Solutions for Physics 1201 Course Review (Problems 10 through 17) 10) a) Here, there is no friction between any of the surfaces in contact, so we do not need to be concerned with the normal forces on the blocks. Since the only horizontal force on m

1 would be due to friction, in this case the upper block is not accelerated; thus, a

1 =

0 . In the absence of friction, the only horizontal force acting on m2 is the applied force F

= 6 N. ; hence, the acceleration of the lower block is

€

a2 = Fm2

=6 N.1.3 kg.

≈ 4.62 m.sec.2 .

The lower block will ultimately be pulled out from under the upper one. b) Since the contact surfaces between the blocks and between the lower block and the tabletop are horizontal, the normal forces on the blocks are simply N

1 = m

1g and N

2

= ( m1 + m

2 ) g . The static frictional force acting between the blocks is then

fs ≤ µ

sN

1 = µ

sm

1g . So the two blocks will move together as a unit provided that this limit

applies. (Since the tabletop is frictionless, N2 is not important to know and contact with

the upper block is the only source of friction.)

The horizontal force equation for the lower block is F – f

s = m

2a

2 , while that for

the upper block is fs = m

1a

1 . If the two blocks are to stay together, their accelerations

must match, implying that

€

a1 =fsm1

= a2 =F − fsm2

. When we apply the maximum

possible value for the static frictional force

€

fsmax = µsN1 , we find that the limit for the

applied force is given by

€

µsm1gm1

=Fmax − µsm1g

m2⇒ Fmax = µs ⋅ (m1 + m2 ) ⋅ g

€

≈ 0.45 ⋅ (0.7 + 1.3 kg.) ⋅ (9.81 m.sec.2

) ≈ 8.83 N.  

Since Fmax

> 6 N. , a horizontal applied force at this level still permits the blocks to stay together. Because they move as a unit, their mutual acceleration is

€

a = Fm1 + m2

=6 N.

(0.7 + 1.3 kg.)≈ 3.00 m.

sec.2 . Note that this means that the static

frictional force here is fs = m

1a = ( 0.7 kg. ) ( 3.00 m./sec.2 ) = 2.10 N. , which is certainly

less than

€

fsmax = µsm1g = 0.45 ⋅ (0.7 kg.) ⋅ (9.81 m.sec.2

) ≈ 3.09 N. As a check, we find

that

€

a2 =F − fsm2

≈6.00 − 2.10 N.

1.3 kg.≈ 3.00 m.

sec.2= a    .

For an applied horizontal force F = 10 N. > F

max , the blocks will not move together

and the kinetic frictional force fk = µ

km

1g must be used in our analysis. Upon replacing

fs with f

k in the horizontal force equations for each block, we now have

€

a1' =fkm1

=µk m1gm1

= µk g ≈ 0.32 ⋅ (9.81 m.sec.2

) ≈ 3.14 m.sec.2

and

€

a2' =F − fkm2

≈10.0 N. − 0.32 ⋅ (0.7 kg.) ⋅ (9.81 m.

sec.2)

1.3 kg.  

 

       

€

≈10.0 − 2.20 N.

1.3 kg.≈ 6.00 m.

sec.2   .

The lower block will be pulled out from under the upper one, but here the upper block will move forward a bit first, which was not the case in the situation we treated in part (a) above. c) Nothing else about this system is changed if we apply the force to the lower block with a spring, instead of a cord. In order for the two blocks to continue moving together, the force applied must not exceed F

max ≈ 8.83 N. Hooke’s Law tells us that the restoring force of a

spring is given by Fsp

= k · Δx , where Δx is the displacement of the spring from equilibrium. A maximum allowed force of 8.83 N. requires that the maximum permitted displacement for

this spring be

€

(Δ x )max =Fmaxk

≈8.83 N.72 N./m.

≈ 0.123 m. This will set the limit on the

amplitude for the pair of blocks oscillating at the end of this spring. 11) a) The three horizontal forces acting on the car are the applied force from the

engine,

€

r F A , the kinetic frictional force between the tires and the road,

€

r f k , and the drag

force from the air,

€

r F D . Since the vehicle is on a level road here, the normal force on it

from the road surface is N = Mg , so the kinetic frictional force is fk = µ

k Mg .

With the car traveling at constant speed, the net horizontal force on the car is

€

r F netx

=r F A +

r f k +

r F D = 0 ⇒ Fnetx

= FA − fk − FD = 0 . The net power applied to

the car will then also bezero:

€

Pnet =r F netx

⋅r v =

r F A ⋅

r v +r f k ⋅

r v +r F D ⋅

r v =r F A v cos 0o +

r f k v cos 180o +

r F Dv cos 180o

€

= FAv − fkv − FDv = 0 . We are told that the power being provided by the engine is P

A = F

Av = 95,000 W . At

a speed of

€

v = 65 mi.hr. ⋅ 1609m.mi. ⋅

1 hr.3600 sec. ≈ 29.05 m.sec. , the portion of this power that is

being used to overcome air drag is

€

PD = FDv = FAv − fk v = PA − µkMgv

€

= 95,000W − 0.025 ⋅ (900 kg.) (9.81 m.sec.2

) (29.05 m.sec. )

€

≈ 95,000 − 6410W ≈ 88,600W .

To maintain the car at this speed,

€

88,600W95,000W ≈ 0.933 of the engine’s power is being used

to overcome air drag. b) If the drag force is proportional to the square of the vehicle’s speed, then at 80

mph, the drag is

€

FD 'FD

=v '2v 2

= 80 mph65 mph⎛

⎝ ⎜

⎞

⎠ ⎟ 2

≈ 1.515 times stronger than it is at

65 mph. The power necessary to overcome air drag at this higher speed, however, is

€

PD 'PD

=FD 'v'FDv

= v '2v 2⎛

⎝ ⎜

⎞

⎠ ⎟ ⋅v'v

= 80 mph65 mph⎛

⎝ ⎜

⎞

⎠ ⎟ 3

≈ 1.864 times greater than at 65 mph. This gives

us PD’ ≈ 1.864 P

D ≈ 1.864 · 88,600 W ≈ 165,200 W . The total power the engine must

now supply is PA’ = P

D’ + P

f ≈ 165,200 + 6410 W ≈ 171,600 W ( 230 hp ) .

c) With the car now moving up an incline, the normal force from the road surface becomes N’ = Mg cos θ , making the kinetic frictional force f

k’ = µ

kMg cos θ .

In addition to the forces already described, the weight force on the vehicle now also has a component parallel to the incline, W

|| = Mg sin θ .

The net force on the car is now

€

r F net || =

r F A +

r F D +

r f k ' +

r W || = 0 , so the net

power becomes Pnet

= FAv − F

Dv − ( µ

k Mg cos θ ) v − ( Mg sin θ ) v = 0 . Since the car is

on a 12% grade, tan θ = 0.12 ⇒ sin θ ≈ 0.1191 , cos θ ≈ 0.9929 . We have seen that

the power exerted by air drag is PD ≈ 88,600 W at 29.05 m./sec. (65 mph) and that it is

proportional to v3 , so we can write

€

PD = 88,600 v29.05⎛ ⎝ ⎜ ⎞

⎠ ⎟ 3W . The power which must be

delivered by the engine in order for the vehicle to maintain a speed v on the 12% upward incline is then

€

PA = FAv = PD + (µk Mg cos θ ) v + (Mg sinθ ) v

€

≈ 88,600 v29.05⎛ ⎝ ⎜ ⎞

⎠ ⎟ 3

+ (0.025) (900 kg.) (9.81 m.sec.2

) (0.9929) v + (900 kg.) (9.81 m.sec.2

) (0.1191) v

€

≈ 3.614 v 3 + 219.2 v + 1052 v ≈ 3.614 v 3 + 1271v . We are going to keep the power from the engine at the same level it had in part (b) above, so we need to solve for v the equation 3.614 v3 + 1271 v = 171,600 . It is not very convenient to solve this last equation directly (we can, of course, use graphing software, as will be discussed below), but we don’t need to resort to trial-and-error completely either. We know that this amount of power on level ground allows the car to travel at 80 mph ( ≈ 35.8 m./sec. ), so we might expect that on a 12% climb, the car might move, say, 10% or so slower. Let’s make a first guess that the solution is v = 32 m./sec. Since the cubic term in the equation, 3.614 v3 , changes much more rapidly than the linear term, 1271v , we’ll simply set this term to 1271 · 32 for the present and solve the following for v :

€

3.614 v 3 + 1271⋅ 32 = 3.614 v 3 + 40670 = 171,600        

€

⇒ 3.614 v 3 = 130,900 ⇒ v 3 ≈ 36,200 ⇒ v ≈ 33.1 m.sec. , which is pretty close to our initial guess. If we adjust the linear term to accommodate this new value, we find

€

3.614 v 3 + 1271⋅ 33.1 = 3.614 v 3 + 42070 = 171,600 ⇒ v ≈ 33.0 m.sec. .  

Our solution has stabilized to one decimal place, so we may stop here. A more precise solution, using graphing software to find the x-intercepts for the function 3.614 v3 + 1271 v − 171,600 , gives us v ≈ 32.98 m./sec. , so our result is acceptable.

As a check, we can calculate the individual power terms: power exerted against drag: P

A ≈ 3.614 · 33.03 ≈ 129,900 W

power exerted against friction: Pf ≈ 219.2 · 33.0 ≈ 7230 W

power exerted against gravity: PW ≈ 1052 · 33.0 ≈ 34,700 W

total power to be delivered by engine: 171,800 W , agreeing with the intended total value to about 0.12% . d) Without the applied force from the engine, and assuming that there is no internal friction in the drive train of the car (a bit unrealistic), the net force on the vehicle acting parallel to the downward incline is

€

r F net || ' =

r W || +

r F D +

r f k ' = 0 ⇒ Fnet || ' = W|| − FD − fk ' = 0 ,

making the net power P

net = ( Mg sin θ ) v − F

Dv − ( µ

k Mg cos θ ) v = 0 .

Each of these terms has the same value as it did in part (c) above, giving us

€

1052 v − 3.614 v 3 − 219.2 v ≈ 832.3v − 3.614 v 3 = 0 , which is much easier to solve for v than the power equation we found in the previous part.

We can factor this as

€

v ⋅ (832.3 − 3.614 v 2) = 0  , so either v = 0 (the uninteresting solution, since the power terms are of course all zero when the car is parked) or

€

832.3 − 3.614 v 2 = 0 ⇒ v2 ≈ 230.3 ⇒ v ≈ 15.2 m./sec. (34 mph) . In a real vehicle, internal friction would make the downhill coasting speed somewhat lower than this. 12) a) The central event in this physical arrangement is the elastic collision between the two blocks. The combination of linear momentum conservation with conservation of kinetic energy leads to the result that the relative velocity of the masses before the collision is the opposite of the relative velocity afterwards, that is, v

1 − v

2 = −( v

1’ – v

2’ ) , which may

be rearranged more conveniently as v1 + v

1’ = v

2 + v

2’ . For our collision, this gives us v

+ v’ = 0 + V . This provides a relationship among the velocities, but tells us nothing else. We will need to uncover the rest by examining the energies of the blocks.

One way to analyze this would be to perform a detailed study of all the energy changes for the blocks at each significant event. However, we can save ourselves some trouble by noticing a feature of the motion of m

1 . Upon leaving the spring launcher at the

top of its ramp, m1 has a kinetic energy equal to the potential energy U

sp that had been

stored in the spring. When it reaches the bottom of the ramp to collide with m2 , m

1 has a

kinetic energy Usp

plus the gravitational potential energy it acquired by descending 35 cm., since this ramp is frictionless (so the angle of this incline or any curvature that it has become irrelevant). After the collision, m

1 has just enough kinetic energy to come to rest

as it reaches its launcher; hence, it has lost an amount of energy Usp

. But because the

collision with m2 is elastic, this must be the amount of energy transferred to m

2 . We can

thus conclude that after the collision, m

2 has a kinetic energy of

½ m

2V 2 = U

sp = ½ k

1 ( Δx )

1

2 ≈ ½ ( 100 N./m. ) ( 0.11 m. )2 ≈ 0.605 J. Returning to m1 , we have said that its kinetic energy before the collision was ½ m

1v2 = U

sp + m

1gH = 0.605 + ( 0.1 kg. ) ( 9.81 m./sec.2 ) ( 0.35 m. )

≈ 0.605 + 0.343 J. ≈ 0.948 J. and after the collision, the kinetic energy became ½ m

1v’ 2 = m

1gH ≈ 0.343 J.

So we can calculate that and

( m1 has reversed direction, so v’

becomes negative). The relative-velocity equation then tells us that v + v’ = 0 + V ⇒ V ≈ 4.36 + ( −2.62 ) ≈ 1.73 m./sec.* * rounding from a larger number of significant figures Since we know the kinetic energy of m

2 after the collision, we can now solve for the value

!

v =2 (Usp + m1gH )

m1

"

# $

%

& '

1/ 2

(2 ) 0.948 J.0.1 kg.

*

+ ,

-

. /

1/ 2

( 4.36 m.sec.

!

v' = "2 #m1gH

m1$

% & '

( )

1/ 2

* "2 # 0.343 J.0.1 kg.

+

, -

.

/ 0

1/ 2

* " 2.62 m.sec.

€

m2 =2K2'V 2 =

2Usp

V 2 ≈2 ⋅ 0.605 J.(1.74 m.

sec. )2 ≈ 0.402 kg.

(This can also be determined using linear momentum conservation.) While much of the information in this part of the Problem may seem to have been extraneous, it will be quite useful for the next portion. b) In this new situation, m

2 comes to rest just as it makes contact with the spring

bumper on its ramp. So after the collision, m2 has only enough kinetic energy to climp the

ramp and overcome its friction. The work m2 must perform against gravity is m

2g · ( h

1 +

h2 ) = ( 0.402 kg. ) ( 9.81 m./sec.2 ) ( 0.025 + 0.05 m. ) ≈ 0.296 J. The length of incline

which exerts frictional force on m2 is given by

€

sin 9o =h2L

⇒ L =h2sin 9o

≈0.05 m.0.156

≈ 0.320 m.

The work which m

2 must perform against kinetic friction is then

µ

kN

2L = µ

k · ( m2

g cos 9º ) · L ≈ ( 0.17 ) ( 0.402 kg. ) ( 9.81 m./sec.2 ) ( 0.988 ) ( 0.320 m. ) ≈ 0.212 J. Thus, m

2 must have departed from the collision with m

1 with a kinetic energy of

0.296 + 0.212 ≈ 0.508 J. , which means that its velocity after the collision in this situation

was

€

V ≈2 ⋅ 0.508 J.0.402 kg.

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1/ 2

≈ 1.59 m.sec. .

We now need to find the speed of m

1 before this collision, in order to determine

how it was launched. The relative-velocity equation tells us v + v’ = 0 + V ≈ 1.59 m./sec. As we are really only interested in v , we will solve this equation for v’ and eliminate it from the linear momentum conservation equation: m

1 v + m

2 · 0 = m

1 v’ + m

2 V

⇒ ( 0.100 kg. ) · v ≈ ( 0.100 kg. ) ( 1.59 – v m./sec. ) + ( 0.402 kg. ) ( 1.59 m./sec. ) ≈ 0.159 − 0.100 v + 0.640 kg.−m./sec. ⇒ 0.200 v ≈ 0.799 ⇒ v ≈ 4.00 m./sec. So the kinetic energy of m

1 before the collision was

½ m

1v2 ≈ ½ ( 0.100 kg. ) ( 4.00 m./sec. )2 ≈ 0.799 J.

As in part (a), this energy is the sum of the spring launcher’s stored potential energy and the gravitational potential energy released as m

1 drops by 35 cm. Since this latter amount has

already been found to be 0.343 J. , the launcher must have held a potential energy of 0.799 – 0.343 J. ≈ 0.456 J. Using the formula for the (ideal) spring potential energy in the launcher, the initial compression of the spring in this situation would have been

€

Usp ' = 12 k1 (Δx ' )

2 ≈ 0.456 J. ⇒ Δx ' ≈ 2 ⋅ 0.456 J.100 N.

m.

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

1/ 2

≈ 0.096 m. or 9.6 cm.

13) a) In a “simple” pendulum, it is idealized that all of the pendulum’s mass is concentrated in a mass-point at one end. For a realistic or “physical” pendulum, however, we must take into account that the mass is distributed over the body of the pendulum. If we deflect the pendulum from its vertical equilibrium by an angle θ , the gravitational torque acting to restore it to equilibrium will be in the opposite direction with a magnitude of r

CM · Mg · sin θ , where r

CM is the distance from the pivot point of the pendulum to its

center-of-mass. Since the pendulum turns rigidly, the torque is

€

τ = I totα = I tot ⋅d 2θdt 2

= −MgrCM sinθ . If the angular displacement from equilibrium is

small ( θ < 0.3 radians ), then sin θ ≈ θ (in radians) and we can write

€

I tot ⋅d 2θdt 2

≈ −MgrCMθ ⇒ d 2θdt 2

+MgrCMItot

θ ≈ 0 ,

which is a differential equation for simple harmonic motion. For this pendulum,

€

ω 2 =MgrCMItot

, so it will oscillate with a period of

€

T = 2πω = 2π ⋅ I tot

MgrCM .

If we apply this formula to the simple pendulum, I

tot = M l2 and r

CM = l , since all the

mass is at one end; we then obtain the familiar result

€

T = 2π ⋅ M l 2Mgl = 2π ⋅ l

g .

For this longcase clock (sometimes called a “grandfather’s clock”) pendulum, the rod has mass m and length L and is pivoted from one end, with the bob being a disk of mass M and radius R centered at a distance d below the pivot. The distance of the center-of-mass from the pivot is then

€

rCM =m ⋅ ( 1

2L ) + M ⋅ dm + M ≈

(2.000 kg.) ⋅ ( 12⋅1.100 m.) + 5.000 kg. ⋅ d

2.000 + 5.000 kg. ≈ 0.1571 + 0.7143 d m.

The total moment of inertia of the pendulum is that of a rod pivoting about one end plus a disk revolving about a point outside its center-of-mass. The moment of inertia for the

uniform rod is then

€

13mL

2 . By the parallel-axis theorem, the moment inertia of the heavy

pendulum bob is ½ MR2 + Md 2 . Thus, the total moment of inertia of the pendulum is given by

€

I tot = 13mL

2 + 12MR2 + M d 2

€

≈ 13 (2.000 kg.) (1.100 m.)

2 + 12(5.000 kg.) (0.080 m.)2 + (5.000 kg.) d 2 ≈ 0.82267 + 5d 2 kg. ⋅m.2  

  We want to adjust this pendulum so that the period is very precisely equal to 2.000 seconds at sea level on the Earth’s equator. So we will need to move the bob to the position for which

€

T = 2π ⋅ I totM totgrCM

Mtot

is the total mass of the pendulum

€

≈ 2π ⋅ 0.82267 + 5d 2 kg.⋅m.2

(2.000 + 5.000 kg.) (9.78033 m.sec.2

) (0.1571 + 0.7143 d m.)

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

1/ 2

≈ 2.000 sec .

€

⇒0.82267 + 5d 2

10.7554 + 48.9026 d≈

2.0002π

⎛

⎝ ⎜

⎞

⎠ ⎟

2

⇒ 0.82267 + 5d 2 ≈ 1.08975 + 4.95487 d

€

⇒ 5d 2 − 4.95487 d − 0.26708 ≈ 0 ⇒ d ≈ 1.0422 m. only positive solution

b) Thermal contraction has caused the linear dimensions of this pendulum to be reduced by 1/1000 ( 0.1% ) . The effect on the bob itself is insignificant, but the effect on the length of the rod and the distance of the bob from the pivot both must be considered. The length of the rod is now L’ = 0.999 · L ≈ 1.0989 m. This brings the bob closer to the pivot by a proportional amount, so d’ = 0.999 · d ≈ 0.999 · 1.0422 m. ≈ 1.0412 m. The expression for the distance from the pivot to the center-of-mass becomes

€

rCM ' =m ⋅ ( 1

2L' ) + M ⋅ d 'm + M ≈

(2.000 kg.) ⋅ ( 12⋅1.0989 m.) + 5.000 kg. ⋅1.0412 m.

7.000 kg. ≈ 0.90070 m.

and that for the total moment of interia is

€

I tot ' = 13mL'

2 + 12MR2 + M d '2

a 0.1% reduction of R has negligible effect on the middle term

€

≈ 13 (2.000 kg.) (1.0989 m.)

2 + 12(5.000 kg.) (0.080 m.)2 + (5.000 kg.) (1.0412 m.)2 ≈ 6.24154 kg. ⋅m.2

This gives the period of the pendulum in this situation as

€

T ' ≈ 2π ⋅ 6.24154 kg.⋅m.2

(7.000 kg.) (9.78033 m.sec.2

) (0.90070 m.)

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

1/ 2

≈ 1.99899 sec.

In a week of 7 days · 86400 sec./day = 604,800 seconds, the pendulum will make

€

604,800T '

≈604,8001.99899

≈ 302,553 oscillations,

corresponding to 2 · 302,553 ≈ 605,106 seconds registered by the clock. The clock

would then be 605,106 − 604,800 ≈ 306 seconds ( 5.09 minutes ) fast.

To restore the pendulum to keeping accurate time, the bob would need to be moved

to a distance D from the pivot given by

€

2π ⋅ 0.82105+ 5D2 kg.⋅m.2

(7.000 kg.) (9.78033 m.sec.2

) (0.15699 + 0.7143D m.)

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

1/ 2

≈ 2.000 sec .

€

⇒ 5D2 − 4.95477D − 0.26794 ≈ 0 ⇒ D ≈ 1.0424 m.   only positive solution

The bob should therefore be moved downward away from the pivot by

D – d’ ≈ 1.0424 − 1.0412 m. ≈ 0.0012 m. ( ≈ 1.2 mm.) .

To make such fine adjustments to the position of the bob, it was often fitted with a

calibrated vernier screw, indicated the number of minutes per day of correction. It is this

sensitivity to environmental changes that led people to see non-mechanical methods of timekeeping.

14) We know that this spring-mass system (ideally) obeys the simple harmonic motion

differential equation,

€

m d 2 (Δ x )dt 2

+ k (Δ x ) = 0 , for which the solution for the

displacement from equilibrium is Δx = A sin( ωt + φ ) , with

€

ω = km . At a moment we

will call t = 0 , the displacement is Δx = −0.37 m. (the block starts, say, to the left of the origin), the velocity is v = 2.64 m./sec. (the block is moving to the right), and the acceleration is a = 1.90 m./sec.2 (the force on the block is to the right). The spring constant is given as k = 64 N/m. If we differentiate the displacement equation with respect to time, we obtain

€

v = dxdt = A ⋅ ω ⋅ cos (ω t +φ ) ,

€

a = dvdt = A ⋅ ω ⋅ω ⋅ [ − sin (ω t +φ ) ] = − Aω 2 sin (ω t +φ ) .

At time t = 0 , this gives us ( Δx )

0 = A sin ( ω · 0 + φ ) = A sin φ = −0.37 m. ,

v

0 = Aω cos ( ω · 0 + φ ) = Aω cos φ = 2.64 m./sec. , and

a

0 = −Aω2 sin ( ω · 0 + φ ) = −Aω2 sin φ = 1.90 m./sec.2 .

We can now calculate

€

a0(Δ x )0

=−Aω 2 sinφA sinφ = −ω 2 ≈

1.90 m.sec.2

−0.37 m. ⇒ ω 2 ≈ 5.135 rad.sec.2

€

⇒ ω ≈ 2.266 rad.sec. ;

period of oscillator:

€

T =2πω ≈

2π rad.2.266 rad.

sec.

≈ 2.77 sec . ;

the mass of the block is given by

€

ω 2 = km ⇒ m = k

ω 2 ≈64 N.

m.

5.135 rad.2

sec.2

≈ 12.5 kg. ;

€

ω ⋅ (Δ x )0v0

=ω ⋅ A sinφAω cos φ = tanφ ≈

(2.266 rad.sec.) (−0.37 m.)

2.64 m.sec.

≈ − 0.318 .

Since ( Δx )

0 < 0 , so sin φ < 0 also, and v

0 > 0 , so cos φ > 0 as well. This tells us

that φ is in quadrant IV , and thus that the phase angle is given simply by the calculator

value to φ ≈ tan−1 ( −0.318 ) ≈ −0.308 rad. (or 5.976 rad.). Finally, we can determine the

amplitude of oscillation from

€

A =(Δ x )0sinφ =

A sinφsinφ ≈

−0.37 m.sin (−0.308) ≈

−0.37 m.−0.313 ≈ 1.18 m.

The equation for the displacement from equilibrium as a function of time is therefore Δx = 1.18 sin ( 2.266 t − 0.308 ) m. The total mechanical energy of this spring-

mass oscillator is given by E = ½ k A2 ≈ ½ ( 64 N./m. ) ( 1.18 m. )2 ≈ 44.56 J .

15) Sphere A can be treated as a sphere of radius R and density ρ , with an added, embedded sphere with radius ½ R and a density of 2ρ (giving that region a total density of 3ρ ). Sphere B has a cavity in it of the same size, so we will work with it as a sphere of radius R and density ρ , with the added sphere having a “density” of −ρ (producing a region with density zero). We will then work out the properties of these two sphere in parallel calculations.

a) total mass – This first calculation will make it clear why we choose to “break up” these asymmetric spheres in the peculiar way described above. We have the “large

sphere” with radius R and density ρ , which has a mass of M =

€

ρ ⋅ 4π3 R3 ; a “small excess-density sphere” with radius ½ R , density 2ρ , and a mass of

mA =

€

2ρ ⋅ 4π3 ⋅ 12 R⎛ ⎝ ⎜ ⎞

⎠ ⎟ 3

= ρ ⋅2 ⋅ 4π3 ⋅8 ⋅ R3 = ρ ⋅ π3 ⋅ R

3 or 14 M ; and a “small negative-

density sphere” with radius ½ R , density 2ρ , and a mass of mB =

€

−ρ ⋅ 4π3 ⋅ 12 R⎛ ⎝ ⎜ ⎞

⎠ ⎟ 3

€

= − ρ ⋅ 4π3 ⋅8 ⋅ R3 = − ρ ⋅ π6 ⋅ R

3 or − 18M . We can now put these results together to find

the total mass of sphere A as MA = M + m

A =

€

ρ ⋅ 4π3 + π3

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ⋅ R3 = ρ ⋅ 5π3 R3

(so M = 4mA and M

A = 5m

A ) and M

B = M + m

B =

€

ρ ⋅ 4π3 − π6⎛ ⎝ ⎜ ⎞

⎠ ⎟ ⋅ R3 = ρ ⋅ 7π6 R3

(so M = −8mB and M

B = −7m

B ). These additional relationships will be helpful in the

parts of the problem to follow. b) moment of inertia about symmetry axis – The axis which passes through both the center of the “large sphere” and the center of the “small sphere” may be regarded as the symmetry axis for both sphere A and sphere B .

Since the centers-of-mass of the large and small sphere lie on this axis, the moment of inertia of the assemblage about this axis will simply be the sum of the center-of-mass moments of inertia for the uniform-density spheres. Thus, for sphere A, we have

€

IA1 = 25M R2 + 2

5mA ⋅ (12 R )

2 = 25 ⋅ (4mA )R

2 + 25mA ⋅ (

12 R )

2

€

= ( 85 + 220 ) ⋅mAR

2 = 1710mAR

2 = 1750 MAR

2 = 0.34MAR2 ,

while for sphere B, we find

€

IB1 = 25M R2 + 2

5mB ⋅ (12 R )

2 = 25 ⋅ (− 8mB )R

2 + 25mB ⋅ (

12 R )

2

€

= (− 165 + 220 ) ⋅mBR

2 = − 3110mBR2 = 31

70 MBR2 ≈ 0.443MBR

2 . We see that sphere A, having a greater density of mass concentrated near this rotation axis, has a lower coefficient of rotational inertia ( 0.34 ) than does sphere B ( 0.443 ), where the cavity causes more of its mass to be distributed farther away from this rotation axis. c) moment of inertia about perpendicular axis through geometrical center – Here, we will use a rotational axis through the center of the large sphere and perpendicular to the line we’ve called the symmetry axis.

We can thus still use the center-of-mass moment of inertia for the large sphere, but we need to invoke the “parallel-axis theorem” to work out the altered moment for the small sphere. This requires the addition of a term which is the mass of the object times the square of the distance from its center-of-mass to the rotation axis. The new result for sphere A is

€

IA2 = 25M R2 + 2

5mA ⋅ (12 R )

2 + mA ⋅ (12 R )

2

"parallel−axis"term1 2 4 4 3 4 4

= 25 ⋅ (4mA )R

2 + 75mA ⋅ (

12 R )

2

€

= ( 85 + 720 ) ⋅mAR

2 = 3920mAR

2 = 39100 MAR

2 = 0.39MAR2 ,

while for sphere B, we find

€

IB1 = 25M R2 + 2

5mB ⋅ (12 R )

2 + mB ⋅ (12 R )

2

"parallel−axis"term1 2 4 4 3 4 4

= 25 ⋅ (− 8mB )R

2 + 75mB ⋅ (

12 R )

2

€

= (− 165 + 720 ) ⋅mBR

2 = − 5720mBR2 = 57

140 MBR2 ≈ 0.407MBR

2 . Relative to this rotation axis, there is far less difference between the coefficients of rotational inertia of spheres A and B . d) location of center of mass – When an object has a uniform density, its center- of-mass will be in the position of its geometrical center. Since we are treating spheres A and B as “composite objects”, made up of two individual spheres of constant density, the centers for A and B will lie along the symmetry axis connecting the centers of the large sphere and the small sphere. So we can identify the symmetry axis as the “x-axis” and make the center of the large sphere x = 0 ; this places the center of the small sphere at x = ½ R . This permits us to calculate the position of the center-of-mass of sphere A as

€

x A =0 ⋅M + ( 12R ) ⋅mA

M + mA=

12 ⋅mA

M AR =

mA2 ⋅ 5mA

R = 110 R ,

and that of sphere B as

€

x B =0 ⋅M + ( 12R ) ⋅mB

M + mB=

12 ⋅mB

M BR =

mB2 ⋅ (−7mB )

R = − 114 R .

e) moment of inertia about perpendicular axis through center-of-mass – We again choose rotation axes perpendicular to the symmetry axes of our spheres A and B, as we did in part (c), but now these axes pass through the centers-of-mass we have just determined.

As before, we apply the “parallel axis theorem” to find this moment of inertia for sphere A to be

€

IA3 = [ 25M R2 + M ⋅ x A2

"parallel−axis"term1 2 3

] + [ 25mA ⋅ (12 R )2 + mA ⋅ (

12 R − x A )

2

"parallel−axis"term1 2 4 4 4 3 4 4 4

]

€

= 25 ⋅ (4mA )R

2 + (4mA ) ⋅ (110 R )

2 + 25mA ⋅

14 R

2 + mA ⋅ (12 R −

110 R )

2 ]

€

= ( 85 + 4100 + 2

20 + 16100 ) ⋅mAR

2 = 1910mAR

2 = 1950 MAR

2 = 0.38MAR2 ,

and for sphere B,

€

IB3 = [ 25M R2 + M ⋅ x B2

"parallel−axis"term1 2 3

] + [ 25mB ⋅ (12 R )2 + mB ⋅ (

12 R − x B )

2

"parallel−axis"term1 2 4 4 4 3 4 4 4

]

€

= 25 ⋅ (− 8mB )R

2 + (− 8mB ) ⋅ (−114 R )

2 + 25mB ⋅

14 R

2 + mB ⋅ (12 R − [−

114 R])

2 ]

€

= (− 165 − 8196 + 2

20 + 64196 ) ⋅mBR

2 = − 19770 mBR2 = 197

490 MBR2 ≈ 0.402MBR

2 . 16)

a) Since the only horizontal force acting on the block is static friction with the turntable surface, this must be the source of the centripetal force that keeps the block turning about the rotation axis of the turntable. So we have

€

Fc =mv 2R = mω 2R = fs ≤ µsN .

Because the surface is horizontal, N = mg ; hence,

€

ω 2R ≤ µs g .

The turntable is rotating at a frequency of f = 45 rev./min. , so its angular speed is

€

ω = 2π f = 2π rad.rev. ⋅ 45rev.min. ⋅

1 min.60 sec. ≈ 4.71 rad.sec. . With the block placed at R = 0.06 m.

from the rotation axis, the coefficient of static friction necessary to hold the block in place

is given by

€

µs ≥ω 2Rg ≈

(4.71 rad.sec. )2 ⋅ 0.06 m.

9.81 m.sec.2

≈ 0.136 .

b) In order for static friction to hold the block in place at the edge of the turntable

platter, where R = 0.15 m. , we now require

€

µs ≥(4.71 rad.sec. )

2 ⋅ 0.15 m.

9.81 m.sec.2

≈ 0.339 .

c) For a fixed coefficient of static friction, we can rearrange the force inequality to find the maximum distance at which an object can be situated from the rotation axis and

still be held in place by static friction,

€

R ≤µs gω 2 . With µ

s = 0.23 for this turntable, the

maximum distance from the rotation axis would be given by

€

R ≈0.23 ⋅ 9.81 m.

sec.2

(4.71 rad.sec. )2

≈ 0.102 m. (10.2 cm.) .

d) While the turntable’s angular speed is increasing, there will be both a centripetal

acceleration

€

aC =vT2R = ω 2R and a tangential acceleration a

T = αR , where v

T is the

instantaneous tangential velocity of the turntable at radius R and α is the angular acceleration; both of the these accelerations are in the plane of the turntable surface. Since centripetal acceleration points along a radius of the circle of motion and the tangential acceleration points along a tangent to that circle (hence the name), a

C and a

T are

perpendicular. So the total acceleration of a point on the turntable surface is the magnitude

of the resultant acceleration vector,

€

atot = aC2 + aT

2 . The static friction must be able to

provide this total acceleration to keep the block moving along with the turntable, so we

require

€

atot ≤ µs g .

The turntable accelerates from rest ( ω

0 = 0 ) to a final angular speed

ωf = 4.71 rad./sec. (45 rpm) at a uniform rate α in a time T , or

€

α =ω f − ω0

T =ω fT .

The tangential acceleration is then

€

aT = αR =ω fT

⎛ ⎝ ⎜

⎞ ⎠ ⎟ R . The greatest total acceleration

will therefore occur as the turntable reaches its final angular speed, at which time

€

atot = (ω f2 R )2 + (

ω f

TR )2

⎡

⎣ ⎢

⎤

⎦ ⎥

1/ 2

. So the requirement for static friction to hold the block in

place on the accelerating turntable gives us

€

(ω f2 R )2 + (

ω f

TR )2

⎡

⎣ ⎢

⎤

⎦ ⎥

1/ 2

≤ µs g ⇒ (ω f

TR )2 ≤ (µs g )

2 − (ω f2 R )2

€

⇒ω f2 R2

T 2 ≤ (µs g )2 − (ω f

2 R )2 ⇒ω f2 R2

(µs g )2 − (ω f2 R )2

≤ T 2  

 

 

€

⇒ T 2 ≥(4.71 rad.sec. )

2 ⋅ (0.06 m.)2

(0.23 ⋅ 9.81 m.sec.2

)2 − ( [ 4.71 rad.sec. ]2 ⋅ 0.06 m. )2

≈ 0.0241 sec.2  

   

€

⇒ T ≥ 0.155 sec. Thus, the minimum time in which the turntable may accelerate from rest to its operating angular rate of 45 rpm without causing the block to slip is 0.155 seconds for uniform angular acceleration. (Note that the mass of the block does not enter into any of the results in this Problem.) 17)

a) The hanging spring scale “reads weight” by registering the tension T in its spring. With the scale connected to the block by a cord, this tension can be determined from the free-body diagram for the block from T + B – Mg = 0 ⇒ T = Mg – B , where B is the buoyancy force, equal to the weight of a volume of water which is the same as the

volume of the block, so

€

B = ρw V g = ρw ⋅Mρb

⎛

⎝ ⎜

⎞

⎠ ⎟ ⋅ g =

ρwρb

⎛

⎝ ⎜

⎞

⎠ ⎟ ⋅Mg , with ρ

w being the

density of water and ρb being the density of the block. So the reading on the spring scale

is

€

T = Mg −ρwρb

⎛

⎝ ⎜

⎞

⎠ ⎟ ⋅Mg = 1 −

ρwρb

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥ ⋅Mg    .

The platform scale “reads weight” by registering the normal force N , which is provided by an internal mechanism of the scale (frequently, a spring system). This force can be determined from the free-body diagram for the scale as N − M

wg − M

Bg − B = 0 ⇒ N = ( M

w + M

B )g + B ,

where M

w is the mass of water and M

B is the mass of the beaker. The buoyancy force

enters into this force equation because the fluid pressure which pushes on the block also pushes on the walls of the beaker. For our problem, the block has mass M = 0.840 kg. , the beaker, M

B = 0.095 kg. ,

and the water, Mw = 1.36 kg. The readings for aluminum and magnesium blocks are

density ( gm./cm.3 ) buoyancy force, B

aluminum 2.65

€

1.00 gm.cm.3

2.65 gm.cm.3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⋅ 0.84 kg. ⋅ 9.81 m.

sec.2≈ 3.11 N

magnesium 1.74

€

1.00 gm.cm.3

1.74 gm.cm.3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⋅ 0.84 kg. ⋅ 9.81 m.

sec.2≈ 4.74 N

spring scale reading, T = Mg − B

aluminum

€

(0.84 kg. ⋅ 9.81 m.sec.2

) − 3.11 ≈ 5.13 N

magnesium

€

(0.84 kg. ⋅ 9.81 m.sec.2

) − 4.74 ≈ 3.50 N

platform scale reading, N = ( M

w + M

B )g + B

aluminum

€

([1.36 + 0.095 kg.] ⋅ 9.81 m.sec.2

) + 3.11 ≈ 17.38 N

magnesium

€

([1.36 + 0.095 kg.] ⋅ 9.81 m.sec.2

) + 4.74 ≈ 19.01 N

b) As this is a physical arrangement in static equilibrium, the net force on the rod is zero; thus, T

L + T

R + B − Mg = 0 .

We are told that the rod has a length of 75 cm. and a cross-sectional area of 8.0 cm.2 , so

its volume is 600 cm.3 The buoyancy force is equal to the weight of an equal volume of

water, hence

€

B = ρw V g = (1 gm.cm.3

) 1 kg.1000 gm.⎛

⎝ ⎜

⎞

⎠ ⎟ (600 cm.3 ) (9.81 m.

sec.2) ≈ 5.89 N .

The force equation then gives us T

L + T

R = Mg − B = ( 2.8 kg. ) ( 9.81 m./sec.2 ) − 5.89 ≈ 21.58 N .

We can also analyze the torques acting on the rod. If we place the pivot point at the left end of the rod, then we can neglect the tension T

L , since it will have a moment-

arm of zero length. As for the other torques,

torque due to weight of rod: τ

W = − ( 25 cm. ) · Mg · sin 90º

torque due to buoyancy force: τ

B = + ( ½ · 75 cm. ) · B · sin 90º

torque due to tension in right-hand cord: τ

TR = + ( 75 cm. ) · T

r · sin 90º

For this static equilibrium, τW + τ

B + τ

TR = −25 Mg + 37.5 B + 75 T

r = 0

€

⇒ TR =(25 cm.) ⋅ (2.8 kg.) ⋅ (9.81 m.

sec.2) − (37.5 cm.) ⋅ (5.89 N)

75 cm. ≈ 6.21 N .

From the force equation above, we find that T

L = 21.58 − T

R ≈ 15.37 N .

Were the right-hand cord disconnected or removed, its tension contribution would become zero and our force equation would now be T

L = Mg − B ≈ 21.58 N .

The center-of-mass of the rod would then need to be in a different location in order for the rod to remain static and level. Let us call x this new distance of the center of mass from the left end of the rod. The torque equation is then

€

τW + τ B = − x ⋅ Mg + 37.5 ⋅ B = 0 ⇒ x =(37.5 cm.) ⋅ (5.89 N)(2.8 kg.) ⋅ (9.81 m.

sec.2)≈ 8.04 cm.

-- G.Ruffa

original notes developed during 2007-09 revised: September 2010