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:
A)0 N∙m
B)6.0 N∙m
C)14 N∙m
D)72 N∙m
E)108 N∙m

Question

Quizplus · Accepted Answer

The torque ($\tau$) about the origin can be calculated using the formula $\tau = rF\sin(\theta)$, where $r$ is the radius of the circle, $F$ is the force acting on the object, and $\theta$ is the angle between the force vector and the radius vector. Here, the radius $r$ is the distance from the origin to the circle, which is 0.75 m. The force $F$ can be found from the centripetal force formula $F = m\omega^2r$, where $m$ is the mass of the stone, $\omega$ is the angular velocity, and $r$ is the length of the string (0.50 m in this case). Substituting the given values, $F = 2.0 \times 12^2 \times 0.50 = 144 N$. Since the circle is parallel to the xy plane and centered on the z axis, the angle $\theta$ between the force vector (which points radially inward) and the radius vector (from the origin to the circle) is 90 degrees, making $\sin(\theta) = 1$. Therefore, the torque is $\tau = 0.75 \times 144 = 108 N\cdot m$.

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