Physics 11 Fall 2012

Practice Problems 6 - Solutions

1. Two points are on a disk that is turning about a fixed axis perpendicular to the diskand through its center at increasing angular velocity. One point is on the rim and theother point is halfway between the rim and the center.

(a) Which point moves the greater distance in a given time?

(b) Which point turns through the greater angle?

(c) Which point has the greater speed?

(d) Which point has the greater angular speed?

(e) Which point has the greater tangential acceleration?

(f) Which point has the greater angular acceleration?

(g) Which point has the greater centripetal acceleration?

————————————————————————————————————

Solution

(a) The point on the rim travels a greater distance in the same amount of time, sinceit has to make a bigger “orbit.”

(b) Both points move through the same angle.

(c) The speed is v = rω. Since point particles change their angle by the same amountin the same time, the point with the bigger radius has the bigger velocity. So, thepoint on the rim moves with the greater speed.

(d) Since the angular speed is ω, and both points travel through the same angle inthe same time, they both have the same angular speed.

(e) The problem states that the disk is turing at increasing angular velocity, andso ω is increasing. Thus, v = rω is increasing. The tangential acceleration isatang = rα. So, the particle with the bigger radius has the bigger tangentialacceleration; i.e., the point on the rim has the bigger tangential acceleration.

(f) The angular acceleration is α, and is the same for both points.

(g) The centripetal acceleration is acent = v2/r = (rω)2 /r = rω2. Since ω is the samefor both points, the point on the rim has the bigger centripetal acceleration, sinceit has the bigger value of r.

1

2. What is the angular speed of Earth in radians per second as it rotates about its axis?

————————————————————————————————————

Solution

The angular speed is ω = ∆θ∆t

. The Earth rotates through an angle of 2π radians in 24hours. 24 hours is 24× 3600 seconds, or 86,400 seconds. So, the angular speed is

ωEarth =∆θ

∆t=

2π

86400= 7.3× 10−5 rad/sec.

2

3. The methane molecule (CH4) has fourhydrogen atoms located at the vertices ofa regular tetrahedron of edge length 0.18nm, with the carbon atom at the centerof the tetrahedron. Find the moment ofinertia of this molecule for rotation aboutan axis that passes through the centers ofthe carbon atom and one of the hydrogenatoms.

————————————————————————————————————

Solution

Because the axis of rotation is through thecarbon and one of the hydrogens, neitherof these atoms contribute. The moment ofinertia is then due only to the remainingthree hydrogens. The moment of inertia is

I =3∑i=1

mir2i = mH

3∑i=1

r2i ,

since all of the masses are identical. Nowthe radii are also all equal is is just thedistance from the central hydrogen to theatoms on the vertex. if the distance be-tween the hydrogens is a, then the radiusis

r =a

2

1

cos 30◦=a

2

2√3

=a√3.

Rotation

859

Substitute to obtain: [ ] 2rod12

12sphere

2sphere5

22 LMhMRMI ++=

Substitute numerical values and evaluate I:

( )( ) ( )( )[ ] ( )( )2

212122

52

mkg0415.0

m0.300kg0.0600m0.200kg0.500m0.0500kg0.5002

⋅=

++=I

The percent difference between I and appI is:

%6.3mkg0.0415

mkg0.0400mkg0.04152

22app =

⋅⋅−⋅

=−

III

(b) The rotational inertia would increase because Icm of a hollow sphere is greater than Icm of a solid sphere. 49 •• The methane molecule (CH4) has four hydrogen atoms located at the vertices of a regular tetrahedron of edge length 0.18 nm, with the carbon atom at the center of the tetrahedron (Figure 9-48). Find the moment of inertia of this molecule for rotation about an axis that passes through the centers of the carbon atom and one of the hydrogen atoms. Picture the Problem The axis of rotation passes through the center of the base of the tetrahedron. The carbon atom and the hydrogen atom at the apex of the tetrahedron do not contribute to I because the distance of their nuclei from the axis of rotation is zero. From the geometry, the distance of the three H nuclei from the rotation axis is

3/a , where a is the length of a side of the tetrahedron. a

C

H

HH

H

Apply the definition of the moment of inertia for a system of particles to obtain: 2

H

2

H

23H

22H

21H

i

2ii

33 amam

rmrmrmrmI

=⎟⎠

⎞⎜⎝

⎛=

++== ∑

Substitute numerical values and evaluate I:

( )( )247

2927

mkg104.5

m1018.0kg101.67

⋅×=

××=−

−−I

So, we find that

I = 3mH

(a√3

)2

= mHa2

= (1.67× 10−27) (0.18× 10−9)2

= 5.41× 10−47 kg m2.

3

4. During most of its lifetime, a star maintains an equilibrium size in which the inwardforce of gravity on each atom is balanced by an outward pressure force due to the heat ofthe nuclear reactions in core. But after all the hydrogen “fuel” is consumed by nuclearfusion, the pressure force drops and the star undergoes gravitational collapse until itbecomes a neutron star. In a neutron star, the electrons and protons are squeezedtogether by gravity until they form neutrons. Neutron stars spin very rapidly and emitintense pulses of radio and light waves, one pulse per rotation. These “pulsing stars”were discovered in the 1960s and are called pulsars.

(a) A star with the mass (M = 2.0 × 1030 kg) and size (R = 7.0 × 108 m) of oursun rotates once every 30 days. After undergoing gravitational collapse, the starforms a pulsar that is observed by astronomers to emit radio pulses every 0.10seconds. By treating the neutron star as a solid sphere, deduce its radius.

(b) What is the speed of a point on the equator of the neutron star? Your answer willbe somewhat too large because a star cannot be accurately modeled as a solidsphere. Even so, you will be able to show that a star, whose mass is 106 largerthan the earth’s, can be compressed by gravitational forces to a size smaller thana typical state in the United States!

————————————————————————————————————

Solution

(a) As the star collapses, it speeds up by the conservation of angular momentum. It’sinitial angular momentum is Li = MviRi = MR2

iωi = 2πfiMR2i , where we have

recalled that v = Rω, and that ω = 2πf , where f is the rotation frequency. Theangular momentum after the collapse is, then, Lf = 2πffMR2

f . Conservation ofangular momentum says that Li = Lf , and so

Rf =

√fiffRi =

√TfTiRi,

where we’ve re-expressed the radius in terms of the rotation period. Now, theinitial frequency is once per 30 days, so Ti = 3600× 24× 30 = 2.6× 106 seconds,while Tf = 0.10 seconds. Thus,

Rf =

√TfTiRi =

√0.1

2.6× 1067× 108 = 1.37× 105 m = 137 km.

(b) The speed of a point on the equator is v = Rfωf = 2πRf/T = 2π(1.37×105)/0.1 =8.6× 106 m/s.

4

5. A pendulum consisting of a string of length L attached to a bob of mass m swings ina vertical plane. When the string is at an angle θ to the vertical,

(a) calculate the tangential acceleration of the bob using∑Ft = mat.

(b) What is the torque exerted about the pivot point?

(c) Show that∑τ = Iα with at = Lα gives the same tangential acceleration as found

in Part (a).

————————————————————————————————————

Solution

(a) The pendulum is seen in the figure to theright. The only tangential force is the com-ponent of gravity along the rotation path,

Ft = mg sin θ.

Since Ft = mat, then we just solve for theacceleration,

at =Ftm

= g sin θ.

(b) The torque is just τ = rFt, where r = L isthe distance from the pivot point. So,

τ = mgL sin θ.

(c) The torque is τ = Iα. Since the stringis massless and all the mass is in the bob,then I = mr2 = mL2. So, τ = mgL sin θ =mL2α. Furthermore, since α = τ/I =at/L, then at = τL

I, or

at =τL

I=mgL2 sin θ

mL2= g sin θ,

which is the same result we found in part(a).

Chapter 9

872

Substitute numerical values and evaluate I:

2

2

mkg 1.19

mkg 10.19

s 120s 60

min 1rev

rad 2minrev600

s 0.20s 60

min 1rev

rad 2minrev600

mN 0.50

⋅=

⋅=××−

+××

⋅= ππI

(b) Substitute numerical values in equation (2) and evaluate τfr:

( ) mN 0.10s 120

s 60min 1

revrad 2

minrev600

mkg 10.19 2fr ⋅−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ ××−⋅=

π

τ

where the minus sign is a consequence of the fact that the frictional torque opposes the motion of the wheel. 61 •• A pendulum consisting of a string of length L attached to a bob of mass m swings in a vertical plane. When the string is at an angle θ to the vertical, (a) Using tt maF =∑ , calculate the tangential acceleration of the bob? (b) What

is the torque exerted about the pivot point? (c) Show that∑ = ατ I with αLa =t gives the same tangential acceleration as found in Part (a). Picture the Problem The pendulum and the forces acting on it are shown in the free-body diagram. Note that the tension in the string is radial, and so exerts no tangential force on the ball. We can use Newton’s 2nd law in both translational and rotational form to find the tangential component of the acceleration of the bob.

θ

Tr

gmr

L

θ

(a) Referring to the free-body diagram, express the component of

gm that is tangent to the circular path of the bob:

θsint mgF =

Use Newton’s 2nd law to express the tangential acceleration of the bob:

θθ sinsintt g

mmg

mFa ===

5

6. A uniform 1.5 m diameter ring is pivotedat a point on its perimeter so that it isfree to rotate about a horizontal axis thatis perpendicular to the plane of the ring.The ring is released with the center of thering at the same height as the axis.

(a) If the ring was released from rest,what was its maximum angularspeed?

(b) What minimum angular speed mustit be given at release if it is to rotatea full 360◦?

————————————————————————————————————

Solution

(a) When the ring is just hanging straight down, it has zero potential energy. Whenit’s released from it’s initial height, h = R, then it has potential energy mgh =mgR. Now, at the bottom of the path, all the energy is kinetic. Since it’s rotating,the energy is all rotational kinetic energy, KE = 1

2Iω2. What’s I? We can look

up the moment of inertia of the ring about an axis through the center to findI = mR2. Since the ring is rotating about it’s rim, then we can use the parallelaxis theorem to write the moment of inertia about the rim as Irim = Icenter+md

2 =mR2 +mR2 = 2mR2. So, the kinetic energy is KE = 1

2Iω2 = mR2ω2.

So, if energy is conserved, then KE = PE ⇒ mR2ω2 = mgR. Solving for ω gives

ω =

√g

R.

Plugging in the numbers gives

ω =

√g

R=

√9.8

0.75= 3.61 rad/sec.

(b) Now we want to figure out how fast we’d need to rotate it around in order for itto rotate to the top. In this case, it still has the same kinetic energy, for some ω,KE = mR2ω2. Now it has to get to the top, changing its height by R, so thatit’s final potential energy is PE = mgR. Solving for KE = PE ⇒ ω =

√g/R,

exactly the same as in part (a).

6

7. A uniform solid sphere of mass M and ra-dius R is free rotate about a horizontal axisthrough its center. A string is wrappedaround the sphere and is attached to anobject of mass m. Assume that the stringdoes not slip on the sphere. Find

(a) the acceleration of the object and

(b) the tension in the string.

————————————————————————————————————

Solution

(a) We can draw the force diagram for the sys-tem as seen in to the right. The weight ofthe block creates a torque on the wheel,causing it to rotate. We can write downthe net forces and torques acting on thesystem. For the sphere,∑

τ = TR = Iα.

While for the hanging mass, calling thevertical direction x,∑

Fx = T −mg = −ma.

Chapter 9

890

77 •• A uniform solid sphere of mass M and radius R is free to rotate about a horizontal axis through its center. A string is wrapped around the sphere and is attached to an object of mass m (Figure 9-58). Assume that the string does not slip on the sphere. Find (a) the acceleration of the object, and (b) the tension in the string. Picture the Problem The force diagram shows the forces acting on the sphere and the hanging object. The tension in the string is responsible for the angular acceleration of the sphere and the difference between the weight of the object and the tension is the net force acting on the hanging object. We can use Newton’s 2nd law to obtain two equations in a and T that we can solve simultaneously.

0

R

x

T

T

mg

F

M

x (a) Noting that T = T ′, apply Newton’s 2nd law to the sphere and the hanging object:

∑ == ατ sphere0 ITR (1)

and ∑ =−= maTmgFx (2)

Substitute for Isphere and α in equation (1) to obtain:

( )RaMRTR 2

52= (3)

Eliminate T between equations (2) and (3) and solve for a to obtain:

mM

ga

521+

=

(b) Substitute for a in equation (2) and solve for T to obtain: Mm

mMgT25

2+

=

78 •• Two objects, of masses m1= 500 g and m2 = 510 g, are connected by a string of negligible mass that passes over a pulley with frictionless bearings (Figure 9-59). The pulley is a uniform 50.0-g disk with a radius of 4.00 cm. The string does not slip on the pulley. (a) Find the accelerations of the objects. (b) What is the tension in the string between the 500-g block and the pulley? What is the tension in the string between the 510-g block and the pulley? By how

Now, since the string isn’t slipping around the wheel, the angular acceleration isjust α = a/R. Furthermore, the moment of inertia for a sphere is I = 2

5MR2. So,

we find the following system of equations

T = 25Ma

T = m (g − a) .

Setting these two expressions equal to each other and solving for the accelerationgives

a =g

1 + 2M5m

.

(b) Now we just substitute this result back in to the expression T = 25Ma to find

T =2

5

Mg

1 + 2M5m

=2Mmg

5m+ 2M.

7

8. A basketball rolls without slipping down an incline of angle θ. The coefficient of staticfriction is µs. Model the ball as a thin spherical shell. Find

(a) the acceleration of the center of mass of the ball,

(b) the frictional force acting on the ball, and

(c) the maximum angle of the incline for which the ball will roll without slipping.

————————————————————————————————————

Solution

(a) We can begin by writing down Newton’slaws for the ball, including the torques:∑

Fx = −mg sin θ + Ff = −ma∑Fy = −mg cos θ + Fn = 0∑τi = FfR = Iα.

Furthermore, since the ball is rolling with-out slipping, we have that a = αR. So, wehave enough to solve this system of equa-tions. Plugging in Ff = Iα/R = Ia/R2 tothe first equation and solving for a gives

a =mg sin θ

m+ I/R2.

Rotation

905

Picture the Problem The three forces acting on the basketball are the weight of the ball, the normal force, and the force of friction. Because the weight can be assumed to be acting at the center of mass, and the normal force acts through the center of mass, the only force which exerts a torque about the center of mass is the frictional force. Let the mass of the basketball be m and apply Newton’s 2nd law to find a system of simultaneous equations that we can solve for the quantities called for in the problem statement.

θ

m r

gmr

nFr

fr

y

x0

(a) Apply Newton’s 2nd law in both translational and rotational form to the ball:

∑ =−= mafmgFx ssinθ , (1)

∑ =−= 0cosn θmgFFy (2)

and ∑ == ατ 0s0 Irf (3)

Because the basketball is rolling without slipping we know that:

ra

=α

Substitute in equation (3) to obtain:

raIrf 0s = (4)

From Table 9-1 we have:

232

0 mrI =

Substitute for I0 and α in equation (4) and solve for fs:

( ) maframrrf 3

2s

232

s =⇒= (5)

Substitute for fs in equation (1) and solve for a:

θsin53 ga =

(b) Find fs using equation (5): ( ) θθ sinsin 5

253

32

s mggmf ==

(c) Solve equation (2) for Fn: θcosn mgF =

Use the definition of fs,max to obtain:

maxsnsmaxs, cosθμμ mgFf ==

The moment of inertia of the spherical shell is just I = 23mR2, and so

a =mg sin θ

m+ 2m/3R2=

3

5g sin θ.

(b) Since Ff = IR2a =

23mR2

R2 a = 23ma, we have

Ff =2

5mg sin θ.

(c) Now, we know that the frictional force is Ff = µsFN = µsmg cos θ. Setting thisequal to our result from part (b) we find

2

5mg sin θ = µsmg cos θ ⇒ tan θ =

5

2µs.

So, the maximum angle of the incline that the ball will roll without slipping is

θ = tan−1

(5

2µs

).

The rolling of the ball increases the angle from the sliding case.

8

9. Released from rest at the same height,a thin spherical shell and a solid sphereof the same mass m and radius R rollwithout slipping down an incline throughthe same vertical drop H. Each is movinghorizontally as it leaves the ramp. Thespherical shell hits the ground a horizontaldistance L from the end of the ramp andthe solid sphere hits the ground a distanceL′ from the end of the ramp. Find theratio L′/L.

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Solution

The distance that each ball goes after leaving the ramp depends on it’s speed uponleaving the ramp. Since both balls take the same amount of time to reach the groundthen L = v∆t, while L′ = v′∆t, where v is the speed of the shell and v′ is the speedof the sphere at the bottom of the ramp. Now, the ratio L′/L = v′/v. So, we need toknow the speed of the balls. Each ball starts off with the same potential energy, mgH.The final energy of each ball at the bottom of the ram is made up of translational androtational kinetic energy,

KEshell = 12mv2 + 1

2Ishellω

2

KEsphere = 12mv′2 + 1

2Isphereω

′2

Now, since the balls are rolling without slipping, ω = v/R, while ω′ = v′/R. Pluggingin these expressions gives

KEshell = 12

(1 + Ishell

mR2

)mv2

KEsphere = 12

(1 +

Isphere

mR2

)mv′2

Since energy is conserved KEshell = KEsphere = mgH. So,

mgH = 12

(1 + Ishell

mR2

)mv2

mgH = 12

(1 +

Isphere

mR2

)mv′2

Taking the ratio v′/v gives

v′

v=

√√√√ 1 + Ishell

mR2

1 +Isphere

mR2

.

Since Ishell = 23mR2, while Isphere = 2

5mR2, we have

v′

v=L′

L=

√1 + 2/3

1 + 2/5=

√25

21= 1.09.

9

10. According to the Standard Model of Particle Physics, electrons are pointlike particleshaving no spatial extent. (This assumption has been confirmed experimentally, andthe radius of the electron has been shown to be less than 10−18 meters.) The intrinsicspin of an electron could in principle be due to its rotation. Let us check to see if thisconclusion is feasible.

(a) Assuming that the electron is a uniform sphere whose radius is 1.00 × 10−18 m,what angular speed would be necessary to produce the observed intrinsic angularmomentum of ~/2?

(b) Using this value of the angular speed, show that the speed of a point on the“equator” of a “spinning” electron would be moving faster than the speed of light.What is your conclusion about the spin angular momentum being analogous to aspinning sphere with spatial extent?

————————————————————————————————————

Solution

(a) The angular momentum can be expressed in terms of the moment of inertia, I,and the angular velocity, ω, as L = Iω. Thus, the angular speed is ω = L/I.Now, for a uniform sphere, I = 2

5MR2, and for L = ~/2, we find

ω =L

I=

5~4MR2

.

Taking M = me = 9.11× 10−31 kg, we have

ω =5~

4MR2=

5× 1.05× 10−34

4× 9.11× 10−31 × (10−18)2= 1.44× 1032 rad/sec.

(b) The speed is v = Rω, so using our results from part (a) gives

v = Rω = 10−18 × 1.44× 1032 = 1.44× 1014 m/s!

This is much faster than light. So, if we require that these “points” on the “equator”can’t spin faster than light, we have to abandon the model of the electron as a tinylittle spinning sphere.

10