Mass density of cylinder?

[Seiryuu]

A

6.70 cm-tall cylinder floats in water with its axis perpendicular to the surface. The length of the cylinder above water is 1.70 cm.

Ok, so there's 5.00 cm in height of cylinder above water. The density of water is [math]\frac{1000kg}{m^3}[/math] or [math]\frac{1g}{cm^3}[/math].

What is the mass density of the cylinder?

Is this even possible to do without the radius?

Edited January 17, 2012 by Seiryuu

[Daedalus]

Yes you can actually solve this problem without knowing the radius of the cylinder. Since I can't actually give you the answer, I will help you set up the equations so you can solve it. First, let's define some variables that will make our equations clear and understandable.

 

The height of the cylinder is:

 

[math]H_{cyl} = 6.70 \, \text{cm}[/math]

 

The height of the cylinder that is submerged or under water is:

 

[math]H_{sub} = 6.70 \, \text{cm} - 1.70 \, \text{cm} = 5.0 \, \text{cm}[/math]

 

You should know that the volume of a cylinder is:

 

[math]V = h\, \pi \, r^2[/math]

 

Now that we have the above variables and equation for the volume of a cylinder, we can begin to solve this problem.

 

The magnitude of buoyancy force may be appreciated a bit more from the following argument. Consider any object of arbitrary shape and volume V surrounded by a liquid. The force the liquid exerts on an object within the liquid is equal to the weight of the liquid with a volume equal to that of the object. This force is applied in a direction opposite to gravitational force, that is of magnitude:

 

[math]B=\rho_f \, V_{disp} \, g[/math]

 

where ρ_f is the density of the fluid, V_disp is the volume of the displaced body of liquid, and g is the gravitational acceleration at the location in question.

 

If this volume of liquid is replaced by a solid body of exactly the same shape, the force the liquid exerts on it must be exactly the same as above. In other words the "buoyancy force" on a submerged body is directed in the opposite direction to gravity and is equal in magnitude to:

 

[math]B=\rho_f \, V \, g[/math]

We can see from the above quote from Wikipedia that the magnitude of the buoyancy force is:

 

[math]B=\rho_f \, V_{disp} \, g[/math]

 

We know that the force of gravity is trying to pull the cylinder down and that the buoyancy force is trying to raise the cylinder up. The force of gravity acting upon the cylinder is:

 

[math]F = m\, g[/math]

 

However, the cylinder is neither sinking or rising. This means that the net force on the cylinder is zero and the magnitude of the buoyancy force is equal to the magnitude of the gravitational force:

 

[math]F_{net} = m\, g - \rho_f \, V_{disp} \, g = 0[/math]

 

We can rearrange the terms to show that the forces are equal in magnitude:

 

[math]m\, g =\rho_f \, V_{disp} \, g[/math]

 

Now that the forces are balanced, we can divide both sides by the gravitational acceleration to obtain the mass of the cylinder:

 

[math]m =\rho_f \, V_{disp}[/math]

 

 

 

You should be able to work the problem from here on because:

 

1.) You know that the density of water is [math]\rho_f = 1\, \text{gram} / \text{cm}^3[/math]

 

2.) You know that the volume of water displaced is equal to the volume of the cylinder that is submerged [math]V_{disp} = H_{sub}\, \pi \, r^2[/math]

 

3.) You know that the mass of the cylinder is [math]m =\rho_f \, V_{disp}[/math]

 

4.) You know that the volume of the cylinder is [math]V_{cyl} = H_{cyl}\, \pi \, r^2[/math]

 

5.) You know that density is equal to [math]m / V_{cyl}[/math]

 

 

Make sure that you take into account the final units that your teacher wants when you work out the answer!!!

Edited January 18, 2012 by Daedalus