Question

You wish to measure the rotational inertia of a tire. The tire has a mass of 5.2 kg, an outer radius of 38 cm, and an inner radius of 27 cm. If you hang the tire on a peg and start it oscillating about the point of suspension, you find that it oscillates with a period of 1.76 s. What is the rotational inertia of the tire?

Answer 1

The rotational inertia of the tire, assuming it to be an annular ring, and using the oscillation method with a period of 1.76 seconds, is approximately 0.389 kg·m^2.

Step-by-step explanation:

To calculate the rotational inertia of the tire, we can model it as a solid cylinder or an annular ring and use the oscillation method. The rotational inertia, I, of an object can be determined using the period of oscillation, T, obtained from a physical pendulum setup:

I = (T^2*g)/(4*π^2) * m * r^2

Here, T is the period of oscillation which is 1.76 s, g is the acceleration due to gravity (9.8 m/s^2), m is the mass of the tire, and r is the radius at which the mass is concentrated. For a tire, r would typically be the mean radius. Since the problem does not provide the exact mass distribution, we will estimate the mean radius (r) as the average of the inner and outer radius. The mean radius r = (0.38 m + 0.27 m) / 2 = 0.325 m,

Therefore, the rotational inertia I = (1.76^2 * 9.8)/(4*π^2) * 5.2 kg * (0.325 m)^2 ≈ 0.389 kg·m^2.

The rotational inertia of the tire is approximately 0.389 kg·m^2.