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A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane. If it is slowing down at a rate of 12 rad/s2, the net torque acting on it is _____ N-m. Round your answer to the nearest whole number.

Respuesta :

akande212

Answer:

Torque = –207.4 Nm

Explanation:

Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)

Torque = I × α

α = angular acceleration

I = moment of inertia

I = MR² for a circular hoop

Torque = 3.2×5.4×(– 12)

Torque = –207.4 Nm

onyebuchinnaji

Answer:

The net torque acting on the circular loop is 1120 N.m

Explanation:

Given;

mass of the circular loop, m = 3.2-kg

radius of the circular loop, r = 5.4 m

angular acceleration, ω = 12 rad/s²

Torque = moment of inertia x angular acceleration

Torque = Iω

moment of inertia, I = mr² = 3.2 x 5.4² = 93.312 kgm²

Torque = Iω = 93.312 x 12 = 1120 N.m

Therefore, when the circular loop is slowing down, the net torque acting on it is 1120 N.m

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