Solve Angular Momentum of Homogenous Rod After Hammer Blow

[stunner5000pt]

A homogenous rod of length L and mass M is initially at rest on a horizontal frictionless table in teh lab.
The rod is then hit with a hammer, at a point A which is at a distance d from teh center point of the rod. The blow is in the direction perpendicular to the rod. The blow transfers momentum P to the rod. It is assumed that the duration of hte blow is so short taht we can neglect the motion of the rod during the blow. The known quantities are L, M, d, and P

1. Determine the velocity of the center of mass of the rod after the blow.
Now it DOESNt say that the point O is a fixed point so the rod is going to be kicked off in some direction with translational motion too.

relative to the rod the hammer has angular momentum L = dp
if you conserve angular momentum
$$d p = \frac{1}{12} ML^2 \omega + Mv$$
well this FAR from what the book says
the book says the velocity = P/ M
but this is assuming that the whole $I \omega$ term is one?! how!?

2.Determine the kinetic energy of the rod after the blow

$$\frac{1}{2} m_{hammer} v^2 = \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$$
$$\frac{1}{2} Pv = \frac{1}{2} \frac{1}{12} ML^2 \omega^2 + T$$
what we need to find is T anyway that's why
$$T = \frac{P^2}{2M} - \frac{1}{2} \frac{1}{12} ML^2 \omega^2$$
but $$\omega = \frac{v}{r} = \frac{P}{Mr}$$ soo

$$T = \frac{P^2}{2M} - \frac{1}{2} \frac{1}{12} ML^2 \frac{P^2}{M^2 r^2}$$
$$T = \frac{P^2}{2M} - \frac{P^2}{2M} \frac{L^2 d}{12}$$
did i go wrong somehere here??


Assume there is a poin t C on the rod which - just after the blow - is at rest relative to the table

3. Determine the distance OC. Under what conditions is C a point on the rod??

leave this for later

 

[xanthym]

A homogenous rod of length L and mass M is initially at rest on a horizontal frictionless table in teh lab.
The rod is then hit with a hammer, at a point A which is at a distance d from teh center point of the rod. The blow is in the direction perpendicular to the rod. The blow transfers momentum P to the rod. It is assumed that the duration of hte blow is so short taht we can neglect the motion of the rod during the blow. The known quantities are L, M, d, and P

1. Determine the velocity of the center of mass of the rod after the blow.
Now it DOESNt say that the point O is a fixed point so the rod is going to be kicked off in some direction with translational motion too.

relative to the rod the hammer has angular momentum L = dp
if you conserve angular momentum
$$d p = \frac{1}{12} ML^2 \omega + Mv$$
well this FAR from what the book says
the book says the velocity = P/ M
but this is assuming that the whole $I \omega$ term is one?! how!?

2.Determine the kinetic energy of the rod after the blow

$$\frac{1}{2} m_{hammer} v^2 = \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$$
$$\frac{1}{2} Pv = \frac{1}{2} \frac{1}{12} ML^2 \omega^2 + T$$
what we need to find is T anyway that's why
$$T = \frac{P^2}{2M} - \frac{1}{2} \frac{1}{12} ML^2 \omega^2$$
but $$\omega = \frac{v}{r} = \frac{P}{Mr}$$ soo

$$T = \frac{P^2}{2M} - \frac{1}{2} \frac{1}{12} ML^2 \frac{P^2}{M^2 r^2}$$
$$T = \frac{P^2}{2M} - \frac{P^2}{2M} \frac{L^2 d}{12}$$
did i go wrong somehere here??


Assume there is a poin t C on the rod which - just after the blow - is at rest relative to the table

3. Determine the distance OC. Under what conditions is C a point on the rod??

leave this for later

In the following, all linear and rotational dynamics are referenced to the rod's Center of Mass "O". Further, since the initial impulse was exactly normal to the rod's central axis, angular quantities are computed from their scalar vector magnitudes.
Question #1:
Conservation of Linear Momentum and of Angular Momentum apply to each quantity separately. Thus, Linear Impulses affect Linear Momentum, and Angular Impulses affect Angular Momentum. In this problem, the hammer impulse provides BOTH a Linear Impulse and Angular Impulse, and will cause BOTH linear translational motion and angular rotation. Thus, for a Linear Impulse {FΔt = ΔP} and an Angular Impulse {TΔt = ΔL} and since the rod was initially totally at rest:
{Linear Impulse} = {Change In Linear Momentum} = P = M*v
::: ⇒ {Linear Velocity of Center of Mass} = v = P/M

Required for next questions because the hammer impulse causes rotational motion in addition to linear motion:
{Angular Impulse} = d*{Linear Impulse} = d*P =
= {Change in Angular Momentum about Center of Mass} =
= I*ω
::: ⇒ ω = d*P/I
::: ⇒ ω = (12)*d*P/{M*L²}

Question #2:
{Total Kinetic Energy} = (1/2)*M*v² + (1/2)*I*ω² =
= (1/2)*M*{P/M}² + (1/2)*I*{d*P/I}² =
= (1/2)*P²/M + (1/2)*d²*P²/I =
= (1/2)*P²/M + (12)*(1/2)*d²*P²/{M*L²} =
= (1/2)*{P²/M}*{1 + (12)*d²/L²}

Question #3:
Just after impact, the rod's center of mass is moving right with velocity v (determined above) while a point at distance "C" BELOW the center of mass is moving left with velocity (-ω*C). When these 2 quantities exactly cancel, the point "C" will instantaneously remain still. Thus:
ω*C = v
::: ⇒ C = v/ω
::: ⇒ C = {P/M}/{d*P/I}
::: ⇒ C = I/{d*M}
::: ⇒ C = (1/12)*(M*L²)/{d*M}
::: ⇒ C = (1/12)*L²/d


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[Doc Al]

1. Determine the velocity of the center of mass of the rod after the blow.
Now it DOESNt say that the point O is a fixed point so the rod is going to be kicked off in some direction with translational motion too.

The rod will end up translating and rotating.

relative to the rod the hammer has angular momentum L = dp
if you conserve angular momentum

No reason to think angular momentum is conserved. After all, you're hitting the rod with a hammer!

$$d p = \frac{1}{12} ML^2 \omega + Mv$$

The quantity "dp" is the increase in angular momentum, the angular impulse. Adding angular and linear momentum together (like you do in this equation) makes no sense--they don't even have the same units.

well this FAR from what the book says
the book says the velocity = P/ M
but this is assuming that the whole $I \omega$ term is one?! how!?

The book is correct. The impact of the hammer does two things: It imparts a linear impulse (P) and an angular impulse (dP) to the beam. The linear impulse determines the motion of the center of mass, which is all you care about in this question.

2.Determine the kinetic energy of the rod after the blow

I'll stop here... it looks like xanthym beat me to it!

 

[stunner5000pt]

so in questions like this angular momentum and linear momentum are two separate entities?? But they only equate when you cross with some radial vector??

also in that first question how do you figure that they are asking for the linear velocity only??
Determine the velocity of the center of mass (CM) of the rod after the blow

thank you very much for the help! I appreciate!

also can you have a look at my other problem which is of similar topic (but different concept) https://www.physicsforums.com/showthread.php?t=69023

 

[xanthym]

so in questions like this angular momentum and linear momentum are two separate entities?? But they only equate when you cross with some radial vector??

also in that first question how do you figure that they are asking for the linear velocity only??
Determine the velocity of the center of mass (CM) of the rod after the blow

thank you very much for the help! I appreciate!

Linear Momentum and Angular Momentum are always treated separately. Remember, they have different units. Never equate Linear Momentum & Angular Momentum, and do not add them together. They are different physical quantities.

In the first Q, they only require Linear Velocity to be computed. However, you are correct in that ω is needed in later questions and is computed in msg #2.

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