Question

In a cam-follower , the follower rises by h as the cam rotates by $\delta$ (radians) at constant angular velocity $\omega$ (radians/s). The follower is uniformly accelerating during the first half of the rise perod of the rise period, and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is

• A. $\frac{4h\omega }{\delta }$
• B. $h\omega$
• C. $\frac{2h\omega }{\delta }$
• D. $2h\omega$

Answer 1

The correct option is C $\frac{2h\omega }{\delta }$

Here, outstroke angle
${\theta }_o=\delta$
$\text{and stroke length}=h$
$\text{Angular velocity}=\omega$

$V=u+at$

$(V_0{)}_{max}=0+a\times \frac{t_0}{2}=a\times \frac{\delta }{2\omega }$ . . . (i)

$\frac{h}{2}=0+\frac{1}{2}a{\left(\frac{t_0}{2}\right)}^2$

$h=\frac{a{t}_0^2}{4}=\frac{a{\delta }^2}{4{\omega }^2}$

$a=\frac{4{\omega }^2.h}{\delta }$. . . (ii)

By equation (i) and (ii), we get

$(V_0{)}_{max}=\frac{a\delta }{2\omega }=\frac{4{\omega }^2.h}{{\delta }^2}\times \frac{\delta }{2\omega }=\frac{2.\omega .h}{\delta }$


Point to remember
. For the case of constant acceleration and retardation , and also for the case of cycloidal motion

$V_{\text{max}}=\frac{2h\omega }{\delta }$

$\text{where},h=0.02m$

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

$\delta ={60}^o=\frac{\pi }{N}$

Maximum speed,

$V_{\text{max}}=\frac{2h\omega }{\delta }=\frac{2\times 0.02\times 10\pi }{\pi /3}$

$=1.2m/s$