Question

A 25 kg circular disk has a diameter of 2.5 feet and a thickness of 2.5 cm. Find the density of the disk in kg/m3. Next, find the weight of the object. Then calculate the buoyant force on the disk if it is submerged under water. Finally, will the object sink or float?

Answer 1

Assume that
`\rm g= 9.81\; N\cdot kg^(-1)`;
`\rho(\text{Water}) = \rm 1000\;kg\cdot m^(-3)`.

Density of the disk: approximately
`\rm 2.19* 10^(3)\; kg\cdot m^(-3)`.

Weight of the disk: approximately
`\rm 245\;N`.

Buoyant force on the disk if it is submerged under water: approximately
`\rm 112\; N`.

The disk will sink when placed in water.

Step-by-step explanation:

Convert the dimensions of this disk to SI units:

• Diameter:
  `d = \rm 25\; inches = (25* 0.3048)\; m = 0.762\;m`.
• Thickness
  `h = \rm 2.5\; cm = (2.5* 0.01)\; m = 0.025\;m`.

The radius of a circle is 1/2 its diameter:

`\displaystyle r = \rm (1)/(2)* 0.762\;m = 0.381\; m`.

Volume of this disk:

`V(\text{disk}) = \pi\cdot r^(2)\cdot h = \pi* 0.381^(2)* 0.025 \approx 0.0114009\; m^(3)`.

Density of this disk:

`\displaystyle \rho(\text{disk}) = (m)/(V) = \rm (25\; kg)/(0.0114009\; m^(3)) = 2.19* 10^(3)\;kg\cdot m^(-3)`.

`\rho(\text{disk}) >\rho(\text{water})` indicates that the disk will sink when placed in water.

Weight of the object:

`W(\text{disk}) = m\cdot g = \rm 25* 9.81 = 245.25\; N`.

The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately
`\rm 0.0114009\; m^(3)` of water. The buoyant force on the disk will be:

`\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1* 10^(3)\; kg\cdot m^(-3)* 0.0114009\; m^(3)* 9.81\; N\cdot kg^(-1)\\ &\approx \rm 112\; N\end{aligned}`.

The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.