Consider a stainless steel annular disk with an outer radius 66 mm and inner radius 7.1 mm. The mass of the disk is 1351 grams. Keep your answers to at least 4 significant digits. The stainless steel annular disk is allowed to rotate on a frictionless table with the rotation axis at its center. The disk has a small cylinder rigidly mounted at the top concentrically. The cylinder's radius is 12.4 mm, and the mass of the cylinder is negligible. A string is wrapped around the cylinder, and a hanging mass of 20.2 g is tied at the other end of the string. When the mass falls under gravity, it causes the stainless steel annular disk to rotate. Ignoring the string's mass, and assuming that the string's motion is frictionless, what is the angular acceleration of the stainless steel annular disk?

To find the angular acceleration of the stainless steel annular disk, we can use the principle of conservation of angular momentum.

The moment of inertia of the annular disk can be calculated using the formula:

I = 0.5 * m * (r_outer^2 + r_inner^2)

Where I is the moment of inertia, m is the mass of the disk, and r_outer and r_inner are the outer and inner radii of the disk, respectively.

In this case, m = 1351 grams = 1.351 kg
r_outer = 66 mm = 0.066 m
r_inner = 7.1 mm = 0.0071 m

Substituting these values into the formula, we have:

I = 0.5 * 1.351 * (0.066^2 + 0.0071^2)

Using a calculator, we get:

I ≈ 0.01462 kg⋅m^2 (rounded to 4 significant digits)

Now, let's consider the hanging mass. The force acting on it due to gravity can be calculated as:

F = m_hanging * g

Where F is the force, m_hanging is the mass of the hanging mass, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

In this case, m_hanging = 20.2 grams = 0.0202 kg

Substituting the values, we get:

F = 0.0202 * 9.8 ≈ 0.198 kg⋅m/s^2 (rounded to 4 significant digits)

Now, we can find the torque exerted on the disk by the hanging mass. The torque is equal to the force multiplied by the radius of the small cylinder:

τ = F * r_cylinder

Where τ is the torque and r_cylinder is the radius of the small cylinder.

In this case, r_cylinder = 12.4 mm = 0.0124 m

Substituting the values, we have:

τ = 0.198 * 0.0124 ≈ 0.0024576 N⋅m (rounded to 4 significant digits)

Finally, we can calculate the angular acceleration using the formula:

τ = I * α

Where α is the angular acceleration.

Rearranging the formula, we have:

α = τ / I

Substituting the values, we get:

α = 0.0024576 / 0.01462 ≈ 0.1678 rad/s^2 (rounded to 4 significant digits)

Therefore, the angular acceleration of the stainless steel annular disk is approximately 0.1678 rad/s^2.