Hints for Problem Set #23

HINTS FOR PROBLEM SET #23

Problem Set #23 – Problem 1

Since we are given information about velocities and time intervals, we will apply the Principle of Impulse and Momentum for the system of two rigid bodies: the bar and the bullet.

Linear momentum of the system at time 1 + Impulses acting on the system during the time interval (t1 to t2) = Linear momentum of the system at time 2

And:

Angular momentum of the system at time 1 + Angular impulses acting on the system during the time interval (t1 to t2) = angular momentum of the system at time 2

t1 is just before the bullet hits the rod.

t2 is when after the bullet is embedded in the rod.

delta t = 0.001 second

To apply the two equations above, draw a diagram consisting of these 3 elements:

1)  A picture of the bullet and the rod with the linear momentum and angular momentum vectors shown for both of them at t1.

2)  A picture of the FBD of the bullet and the rod showing the forces as impulses. 

(Just multiply each force by delta t.)

3)  A picture of the bullet and the rod with the linear momentum and angular momentum vectors shown for both of them at t2.

And of course you should put them together in such a way that it is clear that:

the first picture + the second picture = the third picture

The linear momentum vectors in pictures 1 and 3 is shown by vectors acting at the mass centers of each rigid body and consisting of:

(mass of the body) x (velocity of the mass center).

In this case, the linear momentum vector for the rod at t1 is zero. (It starts at rest.)

The linear momentum vector for the bullet at t1 is:

(mass bullet) x (velocity of bullet at t1) = 4.472 lb sec  

(Be careful of units.  You have to convert weight of the bullet to a mass to get this.)

For t2, the linear momentum vector for the rod and bullet together (since the bullet in now embedded in the rod) is:

(mass rod + mass of the bullet) x (velocity of mass center of rod)

This velocity will be in the direction tangent to a circle around C, since the rod must rotate about point C.

(Notice that I am neglecting the fact that the bullet will change slightly the position of the mass center of the rod.  I could have figured it out, but I concluded that it was too much trouble and the difference is insignificant.)

The angular momentum vectors in pictures 1 and 3 are shown by couples consisting of:

(mass moment of inertia about the mass center of the body) x (angular velocity of the body)

The angular momentum vector for both the rod and the bullet at t1 is 0.  (Neither is rotating.)

The angular momentum vector for the combination of the rod and bullet at t2 is:

(mass moment of inertia about the mass center of the rod) x (angular velocity of the rod)

(Again I am neglecting how the bullet will affect the mass moment of inertia of the rod and bullet combination.)

The forces in the FBD (actually the impulses) will include only the reaction forces at C, because we will assume that the impulses due to the weights are negligible.

Now get the x and y components of the linear momentum equation.

Now get the k component of the angular momentum equation.

Another equation is obtained by noting that:

(angular velocity of the rod) = (velocity of the rod) / (distance from mass center of rod to point C)

These 4 equations will allow you to solve for Cx, Cy, angular velocity of the rod after the impact, and the velocity of the mass center of the rod after the impact.

Problem Set #23 – Problem 2

There are 2 parts to this problem:

1. Determination of the velocity of point A on rod AB just before it hits point C on rod CD.

2. Determination of the velocity of point C after the impact with point A.

Part 1)

To determine the velocity of point A (vA) just before it impacts with C, use the Work Energy Theorem for rod AB.

The only force that does work is the gravity force so the equation looks like:

T1 + V1 = T2 + V2

Position 1 is when AB is just released.

Position 2 is when A is just about to hit C.

V1 = mgL

V2 = mg(L/2)

T1 = 0 (AB is released from rest.)

T2 = ½ m v^2 + ½ I w^2

Remember that the v in this equation must be the velocity of the mass center of rod AB.

Since AB must be rotating about point B, v = w x (L/2)

Plugging all of this into the Work Energy Theorem above, you should get that w = sqrt(3g/L)

Using vA = w x L, you should bet that vA = sqrt(3gL)

Part 2)

We will need both the equation with the coefficient of restitution and the Principle of Impulse and Momentum.

The equation with the coefficient of restitution can only be used for the normal direction and for the points of the rigid bodies that are the points of impact. In this case the points of impact are A and C. The normal direction is along the surface of the floor. Since both A and C only have velocity components along this direction before and after the collision, we won’t need subscripts in the equation. So the equation will look like this:

(vC_prime - vA_prime) = e(vA – vC)

vC = 0 (It is at rest before A hits it)

vA = sqrt(3gL) (from Part 1 of this problem.)

vA_prime = (w_prime) L (Even after the collision, point A must be rotating about point B.)

Plugging these into the equation, you get:

vC_prime = w_prime L + e sqrt(3gL)

We need more to get w_prime .

Use the Principle of Impulse and Momentum. Draw the 3 pictures:

     1)  A picture of the system (the two rods AB and CD) with the linear momentum and angular momentum vectors at t1 (the instant before the impact).

     2)  A picture of the FBD of the two rods showing the forces as impulses.

(Just multiply each force by delta t.)

     3)  A picture of the system with the linear momentum and angular momentum vectors at t2 (the instant just after the impact).

And of course you should put them together in such a way that it is clear that:

the first picture + the second picture = the third picture

The linear momentum vector in pictures 1 and 3 is shown by a vector acting at the mass center and consisting of:

(mass) x (velocity of the mass center).

In this case:

        For rod CD at t1 it is 0.  (It is at rest.)

        For rod AB at t1 it is m vG = m w(L/2)

        For rod CD at t2 it is m vC_prime.

        For rod AB at t2 it is m vG_prime = m w_prime (L/2)

The angular momentum vector in pictures 1 and 3 is shown by the couple consisting of:

(mass moment of inertia about the mass center of the body) x (angular velocity of the body).

In this case:

        For rod CD at t1 it is 0. (It is at rest.)

        For rod AB at t1 it is I x w

        For rod CD at t2 it is 0. (Even after the impact, rod CD has no rotation.)

        For rod AB at t2 it is I x w_prime

and recall that Irod = (1/12) m L^2

The forces in the FBD (actually the impulses) will only include reaction forces at B because we will assume that the impulses due to the weights are negligible.

Now get the k component of the angular momentum equation.  Use point B to sum moments about.

This equation with the coefficient of restitution equation will allow you to solve for w_prime and vC_prime.

        w_prime = sqrt(3g/64L)

        vC_prime =sqrt(27gL/64)