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Arisa [49]
3 years ago
14

A disk with a rotational inertia of 2.5 kg-m2 and a radius 1.1 m rotates on a frictionless fixed axis perpendicular to the disk

faces and through its center. A force of 7.7 N is applied tangentially to the rim. The angular acceleration of the disk is _____ rad/s2. Round your answer to the nearest tenth.
Physics
1 answer:
soldier1979 [14.2K]3 years ago
7 0

Answer:

3.4 rad/sec^2

Explanation:

rotational inertia = 2.5 kg-m^2   radius = 1.1 m   force = 7.7 N

t = rotational inertia * angular acceleration    equation 1

also t = force * radius

therefore to calculate angular acceleration equation 1 becomes

f * r = inertia * angular acceleration   hence

angular acceleration = f * r / inertia = \frac{7.7 * 1.1 }{2.5}   8.47 / 2.5 = 3.388 ≈ 3.4 rad/sec^2

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I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}} (1)

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I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}

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