**Physics Lab Exam Ch 8**

*Read each question carefully, draw a diagram where
appropriate and show your work for partial credit. Remember to put your answers in significant
figures. Good Luck!*

1. For the following variables
from chapter eight please state what each one stands for, and then a
description of what each one is. Try to
explain them as you would to someone who has NOT taken a
physics class: (2 ponts each)

**t **__Torque__ – A measure of the twist
produced by a force around a particular axis.
It is found by multiplying the force applied and the length of the lever
arm that the force is applied perpendicular to.

**a **__Angular acceleration__ is a
measure of the rate of change of an objects rotational velocity. That is, how much the object is speeding up
or slowing down in terms of its spin.

**I **__Moment
of inertia__ is a measure of the resistance to changes in the objects
rotational motion. That is, how hard is
it to get the object spinning, stop it once it is spinning or speed it up or
slow it down.
Moment of inertia depends on not only the mass but how it is distributed
with regards to the axis of rotation.

**L **__Angular
(or rotational) momentum__ is a conserved quantity that is given by
the product of the objects moment of inertia and its rotational velocity. The higher the rotational momentum, the
larger the outside torque needs to be to change it.

**KE _{r}**

2. Both Torque and Work
are products of force and distance. How
are they different? Do they have the
same units? How can you produce the most
torque possible for a
given force? (8 points)

Work is linear whereas Torque is rotational. Also, in the work formula, the distance
refers to the displacement in a direction parallel to the force applied. For torque, the force must be applied
perpendicular to the lever arm for rotation to occur. Work is measured in Joules, but torque is
measured in N m. The maximum torque
occurs when the force is made totally perpendicular to the lever arm and the
length of the lever arm is made as long as possible.

3. What are the two conditions for an
object to be in complete equilibrium? (6 points)

(i) The
Net Force must equal zero

(ii) The Net
Torque must equal zero

4. Suppose you are
designing a car for a coasting race (a pinewood derby or the like) where the
cars have no engines, they simply coast down hill. Do you you want
large wheels or small wheels? Do you
want solid, disk-like wheels or hoop like wheels? Should the wheels be heavy or light? Defend your answers. (8 points)

This is about conservation of energy. The car starts with only potential
energy. At the

bottom ofthe ramp, the PE has been
converted to translational KE of the car and rotational KE of thewheels (disregarding energy generated in the form of
heat from resistive forces).In essence you want the rotational KE to be as
small as possible so more of the PE is turned into translational KE. Since KE_{r} = 1/2Iw^{2} the smaller I (moment of inertia) is made, the smaller KE_{r}
will be.

Therefore, you want ), light wheels
instead of heavy (small M compared to large M), discs instead of hoops (1/2MR^{2} instead of MR^{2} and small wheels (little R compared to big R)

5. Stars originate as
large bodies of slowly rotating gas.
Because of gravity, once stars run out of nuclear fuel they slowly
decrease in size as they collapse on themselves. What happens to the angular speed of the star
as it shrinks? Why? (6 points)

As the star shrinks, it spins faster and faster. This is a matter of conservation of
rotational momentum. L_{i} = L_{f}
Therefore, Iw_{i} = Iw_{f}. Because the moment of inertia of the star
becomes smaller as it shrinks (2/5 MR^{2} where the mass stays constant but R gets smaller) the
angular velocity must increase for momentum to be conserved.

6. A large grinding wheel
coasts for a minute or more after the power is turned off, while an electric
drill coasts only for a few seconds.
Considering that they both spin at about the same speed while in
operation, why is there such a difference in their stopping times? (5 points)

This is a case of the
moment of inertia of the grinding wheel being MUCH larger. It has a larger radius and a greater mass so
that even though both moments of inertia are found using 1/2MR2, the grinding wheel’s is much greater.
Therefore, it resists changes in rotational motion more and doesn’t come
to a stop as quickly.

7. A cylindrical fishing
reel has a moment of inertia of 6.8 x 10^{-4} kg m^{2} and a
radius of exactly 4.0 cm. What is the
mass of the reel? (3 points)

I = ½ MR^{2}.
Therefore, M = 2(I)/R^{2} = 2(6.8
x 10^{-4} kg m^{2})/(0.04m)^{2} = 0.85 kg

When the fisherman gets a bite, the angular acceleration of the wheel
is 66 rad/s^{2}.
What must be the net torque acting on the line? (3 points)

t = Ia = (6.8
x 10^{-4} kg m^{2})(66 rad/s^{2}) = 0.04488 ~ 0.045 N m

There is a friction clutch in the reel which exerts a restraining
torque of 1.3 N m
if a fish pulls on the line. Therefore, what is the torque produced by the __fish__? What is the lever arm for this torque? (3
points)

t_{net} = t_{fish} + t_{friction clutch }so 0.045
N m = t_{fish} + (-1.3 N m)

Therefore, = t_{fish} 1.345 N m

The
lever arm for this torque is the radius of the fishing reel.

What is the Force with which the fish pulls on the line? (3 points)

t = F x l so F = t/l and therefore F = (1.345 N
m)/0.04 m = 33.625 N ~ 34 N

8. A construction worker is
cleaning his hands when his wedding band slips off his finger and lands on a
piece of plywood propped up against a wall at a 60.0 degree angle. It lands upright at the top of the board and
starts rolling down this incline. The
piece of plywood is 1.38 meters long.
The ring has a mass of 18.0 grams and a diameter of 2.00
centimeters. How fast will the ring be
moving when it reaches the bottom and the man tries to grab it? (15 points)

First
find the height of the incline with sin(60) x 1.38m =
1.20 m Now ,conservation of energy

PE_{i} + KE_{Ti}_{ }+ KE_{Ri}_{ }= PE_{f} + KE_{Tf}_{ }+ KE_{Rf}_{ }The ring has no KE of either
kind at the top and no

PE at the bottom.
Therefore, _{i}_{Tf}_{ }+ KE_{Rf}_{ }so mgh =
½ mv^{2} + ½ I w^{2}.

The ring
is a hoop so I = MR^{2} and wr = v_{t} so w = v_{t}/r and we have

mgh = ½ mv^{2} + ½
(MR^{2})(v_{t}/r)^{2} Cancel the mass and the radius ~~m~~gh = ½ ~~m~~v^{2} + ½ (~~MR~~^{2})(v_{t}/~~r~~)^{2}

And we have gh = ½ v^{2} + ½ v^{2}
or gh = v^{2} v = square root of (9.81 m/s2)(1.20m)
= 3.42 m/s

9. A pizza maker tosses up a 5.0 kg piece of dough (a disk) with a radius
of .30 m. What is the moment of inertia
of the dough? (3 points)

It is a disc so I = ½ MR^{2} thus I = (0.5)(5.0 kg)(0.30 m)^{2}
= 0.225 kg m^{2}

If he applies a torque of 4.0 N m, what is the angular acceleration of
the dough? (3 pts)

t = Ia so a = t/I Plugging in we get a = 4.0 (N m)/(0.225
kg m^{2}) = 17.7777 ~ 18
rad/s^{2}

If this torque is applied for a
total of 0.80 s, what will be the angular velocity of the dough? (3 points)

w = aDt so w = (17.8 rad/s^{2})(0.80 s) = 14.24 ~ 14 rad/s

As the dough goes up it spreads
out, reaching a radius of 0.55 m. What
is its final moment of inertia? What is
its final angular velocity? (5 points)

The
dough is still a disc so = ½ MR^{2} thus I = (0.5)(5.0 kg)(0.55 m)^{2}
= 0.756 kg m^{2}

L_{i} = L_{f} so I_{i}w_{i}
= I_{f}w_{f}_{ }Solving for w_{f} we get w_{f} = I_{i}w_{I} / I_{f}.

Thus =
(0.225 kg m^{2})(18 rad/s)
/ (0.756 kg m^{2}) = 5.357 ~ 5.4 rad/s

10. As shown in the picture
below two window washers, Bob and Joe, are on a 3.00 m
long 345 N scaffold supported by two cables attached to its ends. Bob weights 750 N and stands 1.00 m from the
left end. 2 m from the left end is the 500 N washing equipment.
Joe is 0.500 m from the right end and weighs __1000 N__. Given that the scaffold is stationary, what
are the forces on each cable? (16
points) ANSWER IS BELOW THE PICTURE

Use two conditions of equilibrium.

First F_{net}
= 0 N

I will call the left and right cable tensions F_{1} and F_{2} respectively.

F_{net} = F_{1} + F_{2} + (-345 N) + (-750 N) + (-500 N) +(-1000 N) = 0 N

So F_{1} + F_{2} = 2595 N

Second condition says that t_{net} = 0 N m Use left end as the
axis of rotation. Then

t_{net} = (-750 N)(1.00M) + (-345)(1.50m) +
(-500)(2.00m) + (-1000 N)(2.50m) + (F_{2})(3.00m) = 0 N m

So (-4767.5 N m) + F_{2}(3.00m) = 0 N m.
Thus F_{2} = 1589 N or 1590 N with significant figures.

Thus since F_{1} + F_{2} = 2595 N and Thus F_{2} = 1590 N, then F1 must equal 2595 N – 1590 N = 1005 ~ 1010 N with
significant figures.