' Rod BA in Figure 9-4.3 is articulated at point A, and its other…

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Abhishek Sachan · Accepted Answer

Certainly! The linear speed ($v_{BC}$) of rod BC is related to the angular velocity ($\omega_{BA}$) of rod BA and the distance from the axis of rotation to the point where rod BC is attached (denoted as $r_{BC}$). The formula $v_{BC} = \omega_{BA} \times r_{BC}$ essentially states that the linear speed of a point on a rotating object is the product of its angular velocity and the distance from the axis of rotation. In your case: - Angular velocity ($\omega_{BA}$) is given as $4 \ \text{rad/sec}$. - Distance from point B to the center of mass of rod BC ($r_{BC}$) is given as $0.5 \ \text{m}$. Substitute these values into the formula: $$v_{BC} = 4 \ \text{rad/sec} \times 0.5 \ \text{m}.$$ Calculate this expression to find the linear speed ($v_{BC}$) of rod BC. To calculate the linear speed of rod BC, you can use the formula: $$v_{BC} = \omega_{BA} \times r_{BC},$$ where $v_{BC}$ is the linear speed of rod BC, $\omega_{BA}$ is the angular velocity of rod BA, and $r_{BC}$ is the distance from point B to the center of mass of rod BC. In this case, $\omega_{BA} = 4 \ \text{rad/sec}$ and $r_{BC} = 0.5 \ \text{m}$.  Substitute these values into the formula to find $v_{BC}$.

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Abhishek Sachan

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