Video: Finding the Point of Equilibrium When a Particle is Placed in a Uniform Hemispherical Shell

Video Transcript

A uniform hemispherical shell of mass 21 kilograms rests on a smooth horizontal plane. A particle of mass 14 kilograms is placed on the rim of this shell, causing it to tilt so that the plane of the rim is at an angle of 𝛼 to the horizontal when the system is in equilibrium. Find the value of tan of 𝛼.

Let’s begin by drawing a free body diagram. Here is our uniform hemispherical shell. We’re going to let it have a radius of 𝑟 units. A particle of mass 14 kilograms rests on the rim of this shell, causing it to tilt. We know that the downwards force of the weight of this particle is mass times gravity, so 14𝑔. And we also know that the plane of the rim is at an angle of 𝛼 to the horizontal.

But what do we do about the downwards force of the weight of our uniform hemispherical shell? Well, because it’s uniform, we can quote a formula to help us find the center of mass. The center of mass of our uniform hemispherical shell lies on the axis of symmetry at a distance of 𝑟 over two from the point 𝑜. Well, here is the axis of symmetry. And so our downwards force of 21𝑔 must act here.

Now that we have the diagram, we’re going to see what it means for the system to be in equilibrium. Firstly, we know that for a body to be in equilibrium, the sum of all forces acting on that body must be equal to zero and the sum of all moments acting on the body must be equal to zero, where the moment, which is the turning effect of a force, is calculated by multiplying the magnitude of that force by the perpendicular distance of the line of action of that moment from the point about which the object is trying to turn.

And so let’s imagine we’re going to take moments about 𝑜. We have a force of 21𝑔 and a force of 14𝑔. We need to calculate the perpendicular distance from 𝑜 to the line of action of these forces. Let’s begin with the 14𝑔 force. We can add a right-angled triangle in here. Alternate angles are equal, so the included angle is 𝛼. And let’s call the side that we’re trying to find, which is the perpendicular distance from 𝑜 to that downward force 14𝑔, 𝑥. Relative to the included angle, we want to find the length of the adjacent side. And we’ve actually defined the hypotenuse to be equal to 𝑟.

So let’s link these by using the cosine ratio such that cos of 𝛼 is 𝑥 over 𝑟. We’ll find an expression for 𝑥 by multiplying through by 𝑟. And so we’ll find that 𝑥 is equal to 𝑟 cos 𝛼. We add that to the diagram. And now we consider the 21𝑔 force. Here is our triangle. Now the included angle down here is 𝛼. We can convince ourselves that this is true since we know that the edge of the hemisphere and the line of action of the center of the mass to 𝑜 must meet at 90 degrees.

Angles on a straight line sum to 180. So this angle here is 90 minus 𝛼. And then if we subtract 90 minus 𝛼 and 90 from 180 degrees, that’s the interior angle sum of a triangle, we get our included angle to be 𝛼. This time then, we’re interested in the opposite side and the hypotenuse of this triangle. So we’re going to use the sine ratio. This time, sin of 𝛼 is 𝑦 over 𝑟 over two. And we can solve for 𝑦 by multiplying by 𝑟 over two. So 𝑦 is 𝑟 over two sin 𝛼. Let’s add this to the diagram. And we’re ready to take moments.

Now when we take moments about a point, we do need to define a positive direction. Let’s say that the counterclockwise direction here is positive. Thinking about our 21𝑔 force, we multiply force by perpendicular distance. That’s 21𝑔 times 𝑟 over two sin 𝛼. This force is trying to turn the object in a counterclockwise direction. And so its moment is going to be positive. The other force, the 14𝑔 force, is trying to turn the object in a clockwise direction. And so its moment is going to be negative.

Force times perpendicular distance is 14𝑔 times 𝑟 cos 𝛼. And we know that the body is in equilibrium. So let’s set this equal to zero. Remember, we’re trying to find the value of tan of 𝛼. And we have two variables at the moment. We have 𝑟 and 𝛼. Of course, 𝑟 is the radius of our hemisphere. And so it cannot be equal to zero. This means we can divide our entire equation by 𝑟. Similarly, 𝑔 is the acceleration due to gravity. It’s about 9.8. And so we can divide through by 𝑔.

While we’re here, we might also spot that we can divide through by seven. 21 divided by seven is three, and 14 divided by seven is two. And so our equation becomes three over two sin 𝛼 minus two cos 𝛼 equals zero. So we have an equation purely in terms of 𝛼. But how do we link sin 𝛼 and cos 𝛼 to tan 𝛼? Well, we know the trigonometric identity tan 𝛼 is equal to sin 𝛼 over cos 𝛼. And so we need to manipulate our equation. We’re going to do this by adding cos 𝛼 to both sides.

So we get three over two sin 𝛼 equals two cos 𝛼. Now if we divide through by cos 𝛼, on the left-hand side we’re going to have sin 𝛼 over cos 𝛼. So three over two sin 𝛼 over cos 𝛼 equals two. And so we replace sin 𝛼 over cos 𝛼 with tan 𝛼.

We’re nearly finished. To find the value of tan of 𝛼, we just need to divide through by three over two. tan of 𝛼 then is two divided by three over two, which is the same as two times two over three, which is four-thirds. We’re, therefore, able to say then that the value of tan 𝛼 under these circumstances is four-thirds.