Question

The displacement of a follower of a cam in a printing machine is represented by the expression
$x=10\theta +120{\theta }^2+1500{\theta }^3$
$\text{Where}\theta \text{is the angle of rotation of the cam. The jerk given by the system at any position is}$

• A. $9000{\omega }^3+48000{\omega }^3\theta +150000{\omega }^3{\theta }^2$
• B. $9000{\omega }^3$
• C. $240{\omega }^2+9000{\omega }^2\theta +24000{\omega }^2{\theta }^2+50000{\omega }^3{\theta }^3$
• D. None of these

Answer 1

The correct option is B $9000{\omega }^3$

$x=10\theta +120{\theta }^2+1500{\theta }^3$

$\text{Jerk}=\frac{d^{3}x}{d{t}^3}=\frac{d^{3}x}{d{\theta }^3}\times \frac{d{\theta }^3}{d{t}^3}={\omega }^3\times \frac{d^{3}x}{d{\theta }^3}$

$\therefore \frac{dx}{d\theta }=10+240\theta +4500{\theta }^2$

$\frac{d^{2}x}{d{\theta }^2}=240+9000\theta$

$\frac{d^{3}x}{d{\theta }^3}=9000$

$\therefore \text{jerk}={\omega }^3\left(9000\right)$