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A disk with a rotational inertia of $5.68 \mathrm{~kg} \cdot \mathrm{~m}^{2}$ rotates like a merry-go-round while undergoing a torque given by $\tau=(8.44+7.42 \mathrm{t}) \mathrm{N} \cdot \mathrm{m}$. At time $t=1.00 \mathrm{~s}$, its angular momentum is $4.61 \mathrm{~kg} \cdot \mathrm{~m}^{2} / \mathrm{s}$. What is its angular momentum at $t=3.00 \mathrm{~s}$ ?
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Studdy Solution
Calculate the angular momentum at $t = 3.00 \, \text{s}$.
The initial angular momentum at $t = 1.00 \, \text{s}$ is $4.61 \, \text{kg} \cdot \text{m}^2/\text{s}$.
The angular momentum at $t = 3.00 \, \text{s}$ is:
$$L_{3.00} = L_{1.00} + \Delta L$$
$$L_{3.00} = 4.61 + 46.56$$
$$L_{3.00} = 51.17 \, \text{kg} \cdot \text{m}^2/\text{s}$$
The angular momentum at $t = 3.00 \, \text{s}$ is $\boxed{51.17 \, \text{kg} \cdot \text{m}^2/\text{s}}$.