Rotational Inertia and Angular Momentum

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SUMMARY

The discussion focuses on calculating the angular momentum of a disk with a rotational inertia of 8.38 kg·m², subjected to a time-dependent torque τ = (5.03 + 1.01t) N·m. At t = 1.00 s, the angular momentum is given as 6.57 kg·m²/s. The correct calculation for angular momentum at t = 3.00 s involves integrating the torque over time, leading to the formula L = 5.03t + 0.505t² + C, where C is determined to be 0.53. The final angular momentum at t = 3.00 s is 24.71 kg·m²/s, although the initial calculation was incorrect due to a misunderstanding of integration.

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Homework Statement


A disk with a rotational inertia of 8.38 kg·m2 rotates like a merry-go-round while undergoing a torque given by τ = (5.03 + 1.01t) N · m. At time t = 1.00 s, its angular momentum is 6.57 kg·m2/s. What is its angular momentum at t = 3.00 s?

Homework Equations


dL/dt= T L=Iw

The Attempt at a Solution


dL/dt=T (my symbol for touque) -> L=5.03t+1.01t^2+C --> 5.03 (1)+ 1.01(1)^2+C= 6.57 -->C=0.53

L=5.03(3)+ 1.01(3)^2 +0.53 --> 24.71 kg*m/s^2

Apparently that's wrong... Help?
 
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For the sake of any suffering physics student that comes upon this, remember as I did not the integral of x= (x^2)/2 and you should be able to solve this.
 
So, you solved this problem, right?
 

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