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Setler [38]
3 years ago
5

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele

ration of 0.5 rad.s?. The magnitude of the applied torque is: a) 0.30 N.m b) 3.0 x 102 N.m C) 0.15 N.m d) 7.5 x 102 N.m e) 6.0 x 102 N.m.
Physics
1 answer:
Artemon [7]3 years ago
6 0

Answer:

0.06 Nm

Explanation:

mass of object, m = 3 kg

radius of gyration, k = 0.2 m

angular acceleration, α = 0.5 rad/s^2

Moment of inertia of the object

I = mK^{2}

I = 3 x 0.2 x 0.2 = 0.12 kg m^2

The relaton between the torque and teh moment off inertia is

τ = I α

Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia

τ = 0.12 x 0.5 = 0.06 Nm

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