HW 5 Solns
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Transcript of HW 5 Solns
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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
5:50 PM
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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
5:50 PM
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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.
Problem 3Friday, June 01, 2012
5:50 PM
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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.
Problem 4Frida y, June 01, 2012
5:50 PM
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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.
Problem 5Friday, June 01, 2012
5:50 PM
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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
3:56 PM
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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...
Problem 7Friday, June 01, 2012
5:50 PM
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Thrust is found as...
This thrust is a force that we can put into a free-body diagram.
Problem 8Frida y, June 01, 2012
5:51 PM
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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.
Problem 9Friday, June 01, 2012
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This is a problem in rotational kinematics.
This is also a standard problem in rotational kinematics.
Problem 10Frida y, June 01, 2012
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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.
Problem 11Frida y, June 01, 2012
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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
Problem 12Friday, June 01, 2012
5:51 PM
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The height of the door is not needed. We are only interested is
dimensions off the axis.
Problem 13Sunday, October 07, 2012
7:55 PM
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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
8:15 PM
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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
Problem 15Saturday, June 02, 2012
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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.
Problem 16Sunday, October 07, 2012
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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.
Problem 17Saturday, June 02, 2012
8:15 PM
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The total energy is the sum of these two.