Hints for Problem Set #6

HINTS FOR PROBLEM SET #6

Problem Set #6 – Problem 1

Since the problem is defined such that

an angle and a distance from a pole

describes the position of the pin,

you should use radial and transverse coordinates.

Draw the FBD/IVD for the pin using these coordinates.

For the FBD:

The force of friction is neglected so

the only forces that you need to show acting on the pin

are the normal forces from the two slots, P and Q.

You may wonder what happened to gravity.

Here's what:

Slot DE is cut in a horizontal plate.

So gravity acts perpendicular to the motion.
(i.e. perpendicular to the paper.)

It does not have any components in the radial or transverse directions.

For the IVD:

You need to show the transverse and radial components of the acceleration vector.

To get expressions for these components, use the kinematic formulas.

These formulas have the quantities r, rdot, rdotdot, qdot and qdotdot.

You need to express these quantities in terms of q only.

It is given that

           qdot = 12
Hence
           qdotdot = 0

Now use geometry to find a relationship between r and q:

           r = 0.2/cosq

From this you can get rdot and rdotdot in terms of q, qdot and qdotdot.

Plug these into the expressions for the radial and transverse components of the acceleration.

Now you can apply F_net = ma

to the FBD/IVD picture in each component direction

to find the forces you are asked for.

Be careful: You need to differentiate between

the forces acting on the pin from the slots 
and the resultant forces in the radial and transverse directions.

The forces acting on the pin from the slots, P and Q, 
are contact forces and act perpendicular to the slots.

The resultant forces in the radial and transverse directions, R_r and R_q, 
are the vector sums of any forces that have components in these directions.
P and Q have components in these directions.

Problem Set #6 – Problem 2

This is motion under a central force.

So use Conservation of Angular Momentum.

For convenience, use Angular Momentum per unit mass:

          h = H/m

The magnitude of the initial angular momentum, h, (when the particle is at point A) is:

           h = r_o v_o sin(90^o) = r_o v_o

The magnitude of the angular momentum at any other point, B, on the path is:

           h = r v sinf 
where f is the angle between r and v.

Note that  v sinf is the component of the velocity in the transverse direction, v_q.

(To see this look at the picture.)

Furthermore, in this particular problem v_q is equal to v cosq.
(This is not always true, but it is true in this problem.)

Where did this come from?

To see it, draw a line from O to point B on the semi-circle.

Also draw the line from the center, C, of the semi-circle to point B.

Note that the triangle OCB is isosceles.

Hence, the angle between BC and BO must also be equal to q.

Since BC is perpendicular to the velocity, and BO is perpendicular to the transverse direction,

then the angle between the velocity and the transverse direction is also q.

Hence:

          v_q = v cosq

Now use this result and equate the two expressions you have for h,

one at point A and the other at point B along the path of motion.

           r_o v_o= r v cosq

 Solving for v:

            v =  r_ov_o/ r(cosq)

This is equation 1.

From geometry you can relate r_o and r:   
           r = 2 (r_o/2) cosq = r_o cosq

This is equation 2.

Plug the expression for r from equation 2 into equation 1.

            v =  v_o/cos²q

Problem Set #6 – Problem 3

You are looking for the components of velocity in the r and q directions, v_r and v_q, as a function of q.

Since you  have motion under a central force, use Conservation of Angular Momentum.

The magnitude of the initial angular momentum, h, (when the particle is at point A) is:

         h =  r_o v_o sin(90^o) =  r_o v_o

The magnitude of the angular momentum at any other point on the path is:

         h = r v sinf

where f is the angle between r and v.

Note that v sinf is the component of the velocity in the transverse direction, v_q.

(To see this look at the picture.)

Hence the angular momentum at any point is:

         h = r v_q

Now equate the two expressions you have for h, one at point A and the other at point B along the path of motion:

          r_o v_o = r  v_q                          (equation 1)

Solve this for v_q:

          v_q= r_o v_o / r                          (equation 2)

The given relationship between r_o and r is:   
          r = r_o / sqrt(cos2q)                (equation 3)

Substituting equation 3 into equation 2, you get:

           v_q= v_o sqrt(cos2q)

This is the component of velocity in the q direction as a function of q.

Now you need to get v_r, the component of v in the r direction, as a function of q.

From the kinematic equation, you know that:

          v_r = rdot

So you need to get rdot in terms of q.

Start by differentiating equation 3 above.

This expression will have a qdot in it.

So you need to get qdot.

The kinematic equation gives:

          v_q = r qdot

Substitute this into equation 1 to get:

         r_o v_o = r² (qdot)

Solving for qdot you get:

         qdot = (r_o v_o)/r²

Substitute the expression for r from equation 3:

         qdot = (r_o v_o) / ( r_o²/cos2q)         qdot =   v_ocos2q / r_o

Plug this in the expression you got for rdot and you will have

the component of velocity in the r direction as a function of q.