Using Calculus to Find torque around a point.

[BSJ90]

Homework Statement

A cylindrical coffee cup (8 cm in diameter and 10 cm tall) is filled to the brim
with coffee. Neglecting the weight of the cup, determine the torque at the handle
(2 cm from edge of cup 5 cm up from bottom of cup).

The easy way would be to just use the center of mass of the cup but that isn't the way its told to be done. So triple integrals it is

density of water(ρ) = 1g/cm³

Homework Equations

T = r x F

if R is from

The Attempt at a Solution

So to set up my coordinate's I put the origin in the center of the cylinder with x going out of the page y to the right and z up.


T = r x F
= r x mFg (Fg = force of gravity)
dT = r x dmFg (dm = ρdxdydz)

$\int$ $\int$$\int$ |r|ρgdzdydx

I think (tell me if I'm wrong) the bounds of the first integral (from left to right):
-4 to 4, -$\sqrt{}$4-x² to $\sqrt{}$4-x², -5 to 5.

but where I'm having problem is finding an equation that relates r ( the vector from the handle to the cup) for each dm. If someone could help me out that would be great.

 

[mighty2000]

Thank you for your post. I am a scientist and I would like to help you with your problem.

First, let's clarify the variables in this problem. The diameter of the cup is given as 8 cm, so the radius (r) is 4 cm. The height (h) of the cup is given as 10 cm. The handle is located 2 cm from the edge of the cup and 5 cm up from the bottom of the cup. We can use these values to find the position vector (r) from the handle to the cup.

To do this, we can use the Pythagorean theorem to find the distance from the handle to the center of the cup, which is the hypotenuse of a right triangle with sides of 2 cm and 5 cm. This distance is √(2^2+5^2) = √29 cm. Now, we can use this distance to find the position vector from the handle to the center of the cup, which is r = <√29, 0, 0>.

To find the position vector from the handle to each dm, we can use the same method. The distance from the handle to the point (x, y, z) on the cup is given by the Pythagorean theorem as √((x-2)^2+y^2+z^2). Therefore, the position vector from the handle to each dm is r = <√((x-2)^2+y^2+z^2), y, z>.

Now, we can set up the triple integral to calculate the torque at the handle:
T = ∫∫∫ r x dmFg
= ∫∫∫ <√((x-2)^2+y^2+z^2), y, z> x ρgdxdydz
= ∫∫∫ <yρgz, zρg√((x-2)^2+y^2+z^2), -yρg√((x-2)^2+y^2+z^2)>dxdydz

To evaluate this integral, we need to determine the bounds for each variable. Since the cup is symmetric about the x-axis, we can integrate from x = 0 to x = 8 cm. For each value of x, the y-values will range from -√(4^2-x^