I was wondering what shape of cam would produces a sinusoidal displacement over a complete revolution. So, I decided to come up with a sinusoidal function with a period of 360˚, and with a maximum of 1 and minimum of zero. I then plotted the function on a polar co-ordinate system, so that the plot should be the shape of the cam. The function is $\frac{1}{2} ( \cos{\theta} + 1 )$.
Plot of $\frac{1}{2} ( \cos{\theta} + 1 )$ on polar co-ordinate system
I'd like to know why the plot appears to have a "discontinuity" (perhaps the wrong term?) at 180˚. The plot also looks like it has a very steep gradient either side of this point. I was expecting a smooth continuous plot with no steep gradients. Assuming the cam follower is a thin "knife edge" that can follow any cam profile, would it actually produce sinusoidal motion?
A few observations:
A practical cam would usually have a minimum radius at least equal to the radius of the axle. If I modify the function to have a minimum of 0.25 and maximum of 1.25, the function is smooth as I'd expect, and there is no steep gradient around 180˚:
Plot of $\frac{1}{2} ( \cos{\theta} + 1.5 )$ on polar co-ordinate system
Using the differentiate command in the graphing application confirms that both plots have the same gradient. I can only assume there's some peculiar property of polar plots that exaggerates gradients when the radius approaches zero. My best guess is that it's related to the "surface speed" of a rotating object with zero radius being zero.