In the figure below, a wheel of radius 0.275 m is mounted on a frictionless

horizontal axle. A massless cord is wrapped around the wheel and attached to a
1.95 kg box that slides on a frictionless surface inclined at angle θ = 32.7 ° with the
horizontal. The box accelerates down the surface at 2.15 m/s2. Calculate the
rotational inertia of the wheel about the axle.

mg sinθ = ma + Iα

where α = a/r
solve for I

Although without a diagram I can't be certain

Don't forget that torque=I(alpha), so multiply mg sin(theta) and ma both by r so you'll be in the same units.

To calculate the rotational inertia of the wheel about the axle, we can use the concept of torque. The torque acting on an object is equal to the moment of inertia (rotational inertia) times the angular acceleration. In this case, the torque acting on the wheel is due to the force of gravity on the box.

The first step is to find the force of gravity on the box. The force of gravity can be calculated using the formula F = m * g, where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).

F = 1.95 kg * 9.8 m/s^2 = 19.11 N

Next, we need to find the torque acting on the wheel. Torque (τ) is equal to the force (F) multiplied by the distance (r) from the axis of rotation to the point where the force is applied. In this case, the force of gravity is applied at the radius of the wheel, so the distance is equal to the radius.

τ = F * r

τ = 19.11 N * 0.275 m = 5.25 N·m

Finally, we can use the torque and the angular acceleration to find the rotational inertia (I) of the wheel.

τ = I * α

Where α is the angular acceleration (in radians/s^2). Since we know the linear acceleration (a) of the box, we can relate it to the angular acceleration using the formula a = r * α, where r is the radius of the wheel.

α = a / r

α = 2.15 m/s^2 / 0.275 m = 7.818 rad/s^2

Now we can substitute the torque and the angular acceleration back into the torque equation to solve for the rotational inertia (I).

5.25 N·m = I * 7.818 rad/s^2

I = 5.25 N·m / 7.818 rad/s^2

I ≈ 0.671 kg·m^2

Therefore, the rotational inertia of the wheel about the axle is approximately 0.671 kg·m^2.