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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.

 

KEr     Rotational kinetic energy indicates the energy needed (in Joules) to rotate a certain object at a certain rotational speed. 

 

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 KEr = 1/2Iw2 the smaller I (moment of inertia) is made, the smaller KEr will be. 

 

Therefore, you want ), light wheels instead of heavy (small M compared to large M), discs instead of hoops (1/2MR2 instead of MR2 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.  Li = Lf  Therefore, Iwi = Iwf.  Because the moment of inertia of the star becomes smaller as it shrinks (2/5 MR2 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  m2 and a radius of exactly 4.0 cm.  What is the mass of the reel? (3 points)

 

I = ½ MR2.  Therefore, M = 2(I)/R2 = 2(6.8 x 10-4 kg m2)/(0.04m)2  = 0.85 kg

 

When the fisherman gets a bite, the angular acceleration of the wheel is 66 rad/s2.  What must be the net torque acting on the line? (3 points)

 

t = Ia  = (6.8 x 10-4 kg m2)(66 rad/s2) = 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)

tnet = tfish + tfriction clutch so 0.045 N m = tfish + (-1.3 N m) 

Therefore,  = tfish 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

PEi + KETi + KERi =  PEf + KETf + KERf  The ring has no KE of either kind at the top and no

PE at the bottom.  Therefore, PEi + KETf + KERf  so mgh = ½ mv2 + ½ I w2. 

The ring is a hoop so I = MR2 and wr = vt so w = vt/r and we have

mgh = ½ mv2 + ½ (MR2)(vt/r)2  Cancel the mass and the radius mgh = ½ mv2 + ½ (MR2)(vt/r)2 

 

And we have gh = ½ v2 + ½ v2 or gh = v2  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 MR2  thus I = (0.5)(5.0 kg)(0.30 m)2 = 0.225 kg m2

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 m2) = 17.7777 ~ 18 rad/s2

 

 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/s2)(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 MR2  thus I = (0.5)(5.0 kg)(0.55 m)2 = 0.756 kg m2

 

Li = Lf so Iiwi = Ifwf  Solving for wf we get wf = IiwI / If. 

Thus = (0.225 kg m2)(18 rad/s) / (0.756 kg m2) = 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 Fnet = 0 N

 

I will call the left and right cable tensions F1 and F2 respectively.

Fnet = F1 + F2 + (-345 N) + (-750 N) + (-500 N) +(-1000 N) = 0 N

So F1 + F2 = 2595 N

 

Second condition says that tnet = 0 N m  Use left end as the axis of rotation.  Then

tnet = (-750 N)(1.00M) + (-345)(1.50m) + (-500)(2.00m) + (-1000 N)(2.50m) + (F2)(3.00m) =  0 N m

 

So (-4767.5 N m) + F2(3.00m) = 0 N m.  Thus F2 = 1589 N or 1590 N with significant figures.

 

Thus since F1 + F2 = 2595 N and Thus F2 = 1590 N, then F1 must equal 2595 N – 1590 N = 1005 ~ 1010 N with significant figures.