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.
