Topic: Theory of Machines
In the figure shown below, the relative velocity of link 1 with respect to link 2 is 15 m/s. The link 2 rotates at a constant speed of 150 rpm. The magnitude of Coriolis component of acceleration of link 1 is
- A. $201~m/s^{2}$
- B. $239~m/s^{2}$
- C. $418~m/s^{2}$
- D. $471~m/s^{2}$
Answer: D. $471~m/s^{2}$
Explanation: The Coriolis component of acceleration is calculated using the formula $a_c = 2v\omega$, where $v$ is the sliding velocity and $\omega$ is the angular velocity of the rotating link. Given $v = 15$ m/s and $N = 150$ rpm. The angular velocity $\omega = \frac{2\pi N}{60} = \frac{2\pi \times 150}{60} = 5\pi$ rad/s. Substituting the values gives $a_c = 2 \times 15 \times 5\pi = 150\pi \approx 471.2$ m/s², which is approximately 471 m/s².