Question

In a cam-follower mechanism, the follower needs to rise through 20 mm during ${60}^o$ of can rotation, the first ${30}^o$ with a constant acceleration and then with a deceleration of the same magnitude. The initial and final speeds of the follower are zero. The cam rotates at a uniform speed of 300 rpm. The maximum speed of the follower is

• A. 0.60 m/s
• B. 1.20 m/s
• C. 1.68 m/s
• D. 2.40 m/s

Answer 1

The correct option is B 1.20 m/s

Angular velocity,

$\omega =\frac{2\pi N}{60}=\frac{2\pi 300}{60}=10\pi$

Time taken to move ${30}^o$

$=\frac{\frac{\pi }{{180}^o}\times 30}{\omega }=\frac{\frac{\pi }{6}}{10\pi }=\frac{1}{60}s$

During this time, follower moves by distance of 10 mm with initial velocity

$u=0$
$S=ut+\frac{1}{2}a{t}^2$

$0.01=0+\frac{1}{2}\times a\times {\left(\frac{1}{60}\right)}^2$

$\therefore a=0.01\times 2\times (60{)}^2=72m/s^2$

$V_{max}=u+at$

$=0+72\times \frac{1}{60}=1.20m/s$