A spinning flywheel has rotational inertia I = 402.9 kgm2. Its angular velocity decreases from 20.3 rad/s to zero in 266.6 s due to friction. What is the frictional torque acting?
Torque= MomentofInertia*angularacceleration.
You are given the moment of inertia (you named it rotational inertia). Angular acceleration is (20.3/266.3) rad/sec^2
Solve for torque.
To solve for the frictional torque acting on the spinning flywheel, we can use the equation:
Torque = Moment of Inertia * Angular Acceleration
Given:
Moment of Inertia (I) = 402.9 kgm^2
Angular Velocity (ω) = 20.3 rad/s
Time (t) = 266.6 s
First, let's find the angular acceleration (α). We can use the equation:
Angular Acceleration (α) = Change in Angular Velocity / Time
Change in Angular Velocity = Final Angular Velocity - Initial Angular Velocity
= 0 rad/s - 20.3 rad/s
= -20.3 rad/s
Substituting the values into the equation, we can calculate:
Angular Acceleration (α) = (-20.3 rad/s) / (266.6 s)
≈ -0.0761 rad/s^2
Now, we can substitute the values of the moment of inertia and angular acceleration into the torque formula:
Torque = Moment of Inertia * Angular Acceleration
= 402.9 kgm^2 * (-0.0761 rad/s^2)
≈ -30.65 Nm (Newton meters)
The negative sign indicates that the frictional torque is acting in the opposite direction of the angular velocity.