How do I calculate the center of mass of a wheel with a weight attached to it?

Measured from the centre of the wheel the moment of the wheel is zero.

The moment of the mass is 1.5* 22

and the total mass is 8.5

so the position of the centre of mass = total moment / total mass

= 33/8.5

= 3.9 cm from the centre of the wheel.

Here's the method broken down into small steps. It makes the answer long, but it should be easier to follow.

Centre of mass = 'CoM' for brevity.

Call the wheel A, the added mass B and the combination C.

Let point Q be the CoM of C. By symmetry, note that Q lies on the line joining the CoM of A to the CoM of B.

Let d be the distance from the centre of A to Q.

Draw a diagram. Mark Q and d

The moment of an object about a point = (mass of object) x (distance from the point to object's CoM)

For any point P:

(Moment of A about P) + (Moment of B about P) =(Moment of C about P)

A good choice for P is the centre of A, because then the moment of A is zero which simplifies the arithmetic.

We can work in units of kg and cm providing we are consistent.

Moment of A about P = mass x distance of CoM to P = 7 x 0 = 0

Moment of B about P = mass x distance of CoM to P = 1.5 x 22 = 33

Moment of C about P = mass x distance of CoM to P = (7 + 1.5)d = 8.5d

(Moment of A about P) + (Moment of B about P) =(Moment of C about P)

0 + 33 = 8.5d

d = 3.88cm (=3.9cm to 2 sig. figs.)

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Assuming the moment of inertia is required in kgm², do the next part of the calculation in kg and m.

Using the standard formula for the moment of inertia of a uniform disc:

The moment of inertia of A about its CoM =- (1/2)mr² = (1/2) x 7 x 0.32² = 0.3584 kgm²

The moment of inertia of B (point mass) about A's CoM = mr² = 1.5 x 0.22² = 0.0726 kgm²

Moment of inertia of C about A's CoM = 0.3584 + 0.0726 = 0.431kgm²

Using the parallel axis theorem (see link), moment of inertia about Q is

0.431 + (7 + 1.5)x(0.0388)²

= 0.44kgm²