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vg is the speed of girl wrt the ice

vp is the speed of plank wrt the ice

vis the speed of girl wrt the plank

With respect to the ice, we know that there is no initial momentum.

This means there is no total momentum after she stops moving either.

Also, because of the relationships between the velocities...

With these two equations, we have two unknowns and we can solve.

Now that we know the speed of the plank wrt the ice, we can also find

Problem 1Frida y, June 01, 2012

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the speed of the girl wrt the ice.

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If we are dealing with average force, the impulse is...

If we convert this force to pounds, we see this this force is

This claim is obviously nonsense.

Be sure to remember to convert the velocity to SI units.

Problem 2Friday, June 01, 2012

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The impulse delivered is the change in momentum of the ball.

Because it is positive, that means this impulse was delivered in the +x

direction.

The work delivered by the racket is the change in kinetic energy of the

ball.


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We know that there is going to be conservation of momentum for this

closed system. This means...

The negative sign means that this velocity is to the left.

We also know that there is going to be a conservation of mechanical

energy in this problem. Because the wedge doesn't change height, we

can see that potential energy of the block is converted to kinetic energy

in both the block and the wedge.


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This is a standard collision problem. When the wad hits the block, there

will be conservation of momentum.

After the collision, the kinetic energy of the clay+block system is burned

away to friction.


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There are two ways that we could do this problem. A graphicalapproach is the easiest, but I will use equations here.

For the x and y components, we get the equations...

Solving this for v2f, we get.

Problem 6Sunday, October 07, 2012

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From this, he was clearly speeding.

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Because all of these masses are along the y-axis, the center of

mass has to also be on the y-axis.

To find center of mass in the y-direction...


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Thrust is found as...

This thrust is a force that we can put into a free-body diagram.


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Because we know the functional form of the angular acceleration, we

can find the angular velocity.

In general, there could be an integration constant, but this constant

would be the initial angular velocity in this problem. We are told it

starts from rest, so this integration constant is zero.

Now, because we have a functional form for the angular velocity, we

can find the angular displacement.

Once again, there could have been an integration constant in this

process. If we say that the initial angular position is zero, then we don't

have to worry about it.


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This is a problem in rotational kinematics.

This is also a standard problem in rotational kinematics.


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First, lets do this in angular coordinates.

We are not told what the final angular velocity is, but we are told the

final regularvelocity. We just have to make the conversion.

In essence, we have already used this.


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Because the origin is at the center of the rectangle, all of the masses

are the same distance from the z axis.

The moment of inertia of this system about the z axis is then...

Because we know the moment of inertia as well as the angular velocity,

we have a rotational kinetic energy given by


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The height of the door is not needed. We are only interested is

dimensions off the axis.


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As we saw in class, torques are force times the lever arm. In this case,

the lever arms are always perpendicular to the force. The counter-

clockwise torque is...

The clockwise torques are...

The total torque counter-clockwise is then...

Problem 14Saturday, June 02, 2012

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You can certainly use the Newton's Law approach we did in class. The

force and torque equations are

We can plug the force equations into the torque equation and solve for

acceleration. Once we know acceleration, we use kinematics to find the

time. You would probably use


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However, you can also do this with energies.

We do know from kinematics that the final velocity of some blocks that

have accelerated some distance is...

By looking at the energetics, we know that potential energy in m1 is

going to go into potential energy ofm2 as well as kinetic energy in m1,

m2 and the pulley. Because of the moment of inertia for this pulley is...

our energy equation looks like...

From this, we see by comparing with the kinematic equation that

However, what we are looking for is time. Using the kinematic equation

we see that the time is

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From this expression, if the pulley had been massless, our time would

have been...

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Knowing the velocity gets us directly to the angular velocity.


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Translational kinetic energy of the center of mass is...

To find this, we need to get an expression that involves the velocity.


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The total energy is the sum of these two.

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