be negative.

So, my flawed equation is only valid for this range of $$\theta$$ : $$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$$.

You can integrate for these intervals:
1. $$[0,\frac{M}{4}]$$ and $$[0,\frac{\pi}{2}]$$ : For this value of $$\theta$$, the mass of the disk is just a quarter of the total mass since the mass of the disk is proportional to the area of the disk.
2.$$[0,\frac{M}{4}]$$ and $$[-\frac{\pi}{2},0,]$$ : For this value of $$\theta$$, the mass of the disk is just a quarter of the total mass since the mass of the disk is proportional to the area of the disk.
3.$$[0,\frac{M}{2}]$$ and $$[-\frac{\pi}{2},\frac{\pi}{2}]$$ : For this value of $$\theta$$, the mass of the disk is half of the total mass since the mass of the disk is proportional to the area of the disk.

All these intervals will yield the same results.

I choose option #1 to do this :

$$\int_0^\frac{M}{4} dm = \int_0^\frac{\pi}{2} \frac{D}{2}r^5 cos^3\theta d\theta$$

$$cos^3\theta = \frac{1}{4}[cos 3\theta + 3cos\theta]$$

$$[m]_0^\frac{M}{4}=\frac{D}{2}r^5\int_0^{\frac{\pi}{2}} \frac{1}{4}(cos3\theta +3cos\theta) d\theta$$

$$\frac{M}{4}=\frac{D}{2}r^5(\frac{1}{4})[\frac{sin3\theta}{3}+3sin\theta]_0^\frac{\pi}{2}$$

$$M=\frac{D}{2}r^5[\frac{8}{3}]$$

$$M=\frac{4}{3}Dr^5$$

$$D=\frac{3M}{4r^5}$$

$$D=\frac{3(12.3)}{4(0.25)^5}$$

$$=9.4X10^3$$ $$kg/m^5$$