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Chapter 8: Problem 48

A spinning flywheel has rotational inertia $I=400.0 \mathrm{kg} \cdot \mathrm{m}^{2} .\( Its angular velocity decreases from \)20.0 \mathrm{rad} / \mathrm{s}\( to zero in \)300.0 \mathrm{s}$ due to friction. What is the frictional torque acting?

Short Answer

Expert verified
Answer: The frictional torque acting on the spinning flywheel is -400/15 N*m.

Step by step solution

01

Identify the given quantities and the unknown quantity

We are given: Rotational inertia (I) = 400 kg * m^2 Initial angular velocity (ω_i) = 20 rad/s Final angular velocity (ω_f) = 0 rad/s Time (t) = 300 s We need to find the frictional torque acting on the flywheel (τ).
02

Calculate the angular acceleration

Since the flywheel comes to a stop, we'll use the equation: ω_f = ω_i + α * t where α is the angular acceleration. Rearranging for α, we get: α = (ω_f - ω_i) / t Now, we plug in the known values: α = (0 - 20) / 300 = -20 / 300 = -1/15 rad/s^2
03

Calculate the frictional torque

Now, we'll use the equation relating torque (τ), rotational inertia (I), and angular acceleration (α): τ = I * α Plugging in the given values, we get: τ = 400 * (-1/15) = -400/15 N * m
04

State the final answer

The frictional torque acting on the spinning flywheel is -400/15 N * m. The negative sign indicates that the torque is acting in the opposite direction of the initial angular velocity, which is expected since the friction is slowing down the flywheel.

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