Question

the st joseph river swing bridge in st joseph, michigan has a mass of 300 tons (2.72×10⁵ kg) and is 231 ft (70.4 m) long. if the motor produces 563 knm of torque and takes 10 s to accelerate the bridge to 0.05 rad/s, what is the bridge’s moment of inertia?
a spinning ride at a carnival is accelerating at 4 rad/s². if the ride is shaped like a hoop, and the motor is exerting 128000 nm of torque, what is the radius of the 500 kg ride?
images of various objects with their moment of inertia formulas: hoop about cylinder axis (i = mr²), hoop about any diameter (i = ½mr²), solid cylinder (or disk) about cylinder axis (i = ½mr²), solid cylinder (or disk) about central diameter (i = ¼mr² + ¹/₁₂ml²), thin rod about axis through center ⊥ to length (i = ¹/₁₂ml²), thin rod about axis through one end ⊥ to length (i = ⅓ml²), solid sphere about any diameter (i = ²/₅mr²), thin spherical shell about any diameter (i = ²/₃mr²), slab about ⊥ axis through center (i = ¹/₁₂m(a² + b²)), annular cylinder (or ring) about cylinder axis (i = ½m(r₁² + r₂²))

Explanation:

Step1: Solve for bridge's moment of inertia

First, find angular acceleration: $\alpha = \frac{\Delta \omega}{\Delta t} = \frac{0.05\ \text{rad/s}}{10\ \text{s}} = 0.005\ \text{rad/s}^2$
Use torque formula $\tau = I\alpha$, rearrange to $I = \frac{\tau}{\alpha}$
Substitute values: $\tau = 563000\ \text{Nm}$, $\alpha=0.005\ \text{rad/s}^2$
$I = \frac{563000}{0.005}$

Step2: Calculate ride's radius

Use torque formula $\tau = I\alpha$, for a hoop $I = MR^2$, so $\tau = MR^2\alpha$
Rearrange to $R = \sqrt{\frac{\tau}{M\alpha}}$
Substitute values: $\tau=128000\ \text{Nm}$, $M=500\ \text{kg}$, $\alpha=4\ \text{rad/s}^2$
$R = \sqrt{\frac{128000}{500 \times 4}}$

Answer:

Bridge's moment of inertia: $1.126 \times 10^8\ \text{kg·m}^2$
Carnival ride's radius: $8\ \text{m}$