A tangent cam with a base circle diameter of 50 mm operates...

A tangent cam with a base circle diameter of 50 mm operates a roller follower 20 mm in diameter. The line of stroke of the roller follower passes through the axis of the cam. The angle between the tangential faces of the cam is 60°, speed of the cam shaft 200 r.p.m and the lift of the follower 15 mm. Calculate, (a) The main dimensions of the cam (b) The acceleration of the follower at, (i) The beginning of lift (ii) Where the roller just touches the nose i.e., the apex of the circular nose.

Transcript

00:01 In this question we have to find the main dimension of the calm as here this complete this one is the row circle and this one is the base circle here and this point is o here and this point is p here and as this distance is e here and in this direction here is g.

00:15 Similarly in this direction here this one is a and this radius is here row 1 here and this point is o here and this complete distance is small d here.

00:28 So as this one is the target fork here, so as the base circle diameter, base circle diameter is bb is equals to 50 m m or as here base circle radius, base circle radius equals to p1 here.

00:41 So p1 is equals to 50 by 2 which is equals to 25 mm here.

00:46 And h3 is equal to roller radius as here roller radius, h3 is equal to our s3.

00:52 Let us take this one is r3 here roller radius equals to r3 which is equals to 20 by 2 mm or we can write it as here.

00:58 10 m m here for the follower x is equal to 15 m m is equal to 200 rpm here and alpha is equal to 60 degree so w can be written as here w is equal to 2 pi and divided by 60 here this one is equals to here r n equals to 200 so by simplifying this one is 20 .94 radiant per second is that value here now come to the first part in the first part we have to find main dimension of the calm.

01:30 So as main dimension of the calm for this one as the base circle center is o.

01:36 O p plus of p t is equal to where p is the center of row circle here.

01:41 So o p t plus p t plus e plus e plus c t here.

01:46 So o p can be written as a oc plus c t minus of pt here.

01:50 So o p can be written as a oc.

01:52 This one is r1 plus x minus of r2 here.

01:55 So by simplifying this one is equal to here 25 plus 15...

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