00:01
In this problem on the topic of kinetics of a particle, we are given the rate at which the arm in the diagram is rotating, and we know the angular acceleration and the angle theta.
00:13
We then must determine the force that it is exerting on the smooth cylinder if it is confined to move along the slotted path, and we assume that motion occurs in the horizontal plane.
00:26
Now we can see here that r is equal to 2 over theta, which means that dr d theta is equal to minus 2 over theta squared.
00:43
Now the angle si between the extended radial line and the tangent can be determined as follows.
00:52
Tan of si is equal to r over the r d theta, which is true.
01:01
From above is simply 2 over theta divided by minus 2 over theta squared.
01:09
And so this simplifies to minus theta.
01:15
Now we know at theta is equal to 180 degrees, which is pi radiance, the tan of ipsi is equal to minus pi, which gives us the psi to be minus 72 .34 degrees.
01:43
Now the negative sign here indicates that psi is measured from the extended radial line in the negative sense of theta, which is clockwise to the tangent.
01:53
So if we look at the full body diagram of the peg as shown in the figure below, then we can apply newton's second law as follows.
02:03
So if we take the sum of the radial forces, this must equal to the mass of the peg m times the radial acceleration ar.
02:18
And from here we can see that minus n sign of 72 .34 degrees must equal to 0 .5 ar.
02:34
Similarly, the sum of all the angular forces, f theta, is equal to m times a theta.
02:49
And so from here we can see that this along the theta axis is f minus n times the cosine of 72 .34 degrees, and this is equal to 0 .5a theta.
03:09
So now if we use kinematics and use the chain rule for the first and second derivatives, or first and second time derivatives of r, we get, firstly we know that r is 2 theta to the minus one...