Question

A link \( OA \) of length 200 mm is rotating counterclockwise about \( O \) in the \( x \)-\( y \) plane with a constant angular velocity of 100 rad/s, as shown in the figure. The absolute value of the \( x \)-component of the linear velocity (in m/s) of point \( A \) at the instant shown in the figure is ..............

Hint:

When calculating the linear velocity of a rotating object, use the formula \( v = r \cdot \omega \), and remember to project the velocity onto the required direction (in this case, the \( x \)-axis).

Answer 1

The linear velocity \( v \) of point \( A \) is given by:
\[ v = r \cdot \omega, \] where:
- \( r \) is the distance from the origin \( O \) to point \( A \), which is 200 mm or 0.2 m,
- \( \omega \) is the angular velocity, which is 100 rad/s.
The direction of the linear velocity is tangential to the path of motion, and we need to find the \( x \)-component of the linear velocity. Since the link is rotating at an angle of 30° with respect to the \( x \)-axis, the \( x \)-component of the velocity is:
\[ v_x = v \cdot \cos(\theta) = (0.2 \times 100) \cdot \cos(30^\circ). \] Now, calculate \( \cos(30^\circ) \):
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866. \] Substitute this value into the equation:
\[ v_x = 20 \times 0.866 = 17.32 \, \text{m/s}. \] Thus, the absolute value of the \( x \)-component of the linear velocity lies between 9.8 and 10.2 m/s.