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A 2.0-kg stone is tied to a 0.50 m long string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is

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samuelonum1

Answer:

108 Nm

Explanation:

 Given data

mass m= 2 kg

radius r= 0.5 m

angular velocity ω= 12 rad/s

distance d= 0.75 m

we know that

[tex]Force= mass * acceleration\\\ acceleration= w^2r\\\\ Force= m*w^2r\\\\Force =2*12^2*0.5= 144 N[/tex]

we know that torque = F*d= 144*0.75= 108 Nm

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