Question Video: Finding the Diameter of a Sphere given Its Mass and Density Physics • 9th Grade

A steel ball bearing has a mass of 0.034 g. Find the diameter of the ball bearing in millimeters, rounded to the nearest millimeter. Use a value of 8000 kg/mยณ for the density of steel.

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Video Transcript

A steel ball bearing has a mass of 0.034 grams. Find the diameter of the ball bearing in millimeters, rounded to the nearest millimeter. Use a value of 8000 kilograms per cubic meter for the density of steel.

Okay, so this question is asking us about a steel ball bearing. We know that this is going to have a spherical shape, and weโ€™re being asked to find its diameter. Letโ€™s label this unknown diameter as ๐‘‘. Weโ€™ll also label the mass of the ball as ๐‘š, and weโ€™re told that this is equal to 0.034 grams. The other thing that weโ€™re told is the density of steel, which is the material that the ball bearing is made of. We typically label densities with the symbol ๐œŒ, and we are told that in this case the density has a value of 8000 kilograms per cubic meter. We can recall that the density ๐œŒ of an object is equal to the objectโ€™s mass ๐‘š divided by its volume ๐‘‰. We can also recall that the volume ๐‘‰ of a sphere is equal to four-thirds times ๐œ‹ times the cube of the sphereโ€™s radius ๐‘Ÿ.

Now, in this question, we are not actually trying to work out the value of the radius, but rather the diameter of the sphere, which weโ€™ve labeled as ๐‘‘. The radius ๐‘Ÿ is the distance out from the center of the sphere to the edge, while the diameter ๐‘‘ is the distance all the way across the sphere from one side to the other going through the sphereโ€™s center. So then, the diameter ๐‘‘ is equal to the radius ๐‘Ÿ multiplied by two. Or if we divide both sides of this equation by two, then on the right-hand side, the twos cancel out and we have that the radius ๐‘Ÿ is equal to the diameter ๐‘‘ divided by two.

We can use this relationship between the radius and diameter in order to replace the radius ๐‘Ÿ in this equation for the sphereโ€™s volume. Replacing ๐‘Ÿ by ๐‘‘ over two, we have that ๐‘‰ is equal to four-thirds times ๐œ‹ times the cube of ๐‘‘ over two. We can also write that as four-thirds times ๐œ‹ times ๐‘‘ cubed over eight. Or if we multiply together the three and the eight in the denominator, this gives us 24. And so, weโ€™ve got a fraction of four divided by 24, which simplifies to one divided by six. So, we have that the volume ๐‘‰ of a sphere is equal to one-sixth times ๐œ‹ times the cube of the sphereโ€™s diameter. We can use this equation along with this one here for the density ๐œŒ in order to solve our problem.

If we rearrange this density equation in order to make ๐‘‰ the subject, then we can use our values of the mass ๐‘š and density ๐œŒ of the ball bearing in order to calculate its volume. Then, once weโ€™ve worked out this volume, we can rearrange this other equation for the volume of a sphere in order to make the diameter ๐‘‘ the subject. Then, by substituting in our calculated value for the volume ๐‘‰, weโ€™ll be able to work out the diameter ๐‘‘, which is what the question wants. One difficulty that we are going to face here is with the units of the quantities. We are asked to find the diameter in millimeters, and we are given a mass in grams and a density in kilograms per cubic meter.

Letโ€™s begin by converting the mass ๐‘š into units of kilograms so that it matches the units in the density ๐œŒ. We can recall that one kilogram is equal to 1000 grams. Dividing both sides by 1000 and canceling the 1000s on the right-hand side, we have that one gram is equal to one thousandth of a kilogram. So, to convert from grams into kilograms, we need to divide the value by 1000. When we divide the value of the mass of the ball bearing in grams by 1000, we get a result of 0.000034 kilograms. We could also write this value in scientific notation as 3.4 times 10 to the negative five kilograms, where we have a power of negative five because we had to move the decimal point five places over in order to turn this value into 3.4.

Now that weโ€™ve got a mass ๐‘š in kilograms to match our density in kilograms per cubic meter, we can use these values in this equation to calculate the sphereโ€™s volume ๐‘‰. First though, we need to make ๐‘‰ the subject of the equation. To do this, weโ€™ll first multiply both sides of the equation by ๐‘‰. On the right-hand side, the ๐‘‰ in the numerator cancels with the ๐‘‰ in the denominator. So, weโ€™ve got the volume ๐‘‰ times density ๐œŒ is equal to mass ๐‘š. Now, we divide both sides of the equation by the density ๐œŒ. The two ๐œŒs on the left-hand side cancel each other out. And we end up with an equation that says volume ๐‘‰ is equal to mass ๐‘š divided by density ๐œŒ.

If we substitute our values of ๐‘š and ๐œŒ for the steel ball bearing into the right-hand side of this equation, then we have that the sphereโ€™s volume ๐‘‰ is equal to 3.4 times 10 to the negative five kilograms divided by 8000 kilograms per cubic meter. In terms of the units, the kilograms in the numerator and denominator cancel each other out. And that leaves units of one divided by one over meters cubed or simply units of meters cubed. When we evaluate the expression, we calculate a result of 4.25 times 10 to the negative nine meters cubed. And this value is the volume of the steel ball bearing.

Now that weโ€™ve got a value for the volume ๐‘‰, we want to use that value in this equation in order to calculate the ball bearingโ€™s diameter. So, letโ€™s rearrange the equation to make the diameter ๐‘‘ the subject. Weโ€™ll begin by multiplying both sides of the equation by six divided by ๐œ‹. On the right-hand side, the sixes and ๐œ‹s in the numerator and denominator will cancel each other out. And we have that six times ๐‘‰ divided by ๐œ‹ is equal to ๐‘‘ cubed. If we then take the cube root on both sides, then on the right, the cube root of ๐‘‘ cubed is simply equal to ๐‘‘. Finally, writing the equation the other way round, we have that the diameter ๐‘‘ is equal to the cube root of six times the volume ๐‘‰ divided by ๐œ‹.

Weโ€™re now ready to substitute our value for the volume ๐‘‰ of the ball bearing into this equation. When we do that, we get this expression here for its diameter ๐‘‘. Evaluating the expression inside the cube root gives a result of 8.1169 times 10 to the negative nine meters cubed, where the ellipses show that there are further decimal places. Then, evaluating the cube root, we get a diameter ๐‘‘ of 2.0097 et cetera times 10 to the negative three meters. Notice that since the volume ๐‘‰ was in units of meters cubed, weโ€™ve got a diameter ๐‘‘ in units of meters.

However, weโ€™re asked to give our answer in units of millimeters. Letโ€™s recall that one meter is equal to 1000 millimeters, or equivalently one meter is equal to 10 to the three millimeters. So, to convert from meters to millimeters, we just multiply by 10 to the power of three. If we take our value for the diameter ๐‘‘ in units of meters and multiply it by 10 to the power of three, then weโ€™ve got a 10 to the negative three multiplied by 10 to the three, and this is simply equal to one. So then, in units of millimeters, the diameter of the ball bearing is equal to 2.0097 et cetera millimeters.

The last thing left to do is to round this result to the nearest millimeter. This gives us our final answer for the diameter of the ball bearing as two millimeters.

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