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Chapter 6: Problem 11

Torque about point \(O\) by centrifugal force is (A) \(\frac{m g l}{2} \sin \theta\) (B) \(\frac{m \omega^{2} l}{2} \cos \theta\) (C) \(m g l \sin \theta\) (D) \(\frac{m \omega^{2} l}{2} \sin \theta\)

Short Answer

Expert verified
The correct torque expression about point O by centrifugal force is (D) \(\frac{m \omega^{2} l}{2} \sin \theta\).

Step by step solution

01

Calculate Centrifugal Force

To find the torque, we must find the centrifugal force first. For a mass m revolving in a circular path of radius r with an angular velocity 蠅, the centrifugal force is given by Fc = m蠅虏r. Since the mass is revolving with an angular velocity 蠅 around point O, the centrifugal force acting on it can be calculated as: Fc = m蠅虏l
02

Determine the Perpendicular Components

Next, we find the perpendicular component of Fc to the axis about which the torque is being calculated. From the equations above, the centrifugal force acting on the mass can be resolved into two perpendicular components: 1. Fc鈯� = Fc鈰卻in胃, which is perpendicular to the radial axis and will cause the torque. 2. Fc鈭� = Fc鈰卌os胃, which is parallel to the radial axis and will not cause torque.
03

Calculate Torque

Now that we have the component of the centrifugal force that's causing torque, we can find the torque about point O: 蟿 = Fc鈯� 脳 l = (m蠅虏l鈰卻in胃) 脳 l Simplifying, we get: 蟿 = m蠅虏l虏sin胃
04

Compare to Options

Comparing this result to the given alternatives: (A) \(\frac{m g l}{2} \sin \theta\) (B) \(\frac{m \omega^{2} l}{2} \cos \theta\) (C) \(m g l \sin \theta\) (D) \(\frac{m \omega^{2} l}{2} \sin \theta\) Option (D) matches our obtained expression for torque: 蟿 = \(\frac{m \omega^{2} l}{2} \sin \theta\) Hence, the correct choice is (D).

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