A horizontal circular platform (m = 122.1 kg, r = 3.43m) rotates about a frictionless vertical axle. A student (m = 82.3kg) walks slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 2.9 rad/s when the student is at the rim. Find omega when the student is 2.39m from the center.
conservation of angular momentum applies:
initial momentum=final momentum
Studentw+wheelw=studentw' + wheelw'
1/2*82.3*3.43^2*wo+1/2*122.1*3.43^2*wo=
= 1/2*82.3*2.39^2wf+1/2*122.1*3.43^2*wf
solve for wf, knowing wo=2.39
To find the angular velocity when the student is 2.39m from the center, we can use the principle of conservation of angular momentum.
The equation for angular momentum is given by:
L = Iω
Where:
L = Angular momentum
I = Moment of inertia
ω = Angular velocity
The angular momentum is conserved when there are no external torques acting on the system. In this case, we can assume there are no external torques, and the rotational motion is isolated.
The moment of inertia for a circular platform rotating about a vertical axle is given by:
I = (1/2)mr^2
Where:
m = mass of the platform
r = radius of the platform
Now let's calculate the initial angular momentum when the student is at the rim of the platform.
I_initial = (1/2)mr^2
= (1/2)(122.1 kg)(3.43 m)^2
= 713.974 kg·m²
ω_initial = 2.9 rad/s
Since angular momentum is conserved, we can equate the initial angular momentum to the final angular momentum when the student is 2.39m from the center:
L_final = I_final * ω_final
I_final = (1/2)mr^2
= (1/2)(122.1 kg)(2.39 m)^2
= 145.366 kg·m²
Now, we can solve for ω_final:
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
(713.974 kg·m²) * (2.9 rad/s) = (145.366 kg·m²) * ω_final
Solving for ω_final:
ω_final = (713.974 kg·m² * 2.9 rad/s) / (145.366 kg·m²)
= 14.286 rad/s
Therefore, when the student is 2.39m from the center, the angular velocity of the system is 14.286 rad/s.