A flat-faced follower is driven using a circular eccentric cam rotating at a constant angular velocity ω. At time t = 0, vertical position of follower is y(0) = 0, and the system is in the configuration shown below

Then vertical position of the follower face, y(t) is given by

The correct option is e(1 – cos ωt).

Step-by-Step Derivation and Explanation:

Consider a circular eccentric cam of radius $R$ and eccentricity $e$ driving a flat-faced follower. Let the center of rotation of the cam be $O$ and the geometric center of the circular cam be $C$. The distance between $O$ and $C$ is the eccentricity $e$.

Since the follower has a flat face, the face of the follower is always tangent to the circular profile of the cam. The perpendicular distance from the geometric center of the cam $C$ to the contact line (the follower face) is constant and equal to the radius $R$ of the circular cam.

Let the angular position of the eccentric center $C$ relative to the vertical line of action of the follower be denoted by $\theta =\omega t$.

At any time $t$, the vertical displacement of the flat-faced follower from the center of rotation $O$ is given by the vertical projection of the eccentric link $OC$ plus the radius $R$ of the circle:

$$Y(t)=R-e\cos (\omega t)$$

Here, the negative sign indicates that in the initial configuration at $t=0$, the geometric center $C$ lies directly below the center of rotation $O$, resulting in the lowest vertical position of the follower.

At $t=0$, the vertical position is:

$$Y(0)=R-e\cos (0)=R-e$$

The displacement of the follower $y(t)$ relative to this initial position $y(0)=0$ is:

$$y(t)=Y(t)-Y(0)$$

Substituting the expressions for $Y(t)$ and $Y(0)$:

$$y(t)=[R-e\cos (\omega t)]-(R-e)$$

Simplifying the equation gives:

$$y(t)=e-e\cos (\omega t)$$

Factoring out the eccentricity $e$:

$$y(t)=e(1-\cos (\omega t))$$