Question

physics 03-05 moment of inertia
dynamics of rotational motion
newton’s second law for rotation
-

• $\alpha$ is in
• $i = mr^2 \

ightarrow$ moment of of a

• moment of inertia ($i$) measures how much an wants to keep

(or not start )

• use to find $i = \sum mr^2$
• unit:

the st. joseph river swing bridge in st. joseph,
michigan has a mass of 300 tons ($2.72\times10^5$ 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/$s^2$.
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?

Explanation:

First Problem (Swing Bridge)

Step1: Find angular acceleration

$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{0.05\ \text{rad/s} - 0}{10\ \text{s}} = 0.005\ \text{rad/s}^2$

Step2: Rearrange torque formula for $I$

From $\tau = I\alpha$, solve for $I$: $I = \frac{\tau}{\alpha}$

Step3: Substitute values to calculate $I$

$\tau = 563\ \text{kNm} = 563000\ \text{Nm}$, so:
$I = \frac{563000\ \text{Nm}}{0.005\ \text{rad/s}^2}$

Step1: Rearrange torque formula for $I$

From $\tau = I\alpha$, solve for $I$: $I = \frac{\tau}{\alpha}$

Step2: Substitute values to find $I$

$\tau = 128000\ \text{Nm}$, $\alpha = 4\ \text{rad/s}^2$, so:
$I = \frac{128000\ \text{Nm}}{4\ \text{rad/s}^2} = 32000\ \text{kg·m}^2$

Step3: Use hoop moment of inertia formula

For a hoop, $I = MR^2$, rearrange for $R$: $R = \sqrt{\frac{I}{M}}$

Step4: Substitute $I$ and $M$ to find $R$

$M = 500\ \text{kg}$, so:
$R = \sqrt{\frac{32000\ \text{kg·m}^2}{500\ \text{kg}}}$

Answer:

$1.126 \times 10^8\ \text{kg·m}^2$

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Second Problem (Carnival Ride)