Answers to Questions.
Last updated on 28 May/98
A. Questions.
1. If a mass on a string of radius r is subject to an angular
acceleration, a
for a period of t, what is the angular and tangential velocity of
the mass at this time. Let r = 0.5m, and a = 4 /sec^2.
w = at = 4*t
v = rw = 0.5*4*t
2. To achieve a tangential speed of v by your foot when the foot
pivots by your hip joint by a constant
angular acceleration over an angle q, what is the necessary angular
acceleration? The foot and hip are
separated by a distance r. Let v = 10m/s and q = 1.2
(v/r)^2 = w^2=2aq. a = (v/r)^2/q = (10/r)^2/1.2
B. Questions.
1. A manual for tightening the head bolts of a car specify the torque on the bolts for the final tightening. Howcome they don’t specifiy the force? What is the tool used for this and how does it work?
The manual torque wrench applies a measureable torque to a bolt head or nut. The wrench measures the torque by measuring the applied force at right angles to a carefully prepared fixed radius.
2. Make a sketch of the torque produced by your bicep while you hold a mass in your hand with the forarm held horizontally. How much torque must your bicep produce if your hand is holding 10 Kg? How much force on the bicep is required.?
Please see the sketch made in the classroom. For my hand, the approximate radius to the bicep from the elbow pivot point is approximately 1.2" and the radius to my hand is14".
The torque produced by the 10 kg mass is T = fr = 10*10 * 0.36m = 36 N-m
To produce this torque with the bicep, the force from the bicep must be:
F=T/r = 36/0.03= 1200N!!
C. Questions.
1. If an athlete reduces
the moment of inertia of his leg by flexion such that the effect
is to shorten a length of 2/3 of the original leg length, how
much faster can the athlete rotate his leg through Dq?
Shortening the leg by flexing at the knee will reduce the moment of inertia. Since I = mr^2, we have T=Ia = mr^2*a.
a = T/mr^2
and q=1/2 * a *t^2. so t^2 = 2q/a = 2q*mr^2/T. Now take the ratio of t' and t where the T, m and q are the same.
t'/t = r'/r = 2/3
2. To have a quick release of the throw to 2nd base to catch a
steal, how does the catcher flex his arm for the cock and throw?
The catcher should use a
shorted flexed arm to reduce the moment of inertia for a quick
cock and fire. This gives good velocity plus short time to throw
the runner out. See calculation above in 1.
3. Estimate the moment of inertia of a diver in a tuck position
vs an extended position. How does this affect the speed of
rotation. What position should be used for the entry to the water
for a high scoring dive?
The diver in the tuck position is approximately half the length of the extended position. This will reduce the moment of inertia by to 1/4. Since momentum is conserved, Iw = I'w' . We have w'/w = I/I' = 4. The tuck position gives 4x the angular velocity. To enter the water, the diver needs to have a vertical entry to make a very small splash. By extending the body, the rotation slows down and the torque exerted on the hands at the beginning of the entry will slow the rotation even more.
D. Questions.
1.The skater doing a spin will pump herself up [applying torque ]
to a high angular
speed. Then without effort, she will increase the angular speed
for the finale! Is
this violating a law?
The skater produces torque from the
reaction of the ice to pushes with her skates to build up to a
high angular momentum. She will have her arms and leg extended
sideways to produce a larger moment of inertia about a vertical
axis. Rising to her toe so that she minimizes the friction [no
torque condition], she will spin for a long time. By drawing her
arms and leg into the turning axis, she will reduce the moment of
inertia. To conserve angular momentu, she must spin faster. This
is a illustration of the conservation of angular momentum and not
a violation of physical law!
2.A diver takes off from the high board and rotates several times
with high speed
then slows down to enter the water with a vertical entry. Does
she violate a
physical law?
The skater produces torque while pushing
off from the diveboard and has a certain angular momentum as she
leaves. This is preserved. She can increase her rotational speed
by decreasing her moment of inertia by using the Pike or Tuck
position so that her body parts are at a smaller radius to the
cm. To prepare for entry to the water, she would extend herself
fully thereby having a large moment of inertia and having a
resultant small angular velocity. This small angular velocity
will be further diminished by the torque produced by the hands
entering the water. No laws are broken. This is an illustration
of the law of conservation of energy and of rotational dynamics.
3. A cat released from a legs-up position and without any initial
angular rotation will
be able to land feet down on the floor without rotation. How is
this possible? Is
this violating the conservation of angular momentum? Is there
some human sport
maneuver that emulates this? Please describe this.
The cat departs legs up with zero angular momentum. By rotating her tail, she can generate torque on her body to rotate the body. When she stops rotating the tail, the body stops rotating. The net angular momentum of the cat is always zero.
The dives [called the gainer?] is a twisting around axese other than the horizontal one in the usual spin moves using initial angular momentum. This is done by motions of the hand and leg to rotate the body while having no angular momentum about a particular axis.
4. A wheel with mass = 5 Kg , I
= 30 Kg-m^2 is spun with a w = 200/sec.
It is permitted to run up a hill with friction. How high up the
hill can this wheel go? Does the slope of the hill matter?
We use the conservation of energy here. The wheel initially has kinetic energy of rotation. This can do work to climb the hill to reach a height where it has potential energy equal to the original kinetic energy much like the pole vault calculation. The slope of the hill doesn't matter if the wheels roll without slipping.
KE = PE =1/2 I w^2 = mgh.
h= 1/2 Iw^2/mg = 1/2*30 *200^2/(5*10)=12000m.
The spinning wheel has lots of kinetic energy! There are buses that have been made with the flywheels spun up at the base and then they run around the city using the stored flywheel energy. The chief limitation is that the wheels are rather heavy so that going up hills is a problem and with high speeds, they might fly apart with great danger to people and property.
Extra Questions.
If you are about to fall-off a balance beam, which way should you move your arms to prevent falling? Make a sketch to illustrate the sense of all of this.
This is called the propellor maneuver by our dancing student. If you were falling off the beam while facing forward, you should rotate your arms with hands going forward at the top, then down, and backwards at the bottom then up to the top. The torque reaction will help rotate your body backwards.
A sprinter after the finish line sometimes will swing arms in a circle to keep his balance. Which way should the arms be swung? Which way would you tend to fall if you didn't swing the arms?
In this case, the cm of your body tends to continue to go forward while your legs being tired are slowing down so you would tend to fall face forward. This is the same situation as above and you should swing your arms in a circle in the same was as described above.
Angular Momentum.
A skater obtains a rotational speed with her arms and legs held
away from the
vertical axis of ration. As she draws her limbs inward, the mass
doesn't change but
the moment of inertia gets smaller. The result is that the
angular velocity increases.
This happens so that the angular momentum must be conserved.
Calculations.
L = I w where w = is the angular velocity = Dq/Dt, and I is the
moment of inertia.
The moment of inertia of a point mass rotating at a distance r
from a fixed point is
I = mr^2.
eg. a skater spinning at 90 revolutions per minute has an angular
velocity of 90
rpm. To carry out calcuations in international units, we must
convert this to
radians/sec. 1 revolution = 6.4 radians and 1 minute = 60 sec.
w = 90 rpm * 6.4/rev *min/60sec =9.6 radians/sec
The foot of a skater would have a moment of inertia [around the
vertical axis of
about: I = mr^2 = 1 * .5^2 = .25 kg-m^2. When the foot is brought
close to the
axis of rotation, the I becomes close to zero.
Questions on Conservation of Energy in fluids.
1. A 1.3 kg sphere is dropped through a tall
column of liquid. When the sphere has fallen a distance of 0.5m,
it is observed to have a velocity of 2.5m/s. [First you should
identify all the energy sources and where the energy can go.]
a) How much work was done by the frictional viscosity of the
liquid?
b) What was the average force of friction?
c) If the ball were to be doubled in radius [but without changing
it's mass] describe quantitatively the change in passage of the
ball. eg it's velocity after falling .5m; how much energy is lost
to the liquid. d) What was the approximate radius of the ball in
the original problem.
a) The potential gravitational energy given up by the ball is mgh = 1.3*10*.5=6.5J
The kinetic energy of the ball after having fallen this far is 1/2 mv^2 = 1/2*1.3*2.5^2=4J
Since the total energy must be accounted,
TE = PE = KE + H. H = PE -KE = 6.5-4=2.5J
b) Force of friction of the fluid . W = F*d, F = W/d = 2.5J/.5m = 5 N
c) What happens if the radius is doubled but mass is still the same .
F=Cd A 1/2 rv^2.
If the radius is doubled, the area is quadrupled. The frictional force would be quadrupled.
This means that the force would be 20N which would be more than the weight of the ball [13N] This can't happen, the ball would be floating and not fall at all.
d) The radius of the ball for the original problem.
A = F*2/(Cd rv^2) = 5*2/(.5*1000*2^2)= .005 m^2 = pi*r^2
r = 0.04m
Stirring up the fluid. Putting energy into the fluid.
form drag
skin drag
wave drag
Reduce Cd, reduce whetted area, reduce form area.
Is Momentum conservation still hold true for lift??
Absolutely. the momentum conservation law is an absolute one. The lift force is produced by a momentum transfer to the air flow. The air must flow downwards after the passage of the airplane. This usually happens in the form of vortices [whirlpools of air.] The vortices produced by the passage of a large aircraft is dangerous to the flights of small planes in their wake.. and often causes crashes.