A force of $16.88 \mathrm {~N}$ is applied tangentially to a wheel of radius 0.340 m and causes an angular acceleration of 1.20 rad/s². What is the moment of inertia of the wheel? A)4.78 kg ? m² B)3.59 kg ? m² C)5.98 kg ? m² D)7.17 kg ? m²

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Quizplus · Accepted Answer

The torque $\tau$ is given by $\tau = r \times F = 0.340 \, \text{m} \times 16.88 \, \text{N} = 5.7392 \, \text{Nm}$. The moment of inertia $I$ is calculated using $\tau = I \times \alpha$, so $I = \frac{\tau}{\alpha} = \frac{5.7392 \, \text{Nm}}{1.20 \, \text{rad/s}^2} = 4.7827 \, \text{kg} \cdot \text{m}^2$, which is approximately 4.78 kg ? m².

A Force Of 16.88 N16.88 \mathrm {~N}16.88 N Is Applied Tangentially to a Wheel of Radius 0

A Force Of $16.88 \mathrm {~N}$ Is Applied Tangentially to a Wheel of Radius 0