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course phy 201
9/13 9:00
Applying basic definitions of motion, force, work and energy to some of our fundamental systems--------------------------------------------------------------------------------
A lot of this document can be regarded as sort of a reading comprehension exercise.
If you interpret the words right you can get the right answers, even if you're not completely sure what some of the answers mean.
The main task is to figure out what to do with the given information, in terms of the relationships you have been given, and do the appropriate calculations (being sure to include the units; however you aren't asked to interpret the units and you may simply express the units as they are given).
Another important task is to get used to the ideas of motion, force, work-energy, etc. in the context of some of the systems we have observed.
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Information from preceding assignments:
Rate of change:
The average rate of change of A with respect to B is (change in A) / (change in B).
Average velocity is average rate of change of change of position with respect to clock time.
Average acceleration is average rate of change of position with respect to clock time.
Graph Trapezoids
The slope of a 'graph trapezoid' is its rise / run.
The 'graph altitudes' of a 'graph trapezoid' are the quantities represented by its vertical sides.
The area of a 'graph trapezoid' is its average 'graph altitude' multiplied by its width.
Force, work-energy, momentum
When an object of mass m is moving with velocity v, it has the following properties
its kinetic energy is KE = 1/2 m v^2
its momentum is p = m v
Forces acting on objects can change their velocity, momentum and kinetic energy.
When an object of mass m changes its velocity, with respect to clock time, at rate a, then the net force acting on it (i.e., the sum of all the forces acting on it) is F_net = m * a.
If a force F acts through a displacement `ds along the line of the force, then the force does work `dW = F * `ds.
If F in the above happens to be the net force acting on an object, then the KE of that object changes by an amount equal to `dW.
If a net force F_net acts on an object for time interval `dt, then the momentum of that object changes by `dp = F_net * `dt.
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As on the preceding assignment, some questions can be answered fairly directly, while others are more challenging. Don't let yourself get bogged down on any one question before moving on to another. Come back on another day to questions you can't answer on your first try.
`q001. On a graph of velocity v (in cm/sec) vs. clock time t (in sec):
What are the velocity and clock time corresponding to the point (4, 12)?
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clock time is x and velocity is y. rise/run = (12cm/s)/(4s)= 3cm/s^2
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What are the velocity and clock time corresponding to the point (9, 32)?
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velocity is 32, clock time is 9.....(32cm/s)/(9s) = 4cm/s^2
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@& You've identified these quantities correctly, but you wouldn't divide them.*@
@& You've identified these quantities correctly, but you wouldn't divide them.*@
If these points correspond to the velocity of a ball rolling down an incline, describe as fully as you can what you think happens between the first event (corresponding to the first point of the graph) and the second event (corresponding to the second point of the graph).
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the ball goes from 12cm/s to 32cm/s in 5 seconds. (32cm/s) - (12cm/s) / (9s)-(4s) = 4cm/s^2
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@& Right. This is where you can divide.*@
What is the change in velocity between these two events?
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(change in A) / (change in B) = (32cm/s)-(12cm/s)/ (9s)-(4s) = 4cm/s^2
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@& The change in velocity is just 32 cm/s - 12 cm/s. You're a step ahead of the question, which is actually no problem.*@
What is the change in clock time between these two events?
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change in clock time = 9s-4s= 5 seconds
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What is the average velocity for the interval between these two events?
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(average roc of postion) / (clock time) = (32cm/s)+(12cm/s)/2= 22cm/s (22cm/s) / (5 s)= 4.4cm/s^2
@& Best to clarify the terminology.
It's
ave rate of change of position wrt clock time =
(change in postion) / (change in clock time) =
average velocity =
(32cm/s)+(12cm/s)/2= 22cm/s
Then
change in position = 22 cm/s * change in clock time = 22 cm/s * 5 s = 110 cm.*@
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What is the average rate at which the velocity changes, with respect to clock time, between these two events?
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(32cm/s)-(12cm/s) / 5s = (20cm/s) / (5 s)= 4cm/s^2
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What is the displacement of the object between these two events?
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(32cm/s)- (12cm/s)= 20 cm
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@& Displacement is 110 cm. See my previous note to clarify this.*@
`q002. A 5 kg mass accelerates at 2 m/s^2. What is the net force acting on the object?
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F_net=m*a 5kg*(2m/s^2)= 10 kg m/s^2
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`q003. A net force of 5000 kg m/s^2 acts on a 100 kg mass. What is the acceleration of the mass?
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acceleration= (5000 kg m/s^2)= (100kg)X= 50 m/s^2
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`q004. A force of 400 Newtons is exerted on an automobile as it is pushed through a distance of 100 meters. How much work is done on the automobile?
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dW=F*ds 400 Newtons * 100 meters = 40000 Newton meters
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`q005. A certain pendulum requires .5 Newtons of force for every centimeter it is pulled back (recall pulling back the pendulum hanging from the tree limb, using the rubber band).
How much force would be required to pull the pendulum back 5 cm, 10 cm and 15 cm?
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0.5Newtons*5cm= 2.5 Newton cm 0.5Newtons*10cm= 5 Newton cm 0.5Newtons*15cm= 7.5 Newton cm
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What is the average force required between pullbacks of 5 cm and 15 cm?
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((7.5newtoncm + 2.5newtoncm)/2) / 1.5s = 10cm/s
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@& The forces are in Newtons. Just (7.5 N + 2.5 N) / 2 = 5 N would be the average force.*@
How much work is done between these two positions?
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7.5newtoncm-2.5newtoncm= 5 Newton cm
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@& This is where the cm come in. Multiply the average force (see previous) by the displacement.*@
`q006. A system rotates through 12 rotations in 4 seconds, first coming to rest at the end of this interval.
How quickly is it rotating, on the average? (The answer is as obvious as it should seem, but also be sure to interpret this as a rate of change with respect to clock time, and carefully apply the definition of average rate)
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Vf=12rotations/sec V0= 0 rotations/sec V ave= 15. rotations/sec
@& 12 rotations means just 12 rotations, not 12 rotations / second. The fact that the system rotates through 12 rotations doesn't tell you anything about how fast it's rotating; to get that you need to know something about the time interval, which in this case is 4 seconds.
Your calculation in the next line gives a good answer to this question.*@
(12-0 rotation) / 4 sec= (12rotations)/(4sec)= 3 rotations/sec for average roc
@& This is the average rate of change of angular position with respect to clock time, with position measured in rotations.
This would be the average rate at which the system is rotating.
On the average, it rotates at 3 revolutions / second.*@
average = ((12+0)/2)/(4sec)= (6rotations)/(4sec)= 3/2 rotations/sec or 1.5 rotations/sec
@& The calculation in this line isn't necessary to answer the question. *@
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Is it speeding up or slowing down?
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It first speeds up from zero then it slows back down and stops (ending at 0)
@& *@
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At what rate is it doing so? (Again attempt to interpret as an average rate of change of an appropriate quantity with respect to clock time).
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Average rotation was found to be 3 rotations/sec so, 3rotations/sec=ave 0rotations/sec=final 6 rotations/sec must be initial
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`q007. At the initial point of an interval a 7 kg mass is moving at 5 meters / second. By the end of the interval it has gained an additional 200 kg m^2 / s^2 of kinetic energy.
How much kinetic energy does it therefore have at the end of the interval?
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KE=(1/2)m*V^2 ((1/2)*7kg)*(5m/s^2)= 3.5kg*(25m/s^2)= 87.5 kgm/s at end of interval= (200kg m^2/S^2)+(87.5 kg m/s)= 287.5 kg m^2/s^2
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Slightly more challenging question: How fast is the mass therefore moving at the end of the interval?
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???
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`q008. An object begins an interval with a kinetic energy of 20 000 kg m^2 / s^2, and ends the interval with a kinetic energy of 15 000 kg m^2 / s^2.
By how much did its kinetic energy change on this interval? (The answer is as obvious as it might seem, but be careful about whether the answer is positive or negative).
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initial= 20000 kgm^2/s^2 final=15000 kgm^2/s^2 -5000 kg m^2/s^2 was the change in KE
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More challenging: During this interval, how much work was done on the object by the net force?
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F_net=m*a KE='dW= -5000 kgm^2/s^2
'dW=F*'ds
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Also more challenging: If the average force on the object during this interval had magnitude 200 Newtons, then what was its displacement during this interval?
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'dw=F*'ds -5000 kgm^2/s^2=200newton*'ds= -48000kgm^2/s^2
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@& Your statement
-5000 kgm^2/s^2=200newton*'ds
is correct. However to find `ds you need to divide both sides by 200 Newtons:
( -5000 kg m^2 / s^2 ) / (200 Newtons) = (200 Newtons * `ds) / (200 Newtons)
so that
`ds = (-5000 kg m^2 / s^2) / (200 Newtons) = -25 (kg m^2) / (Newton s^2).*@
`q009. A mass of 200 grams hangs from one side of a pulley, and another mass from the other side. The gravitational force pulling down on this mass is about 200 000 gram cm / s^2, and the tension in the string pulling it upward is about 180 000 gram cm / s^2.
Pick either upward or downward as the positive direction.
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up as positive (+) and down as negative (-)
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Using + for your positive direction and - for your negative direction, what is the gravitational force on this object?
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-200000 grams cm/s^2
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Using + for your positive direction and - for your negative direction, what is the tension force on this object?
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+180000 gram cm/s^2
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Using + for your positive direction and - for your negative direction, what is the net force on this object?
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- Net Force (-200000 gram cm/s^2) + (+180000 gram cm/s^2)= -20000 gram cm/s^2
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Using + for your positive direction and - for your negative direction, what is the acceleration of this object?
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- acceleration because more gravitational forc pulling down than tension force pulling up
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If the object's displacement during a certain interval is +30 cm, then according to your choice of positive direction, is the displacement upward or downward?
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+30 cm would be upward
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When you multiply the displacement by the gravitational force, what is your result? Be sure to indicate whether the result is positive or negative.
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30 cm * (-200000 gram c/s^2)= -6000000 gram cm^2/s^2
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When you multiply the displacement by the tension force, what is your result? Be sure to indicate whether the result is positive or negative.
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30 cm * (180000 gram cm/s^2)= 5400000 gram cm^2/s^2
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When you multiply the displacement by the net force, what is your result? Be sure to indicate whether the result is positive or negative.
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30 cm * (-20000 gram cm/s^2)= -600000 gram cm^2/s^2 or -600000 erg
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Does gravity do positive or negative work on this object?
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negative
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Does the tension force do positive or negative work on this object?
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positive
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Does the net force do positive or negative work on this object?
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negative
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Does the object speed up or slow down?
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slows
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How would your answers have changed if you had chosen the opposite direction as positive?
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the object would speed up if (+) was chosen as grav forc because displacement and grav force would be greater than tension
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@& Really good work on this problem.*@
`q010. A pendulum of length 2 meters and mass 3 kg, pulled back a distance | x | from its equilibrium position, experiences a restoring force of magnitude k | x |, where k = 15 kg / s^2 * | x |. [Note that for convenience in calculation we are making some approximations here; the actual value of k for this pendulum would actually be closer to 14.7 kg / s^2, and this is so only for values of | x | which are a good bit smaller than the length. These are details we'll worry about later.]
How much force does the pendulum experience when x = .1 meter?
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15 kg/s^2 * 0.1 meter= 1.5 meter kg/s^2
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How much force does the pendulum experience when x is .05 meter, .1 meter, .15 meter and .2 meter?
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(15 kg/s^2)*(0.05m)= 0.75 m kg/s^2 (15 kg/s^2)*(0.15m)= 2.25 m kg/s^2
(15 kg/s^2)*(0.1m)= 1.5 m kg/s^2 (15 kg/s^2)*(0.2m)= 3 m kg/s^2
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What do you think is the average force between | x | = .05 meter and x = .2 meter?
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1.9 m kg/s^2
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How much work do you think would be done by this force between | x | = .05 meter and | x | = .2 meter?
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??? (3 m kg/s^2)-(0.75m kg/s^2)= 2.25 m kg/s^2
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@& To get the work, you would multiply your average force, found in the preceding, by the displacement.
The displacement from .05 meter to .2 meter is .15 meter.*@
How fast would the pendulum have to be going in order for its kinetic energy to equal the result you just obtained for the work?
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15 kg/s^2
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@& This is actually a tough quesiton at this point. We'll discuss it in class.*@
If the pendulum moves from position x = .05 meter to x = .2 meter, is the direction of the force the same as, or opposite to the direction of the motion?
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??? the same
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@& The force acts back toward the x = 0 equilibrium point, and the motion from .05 m to .2 m is in the direction away from equilibrium, so the two directions are opposite.*@
If the pendulum moves from position x = .20 meter to x = ..05 meter, is the direction of the force the same as, or opposite to the direction of the motion?
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the opposite
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@& The force acts back toward the x = 0 equilibrium point, and the pendulum is moving toward the equilibrium point, so the directions are the same.*@
If the pendulum string was cut, what would be the acceleration of the 1 kg mass?
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(change in v)/(change in t)
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What is the magnitude of the force exerted by gravity on the pendulum's mass? For simplicity of calculation you may use 10 m/s^2 for the acceleration of gravity.
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@& Use F = m a.
m is the mass of the pendulum, a the 10 m/s^2 acceleration.*@
When x = .1 meter, what is the horizontal displacement from equilibrium as a percent of the pendulum's length?
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@& Length is 2 meters, horiz. displacement is .1 meter. What percent is .1 meter of 2 meters?*@
When x = .1 meter, what is the restoring force as a percent of the pendulum's weight?
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What is the magnitude of the acceleration of the pendulum at the x = .15 meter point?
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`q011. The force exerted on a mass has magnitude | F | = 15 Newton / meter * x, for 0 <= x <= .25 meter.
Sketch a graph of | F | vs. x. (You might wish to start by making a table of | F | vs. x, for some appropriate values of x between 0 and .25 meter). Note the convention that a graph of y vs. x has y on the vertical axis and x on the horizontal, so that for this graph | F | will be on the vertical axis and x on the horizontal.
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Verify that the points (.06 meter, .9 Newton) and (.16 meter, 2.4 Newtons) lie on your graph.
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What is the rise between these points?
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rise (y axis)= 2.4Newtons-0.9Newtons= 1.5 Newtons
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What is the run between these points?
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run (x axis)= 0.16meter-0.06meter=0.1 meter
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What is the average slope between these points?
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(2.4-0.9N)/(0.16-0.06m)= (1.5/0.1)=slope
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@& That would be 1.5 Newtons / .1 meter or 15 Newtons / meter.*@
What is the average 'graph altitude' of the 'graph trapezoid' formed by these points?
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(2.4N+0.9N)/2= 1.65Newton*0.1m= 0.165 Newton meter
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What is the width of the trapezoid?
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(0.16 meter) -(0.06 meter)= 0.1 meter
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What therefore is the area of the trapezoid?
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(1.65Newton)*(0.1m)= 0.165Newton meter
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What are the graph points corresponding to x = .05 meter and to x = .20 meter?
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(0.05, 0.75) and (0.20, 3)
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What is the area of the 'graph trapezoid' defined by these points?
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(0.20+0.05)/(2)= 0.125 m 3-0.75=2.25 Newtons 0.125m*2.25Newtons=0.28125 Newton meters
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@& Good, except that the width is .20 meter - .05 meter = .15 meter.*@
What is the meaning of the altitude of this trapezoid?
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the quantities represented by its vertical sides, rise of the trapezoid.
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@& The vertical sides represent forces, so the average altitude represents average force.*@
What is the meaning of the width of this trapezoid?
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distance between the 2 x values, width in meters of the trapezoid.
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@& The width does represent meters.
More specifically it represents meters of displacement.*@
What therefore is the meaning of the area of this trapezoid?
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area=average altitude*width
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@& That's how the area is calculated, but the meaning depends on the meaning of the graph.
In this case the calculation represents average force * displacement.*@
`q012. For the rotation data you took in class:
What was the average rate of rotation in the trial where the added masses were at the end of the rotating beam?
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(change in A)/(change in B) 1.25+1=2.25/2= (1.125 rotations ave)/(2 sec)=1.5625 rotations/sec
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What would then have been the initial rate of rotation (at the instant your finger lost contact with the system)?
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to get 0.5625 rotations/sec for average with final as 0 rotations/sec must have 1.125 rotations/sec as initial
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Assuming that those masses were 14 cm from the center of rotation, how fast were they moving, in cm / second, at that initial instant?
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@& The mass on the end of the beam traveled in a circular path, traveling one circumference of the circle with every rotation. What is the circumference of the circle, and what does this tell you about how fast the mass was actually moving?*@
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Assuming that the masses were each 60 grams, what was the kinetic energy of each mass?
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For University Physics Students only:
`q013. For the v vs. t trapezoid whose width is `dt and whose altitudes are v0 and vf:
What is the slope of the graph and what does the slope mean? Be sure to explain the entire interpretation of the slope.
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What is the area of the graph and what does the area mean? Be sure to explain the entire interpretation of the area.
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In terms of v0, vf, `dt, a and `ds what two equations do we get from the expressions for the slope and the area?
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What equation do we get when we eliminate vf from these two equations? Verify that you know how to do the algebra of this elimination.
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What equation do we get when we eliminate `dt from these two equations? Verify that you know how to do the algebra of this elimination.
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`q014. The following allow us to define work (F * `ds), kinetic energy (1/2 m v^2), impulse (F `dt) and momentum (m v):
If we solve F_net = m a for a and plug the result into the second of the equations obtained above, then solve this equation for F `dt, what is the result? Show the algebra of your solution, or verify that you can do the algebra easily.
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If we solve F_net = m a for a and plug the result into the second of the equations obtained above, then solve this equation for F `ds, what is the result? Show the algebra of your solution, or verify that you can do the algebra easily.
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@& You are using a very good reasoning process and progressing really well. Check my notes carefully for clarification of some important details, and keep up the good work.*@