PHYSICAL SCIENCES 2Lecture 5 Review

Fall 2017

1 Concepts

1. Rotation axis: axis in which rigid body rotates about. It is perpendicular to the plane of rotation.

2. Angle θ:

The angle at which the rigid body rotates throughabout the rotational axis. We typically use θ = 0rad at the x-axis (i.e. θ is the standard angle).We define counterclockwise rotations from θ = 0rad to be positive. For rotational motion, theunits of θ are radians, where 2π rad = 360◦.

3. Angular displacement ∆θ = θf − θi:

• Counter-clockwise displacements are positive ∆θ > 0.

• Clockwise displacements are negative ∆θ < 0.

4. Arc-Length s: is the length of a circular arc. For a circular arc with radius r, the arc-lengththat spans an angular displacement ∆θ is

s = r∆θ. (1)

5. Angular Velocity: defined as the rate of angular displacement in time,

ω =dθ

dt→ Without Calculus: 〈ω〉 =

∆θ

∆t(2)

The sign of ω follows the same sign of ∆θ. All points on a rigid rotating body share the same ω.

6. Tangential speed vt: defined as the magnitude of the velocity of an object (or a fixed point onthe rigid object) along the curve. It is given by the rate of change of the arc-length s in time,

vt =ds

dt= rω Without Calculus: vt = rω (3)

where r is the radial distance between the object (or fixed point) and the axis of rotation. Whileall points on a rotating rigid body share the same ω, their tangential speed will increase withits radial distance from the center.

7. Angular Acceleration α: defined as the rate of change of angular speed in time,

α =dω

dt→ Without Calculus: 〈α〉 =

∆ω

∆t(4)

The sign of α is positive for rotations speeding up in counter-clockwise direction and negativefor rotations speeding up in the clockwise direction.

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8. Tangential acceleration at: describes the acceleration of an object (or point on object) aroundthe arc-length of the curve. A prime example is a person accelerating along a circular track.The relation between the tangential and angular acceleration an object (or fixed point on a rigidobject) at constant r is,

at =dvtdt

= rdω

dt= rα→ Without Calculus: at = rα (5)

If the tangential acceleration is nonzero, the object (or fixed point) is not executing uniformcircular motion. Instead, the object will be executing nonuniform circular motion and thetangential speed vt(t) will be changing in time.

9. Uniform Circular Motion: motion that requires for an object to move around a circle at fixedradial distance from the center r and constant tangential speed vt. In this case, the object hasonly centripetal acceleration,

ac =v2tr

= ω2r. (6)

which only changes the direction of the object’s velocity.

10. Centripetal ac Acceleration versus Tangential Acceleration at: Any object (or fixed point on arigid body) moving along a curve with radius of curvature r and a tangential speed vt experiencesa centripetal acceleration,

ac =v2tr

= ω2r. (7)

This holds whether the object is executing uniform circular motion or nonuniform circular mo-tion. If the object is additionally accelerating along the curve, the object will have a tangentialacceleration, at. In the case of nonuniform circular motion, both the tangential speed vt, and

thereby the centripetal acceleration, ac =v2t (t)r , will change in time.

The total acceleration ~a of an object (or fixed point) alonga curve of radius r is,

~a = ac (inward) + at (along curve) (8)

The magnitude of the total acceleration is just,

a =√a2c + a2t (9)

11. Rotational Kinematics for Constant α:

Angle θ θ(t) = θ0 + ω0t+ 12αt

2

Angular Velocity ω ω(t) = ω0 + αt

Relation of ω and ∆θ ω2f − ω2

0 = 2α∆θ

2

When using these solutions for ω and θ, it is important to use radians (not degrees), and to in-corporate the appropriate sign conventions. For example, a disk that is rotationally deceleratingin the clockwise direction is slowing down in the clockwise direction and hence, has a positiveangular acceleration α > 0. In contrast, a disk that is rotationally accelerating in the clockwisedirection has negative angular acceleration α < 0.

12. Moment of Inertia I: a physical measure of how the mass of the rotating body is distributedabout its axis of rotation. It describes an object’s resistance to rotational motion.

For a point mass, the moment of inertia is

Ipm = mr2 (10)

where r is the distance of the mass from the rotational axis.

For a system of N point masses, the moment of inertia of the system is just the sum of in-ertias,

Isys =∑j

Ij,pm = m1r21 +m2r

22 +m3r

23 + · · ·+mNr

2N (11)

where r1, r2, · · · , rN is the distance between each mass and the axis of rotation.

13. Angular momentum L: a physical quantity defined as a product of a moment of inertia I andan angular velocity ω of an object,

L = Iω. (12)

It quantifies the amount of rotational motion of the object.

14. If a system is isolated from the environment, the total angular momentum of a system is con-served,

∆Ltot = 0 ⇒ Li,tot = Lf,tot (13)

∆Ltot∆t

= 0 With calculus:dLtotdt

= 0 (14)

15. Torque τ : is a physical measure of how much a force acting on an object causes that objectto rotate. It is the rotational analog to force. In this class, we will only be working with themagnitude and sign1 of torque.

τ = F (r sinφ) = Fr⊥ (16)

τ = (F sinφ)r = F⊥r (17)

where r⊥ is called the moment arm and is perpendicular to ~F and Ft is called the tangentialforce and is perpendicular to ~r. See figures below.

1Torque is a vector quantity and mathematically defined as the cross-product of the position vector to the forcerelative to the pivot and force, ~τ = ~r × ~F . The cross-product takes two vectors, ~r and ~F , and creates another vectorthat is perpendicular to both vectors and has a magnitude of rF sinφ. If ~r and ~F lie in the xy-plane, the torque willbe in the ±z. For this class, we just call the torque positive or negative. However, what we really mean is that torquepoints in either the positive z direction or the −z direction.

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The magnitude of torque is,

|~τ | = rF sinφ (15)

where r = |~r| is the distance between the pivotand point of force application, F is the force ap-plied, and φ is the angle between ~r and ~F . Onecan intuitively think of the sin θ term in torque asa measure of how perpendicular ~r and ~F are toeach other.The more perpendicular ~r and ~F are toeach other, the larger the torque will be. Hence,we can alternative write the torque as,

16. Sign of Torque:

• τ > 0 for forces that lead to counter-clockwise rotations about the pivot.

• τ < 0 for forces that lead to clockwise rotations about the pivot

The sign convention can be summarized with the help of the right-hand rule:

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17. Rotational Analog for Newton’s 2nd Law: A net torque on an object changes an angular mo-mentum of a system:

dL

dt= τnet =

∑j

τj . (18)

In case the moment of inertia I of the object remains constant, it is equivalent to say that thenet torque causes the object to rotationally accelerate,

Iα = τnet, (19)

where α is the angular acceleration.

18. Summary of Angular VS Linear Quantities

Pure Translation (Fixed Direction) Pure Rotational (Fixed Axis)

Position: x Angle: θ

Velocity: v = dxdt Angular Velocity: ω = dθ

dt

Acceleration: a = dvdt Angular Acceleration: α = dω

dtMass: m Moment of Inertia: IMomentum: p Angular momentum: L

Newton’s 2nd Law: dpxdt = Fnet,x (max = Fnet,x) Newton’s 2nd Law: dL

dt = τnet (Iα = τnet)Kinetic Energy: K = 1

2mv2 Kinetic Energy: K = 1

2Iω2

19. Conditions for Equilibrium:

• No net force acting on the body: ~Fnet =∑

j~Fj = ~0

• No net torque acting on the body: τnet =∑

j τj = 0

These two conditions must hold for any choice of pivot in order for the object to be inequilibrium. Static equilibrium occurs when an object is at rest. Dynamic equilibrium occurswhen an object moves at constant velocity and angular velocity.

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20. Moment of Inertia of Rigid Objects:

Rigid objects are made of a bunch of infinitesimal point masses, δmj . We can find the mo-ment of inertia of the object by summing over the moment of inertia for all these tiny masses,

Isys =∑j

δmjr2j (20)

where r is the radial distance from the rotational axis. One can use calculus,2 to find the momentof inertia for various shapes shown in the table below.

2We can convert the sum into an integral,∑

j δmjr2j ⇒ Isys =

∫dm r2. For this class, you will never need to

do this. Always refer to the table.

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2 Selected Problems

The difficulty of each problem is rated on a scale of 1-5, with 1 being easy (novice level) and 5being the most difficult (expert level). For context, Sapling problems generally have a difficultylevel of 1-23, and offline HW problems generally have a difficulty level of 3-4. Bonus questions inlecture have a difficulty level of 4-5.

1. (3/5) A bead is given an initial velocity and then circles indefinitely around a frictionless verticalhoop. Only one of the vectors in the figure below is a possible acceleration vector at the givenpoint. Which one? Circle all the correct choices below.

(a) Point (a)

(b) Point (b)

(c) Point (c)

(d) Point (d)

(e) Point (e)

2. (2/5) Torque Square

In the overhead view of the figure shown below, fiveforces of the same magnitude F act on a strange merry-go-round; it is a square that can rotate about point P ,at midlength along one of the edges. Rank the forcesaccording to the magnitude of the torque they createabout point P, greatest first. Hint: It might be easierto use |~τ | = Fr⊥ = Fr sin θ.

3. In the figure below, a disk, a hoop, and a solid sphere are made to spin about fixed centralaxes by means of strings wrapped around them, with the strings producing the same constanttangential force ~F on all three objects. The three objects have the same mass M and radius R,and they are initially stationary. The corresponding moment of inertias for each of the objectsis, Idisk = 1

2MR2, Ihoop = MR2, and Isphere = 25MR2.

(a) (3/5) Rank the objects according to the magnitude of their angular acceleration αhoop, αsphere,and αdisk about their central axes.

(b) (2/5) Rank their angular speeds ωhoop, ωdisk, and ωsphere, greatest first, when the stringshave been pulled for a certain time ∆t. Hint: What can we say about ω(t) when the angularacceleration is constant?

3Note that some of the Sapling problems have a difficulty level of 3-4.

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4. (3/5) Rank in order, from largest to smallest, the angular accelerations αa to αe of the rigidobjects in cases (a)-(e) shown below.

5. The flywheel of a steam engine turns counter-clockwise with a constant angular velocity of 5πrad/s. When steam is shut off, the friction of the bearings and of the air stops the wheel in 7200s.

(a) (3/5) What is the constant angular acceleration of the wheel during the slowdown?

(b) How many revolutions does the wheel make before stopping?

(c) (3/5) At the instant the flywheel is turning at 2.5 π rad/s, what is the tangential componentof the acceleration of a point on the flywheel that is 50 cm from the axis of rotation?

(d) (3/5) What is the magnitude of the net linear acceleration ~a of the particle in (c)?

(e) (3/5) The flywheel has a total mass of M = 100 kg and is a solid cylinder with radiusR = 1 m, Calculate the net work done by friction on the flywheel.

6. Starting from rest, a race car accelerates along a circular racetrack with radius r at constanttangential acceleration at.

(a) (2/5) Write an expression for the tangential speed vt of thecar as a function of time.

(b) (3/5) Write an expression for the magnitude of the centripetal

acceleration as a function of time, |~ac| = ac =v2tr .

(c) (3/5) Find the magnitude of the car’s total acceleration ~anet.

(d) (5/5) If the coefficient of static friction between the road andthe car’s tires is µs, at what time will the car start to slip?

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7. (4/5) You pull on a roll of toilet paper with a given force F and observe the angular accelerationα. A few days later when the radius is half of what it was, you pull with the same force F andangle. (For simplicity, ignore the hollow tube; assume that the toilet paper goes all the waydown to zero radius.) The ratio of the new α to the old is...

(a) 1

(b) 2

(c) 4

(d) 8

(e) 16

8. (2/5) The figure below shows three situations in which the same horizontal rod is supported bya hinge on a wall at one end and a cord at its other end. Without many written calculation,rank the three scenarios according to the magnitudes of the specified forces, greatest first.

(a) the force on the rod from the cord

(b) the vertical force on the rod from the hinge

(c) the horizontal force on the rod from the hinge

9. (3/5) The figure below shows a simplified model of the force needed to do a push-up. We assumethat a persons body is a bar of mass M with a moment of inertia about the pivot (at the toes)of IT . The center of mass is a distance L from the pivot and that the arm force, ~Farm is purelyvertical and applied at a distance of 4L/3 from the pivot. The height h is given by the person’sarm extension and varies throughout the entire push-up.

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(a) Find the value of the applied force arm force, ~Farm, needed for the object to be in equi-librium. Show that it is independent of the angle θ and thus, independent of the height,h.

(b) What is the value of the normal force ~FN exerted on the person by the ground at the person’stoes?

10. (4/5) Stick on A Corner:

A stick with mass m and length l leans against a frictionlesswall, with a quarter of its length hanging over a corner, asshown in the figure. It makes an angle θ with the horizontal.Assuming that there is sufficient friction at the corner tokeep the stick at rest, what is the total force that the cornerexerts on the stick? Hint: We suggest that you break thetotal corner force into x and y components ~Fc = (Fcx, Fcy).

11. (4/5) Tip or Slip?

A uniform rectangular wood block of mass M , with length band height a, rests on an incline as shown. The incline andthe wood block have a coefficient of static friction, µs. Theincline is moved upwards from an angle of zero through anangle θ. At some critical angle the block will either tip overor slip down the plane. Determine the relationship betweena,b, and µs such that the block will tip over (and not slip)at the critical angle. The box is rectangular, and a 6= b.

12. (5/5) Balancing Disk

Ultimate Challenge: A stick with mass m andlength 2R is pivoted at one end on a vertical wall.It is held horizontal, and a disk with mass m andradius R is placed beneath it, in contact with bothit and the wall, as shown right. The coefficient offriction between the disk and the wall is µw, andthe coefficient of friction between the disk and thestick is µs. If the objects are released, what arethe minimum values of µw and µs for which thesystem doesn’t fall?

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3 Solutions to Selected Problems

1. Answer: Point (d)

The only two forces acting on the bead are the normal force, which points radially inward,and gravity ~Fg = −mgy, which always points downward. Summing these two vectors up ateach point gives the net acceleration of the bead. Point (d) is the only place where the drawnacceleration arrow is possible.

Remark: Since the radial ar = v2

r component always points radially inward, the accelerationvector in any arbitrary circular motion can never have a radially outward component. Thisimmediately rules out choices (a) and (c).

2. Answer:τ5 > τ4 > τ2 > τ1 > τ3

The figure right shows the corresponding position vectorsthat point from the pivot to where each force is applied.The side-length of the square is d. Since forces ~F1, ~F2, and~F5 lie along the x and y axes, the best way to calculate thetorque from each of these forces is to use, τ = Fr⊥. Doingso gives,

τ1 = F (0) = 0 (21)

τ4 = F(d

2

)=Fd

2(22)

τ5 = Fd (23)

where |~F1| = |~F2| = |~F3| = F .

The position vector of forces ~F1 and ~F2 is the same and points in the −x direction. Thus, thebest way to calculate torque is τ = F⊥r, where F⊥ = F sin θ and r = d

2 . While we don’t know

the angle θ between these two forces and the position vector, we can see that ~F1 is almost parallelto ~r (i.e. arrow in orange), and thus, ~F2 has a larger perpendicular component than ~F1. Thismeans that,

τ1 = F1,⊥d

2=

(F sin θ1)d

2< τ2 (24)

τ2 = F2,⊥d

2=

(F sin θ2)d

2< τ4 (25)

where sin θ1 < sin θ2 ≤ 1. Putting this together gives, τ5 > τ4 > τ2 > τ1 > τ3.

3. Disk, Hoop, and Sphere

(a) Answer: αsphere > αdisk > αhoop

According to Newton’s 2nd Law for rotational motion,

τnet = Iα⇒ α =τnetI

(26)

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Since the torque on each of the three rigid objects is the same, τnet = FR, the correspondingangular accelerations are,

αdisk =FR

Idisk=

FRmR2

2

=2FR

mR(27)

αhoop =FR

Ihoop=

FR

mR2=FR

mR(28)

αsphere =FR

Isphere=

FR2mR2

5

=5FR

2mR. (29)

This gives us, αsphere > αdisk > αhoop. We should expect this because objects with masscloser to their center of mass, such as the sphere, are easier to rotate than objects with massfurther from their center of mass, such as the hoop. Also note how the torque, and therebythe angular acceleration, is positive since the force causes counter-clockwise rotations.

(b) Answer: ωsphere > ωdisk > ωhoop

Since the angular acceleration for each of the three objects is constant, we can use thekinematic equations for constant α to find ω after time ∆t. Since each object starts fromrest, ω0 = 0 for all three objects. This gives us,

ωdisk = αdisk∆t =2FR∆t

mR(30)

ωhoop = αhoop∆t =FR∆t

mR(31)

ωsphere = αhoop∆t =5FR∆t

2mR(32)

Therefore, ωsphere > ωdisk > ωhoop.

4. Answer: αb > αa > αc = αd = αe

Each of the five cases consists of two point masses connected by a massless rod and subjectto a constant force ~F . According to Newton’s 2nd Law for Rotational Motion, the angularacceleration is,

α =τnetIsys

. (33)

The moment of inertia for each system is, Isys,j = 2mjr2j , where j = {a, b, c, d, e} and rj is

the distance from the center of the rod to each point mass. The net torque for each system isτnet = 2F⊥,jrj , where Fperp,j is the force perpendicular to the rod. Putting these two togethergives us a general expression for the angular acceleration of the system,

αj =τnetIsys

=F⊥,jmjrj

(34)

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Using equation 34 to calculate the angular acceleration in each case gives,

αa =F⊥,amara

=1 N

(2 kg)(1 m)=

1

2rad/s2 (35)

αb =F⊥,bmbrb

=2 N

(2 kg)(1 m)= 1 rad/s2 (36)

αc =F⊥,cmcrc

=1 N

(4 kg)(1 m)=

1

4rad/s2 (37)

αd =F⊥,dmdrd

=1 N

(2 kg)(2 m)=

1

4rad/s2 (38)

αe =F⊥,emere

=1 N

(2 kg)(2 m)=

1

4rad/s2 (39)

5. Rotating Flywheel

(a) Answer: α = − π1440 rad/s2 = −7× 10−4 rad/s2

In this problem, we are told the initial angular velocity, ω0 = 5π rad/s, the final angu-lar velocity, ωf = 0 rad/s, and the time for the friction of the bearings to bring the wheelto rest, tf = 7200 s. Since the wheel undergoes constant angular acceleration α, we canuse the kinematic equations for angular velocity to solve for α.

ωf = ω0 + αtf ⇒ α =ωf − ω0

tf= −5π rad/s

7200 s= − π

1440rad/s2 (40)

It is important to note that the negative sign on the angular acceleration describes thewheel slowing down in the counter-clockwise direction.

(b) Answer: 9000 revolutions

Now that we know the angular acceleration of the wheel, we can calculate the numberof revolutions ∆θ the wheel rotates through before coming to a stop. There are two equa-tions in which we can use to solve for ∆θ,

ω2f − ω2

i = 2α∆θ (41)

∆θ = ω0tf +1

2αt2f (42)

While either will work, the author of this guide personally prefers equation 41, so we willgo with this one. Solving for ∆θ from equation 41 gives,

∆θ =ω2f − ω2

i

2α=

25π2 rad2/s2

π1440 rad/s2

= 18000π rad (43)

Since the question asked for revolutions as opposed to radians, we need to convert fromradians to revolutions. For every revolution, there is 2π radians. Using this conversionfactor, we can solve for how many revolutions the wheel makes,

∆θ = 18000π rad( 1 rev

2π rad

)= 9000 rev (44)

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(c) Answer: at = − π2880 m/s2 = −3.5× 10−4 m/s2

Although we are given ω = 2.5π rad/s, the tangential acceleration of a point on the flywheelonly depends on α and the radial distance from the center r,

at = αr (45)

Given that the radial distance of the point from the center is r = 12 m, the corresponding

tangential acceleration is,

at = − π

1440rad/s2

(1

2m)

= − π

2880m/s2 = −3.5× 10−4 m/s2 (46)

Note how the “rad” unit disappeared in our final expression for at because it isn’t a trueunit such as meters or seconds. The negative sign indicates the tangential accelerationpoints against the curve and that the point is decelerating along the tangential direction.

(d) Answer: ~a =[25π2

8 m/s2](inward)−

[π

2880 m/s2](along)

The total acceleration of the point on the wheel has both a centripetal and radial com-ponent,4

~a = ac (inward) + at (along) (47)

The centripetal acceleration changes quadratically with ω,

ac = ω2r =(5

2π rad/s

)2(1

2m)

=[25π2

8m/s2

](48)

With this and the tangential acceleration from part (c), we can find the total accelerationof the point at this time,

~a =[25π2

8m/s2

](inward)−

[ π

2880m/s2

](along) (49)

(e) Answer: Wfriction = −(25π)2kg m2/s2 = −6168.5 J

According to the Work-Kinetic energy theorem,

Wfriction = ∆K =1

2Iwheel

(ω2f − ω2

0

)(50)

where Iwheel is the moment of inertia of the wheel. Since the wheel is a solid cylinder withmass M = 100 kg and R = 1 m, its moment of inertia is,

Iwheel =1

2MR2 = 50 kg m2 (51)

The initial and final angular speeds are, ω0 = 5π rad/s and ωf = 0 rad/s, respectively.Substituting these values into equation 50 gives,

Wfriction =1

2Iwheel

(ω2f − ω2

0

)=

1

2(50 kg m2)(−25π2 rad2/s2) = −(25π)2kg m2/s2 (52)

4Expert Note: One can more efficiently describe the inward radial direction by −r = −(

cos θx + sin θy)

, which

points inward for all points along the unit circle, and the tangential (along) direction by θ = − sin θx + cos θy. Forthis class, you will not need to use polar coordinates axes.{r, θ, z}, to describe the net acceleration.

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6. Race Car

(a) Answer: vt = att

Since the car has constant tangential acceleration and starts from rest (i.e. v0 = 0), thetangential velocity will be,

vt = v0 + att = att (53)

(b) Answer: ac(t) = (att)2

r

As the car accelerates along the track with radius r, it has a tangential speed vt = att,which increases in time. The centripetal acceleration at every point in time is,

~ac =v2tr

=(att)

2

r. (54)

From this expression, we see that the centripetal acceleration increases quadratically withtime.

(c) Answer: a = |~a| =√

(att)4

r2+ a2t

The total acceleration of the car has a centripetal component, which points inward, and tan-

gential component, which points along the circle. Therefore, ~a = (att)2

r (inward)+at (along).The magnitude of the total acceleration is just the square root of the sum of these two com-ponents squared,

a = |~a| =√a2c + a2t =

√[(att)2

r

]2+ a2t (55)

(d) Answer: t = 1at

(r2[(µsg)2 − at2

])1/4The free-body diagrams from the car from the top and side-view are shown below. The

only force responsible for accelerating the car is the static friction force between the tiresand the road, Fs ≤ µsFN . The constraints on the motion are that the car accelerates inthe xy plane and does not accelerate in the z-direction (i.e. does not levitate off the racetrack). This means there is no net force acting in the z-direction, such Fnet,z = maz = 0,

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and the normal force on the car is the same magnitude as gravity, FN = mg.

According to Newton’s 2nd Law,

~Fnet = ~FS = m~a (56)

The force of static friction is parallel to the total acceleration, and hence, has a changingmagnitude and direction in time. 5 We know that the magnitude of static friction forceis bounded by, |~Fs| ≤ µsmg. We can use this to find an upper bound on the acceleration|~a| = a =

√a2c + a2t . Taking the magnitude of Newton’s 2nd law give,

|~Fnet| = |~Fs| ⇒ µsmg = ma⇒ µsg =

√[(att)4

r2

]+ a2t (57)

It is important to notice that the maximum possible value of the car’s acceleration isindependent of mass and only depends on g and the coefficient of static friction for yourtires µs ∼ 1 for a dry road (or µs ∼ 0.2 for a wet road). We can now solve for the time tfor the car to slip using equation 57,

a2t +(att)

4

r2= (µsg)2

(att)4

r2= (µsg)2 − a2t

(att)4 = r2

[(µsg)2 − a2t

]t =

1

at

(r2[(µsg)2 − a2t

])1/4(58)

Unitwise this makes sense because we have [t] = s2

m

(m4

s4

)1/4= s. However, let’s substitute

in a few numbers to see if our result is reasonable. If the car has a typical acceleration ofat = 4 m/s2, is on a dry road with µs = 1, and rounding a race car track bend with r = 200m, the time for it to slip would be t = 10.58 s. The car’s corresponding tangential velocitywould be vt = att = 93.1 mph. This seems reasonable.

How does this compare with if the maximum speed the driver decided to not acceler-ate around the curve and instead travel with a constant speed? At constant speed, theacceleration only has a inward centripetal component,

ac =v2maxr

= µsg ⇒ vmax =√µsgr (59)

Given the same specs as before, the driver would be able to do so safely at a slightly higherspeed vmax = 97.4 mph. The moral of the story is that it is slightly safer to coast around acurve than to accelerate. It is even safer to drive on a banked (inclined) road, but we willleave this up to the student to read about!

5This problem contrasts to your Offline HW problem with Javier riding on the Merry-go-Round at constanttangential speed. In this case, the acceleration only had a centripetal component (inward) since Javier was notaccelerating along the curve.

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7. Answer: (d)

The acceleration α is caused by the applied torque by ~τ = I~α, so

α =rF

mr2.

For our toilet paper roll, the mass depends on the radius, since m = ρ × Volume ∝ r2. Thismeans that

α ∝ rF

r4∝ 1

r3

so if r is halved the angular acceleration will go up by a factor of 8.

8. Rod attached to ceiling

(a) Answer: FT,1 = FT,3 > FT,2

The free-body diagram for the three rods is shown below, where FH is the hinge force6,

Fg is the gravitational force, and FT is the tension force. The perpendicular component ofthe tension force to the rode is FT,⊥ = FT cos θ, where θ = 50◦ for cases (1) and (3) andθ = 0◦ for case (2). We aren’t given any specific details of the system, but we will assumethe rod is of length L. In all three cases, gravity Fg acts downward on the rod at its centerof mass, L

2 . Since the rod is in static equilibrium, we have

τnet = 0 (60)

Fnet,x = 0 (61)

Fnet,y = 0 (62)

We can relate the force of gravity to the tension force by calculating the net torque aboutthe hinge,

τnet = τT − τg = 0⇒ τT = τg ⇒ FT,⊥L =FgL

2⇒ FT =

FgL

2 cos θ(63)

With this expression, we find that FT,1 = FT,3 > FT,2 since cos 50◦ < 1.

6The hinge force is simply the normal force that the hinge exerts on the rod because the rod cannot penetrate thewall.

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(b) Answer: FH,y,1 = FH,y,2 = FH,y,3

The vertical component of the hinge force Fh,y is simply the perpendicular componentof the force to the rod. This means that we can relate Fh,y to Fg by finding the net torqueabout the end of the rod,7

τnet = τg − τH = 0⇒ τH = τg ⇒ FH,yL =FgL

2⇒ FH,y =

Fg2

(64)

This expression tells us that the vertical component of the hinge force is the same in allthree cases.

Remark: It is important to note that we could have come to the same conclusion usingequation 62,

Fnet,y = FH,y − Fg + FT,⊥ = 0⇒ FH,y = Fg − FT,⊥ (65)

However, we would need to solve for FT,⊥ = FT cos θ, as we did in part (a).

(c) Answer: FH,x,1 = FH,x,3 > FH,x,2

We can solve for the X-component of the hinge force FHx using equation 61. Beforejumping into the calculation, though, it is important to notice the x-component of thehinge force to be zero in case 2 because both the tension and gravitational forces pointin the ±y-direction. Hence, the rod is not being pushed into the wall in the horizontaldirection and therefore, FH,x,2 = 0. Now for the other two hinge forces,

Fnet,x,1 =∑j

Fj,x = FH,x,1 − FT,x,1 = 0⇒ FH,x,1 = FT sin θ (66)

Fnet,x,3 =∑j

Fj,x = −FH,x,3 + FT,x,2 = 0⇒ FH,x,3 = FT sin θ (67)

This gives us FH,x,1 = FH,x,3 > FH,x,2.

9. Push-up

(a) Answer: F = 34mg

The free-body diagram for the person is shownright. In this problem, there are two unknowns:the normal force acting on the person’s toes, ~FN ,and the arm force, ~Farm = F y. Since the per-son is in static equilibrium, we have the followingconstraints,

~Fnet,x = 0 (68)

~Fnet,y = 0 (69)

~τnet = 0 (70)

7For static equilibrium, the net torque about any pivot must be zero.

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For the first part of the problem, we just need to calculate the arm force ~Farm = F yand show that it is the same for any arm length h (i.e. is independent of θ). Since we don’tknow the normal force, we can place the pivot the person’s toes to ensure that the torquedue to the normal force is zero (i.e. τN = 0 since the moment arm is zero). The constraintof the net torque then gives us,

τnet = τarm − τg = 0⇒ τarm = τg ⇒ Frarm,⊥ = Fgrg,⊥

F4L cos θ

3= mgL cos θ

F =3

4mg (71)

Since our expression does not depend on θ, the required arm force in a plank is independentof arm length, h.

(b) Answer: FN = 14mg

We can calculate the normal force acting on the person’s toes using the constraint of thenet force in the y-direction,

Fnet,y = FN + F −mg = 0⇒ FN = mg − F =mg

4(72)

10. Answer: ~Fc =(

mg3 tan θ ,mg

)

The free-body diagram is shown for the stick. In order forthe stick to be in static equilibrium,

τnet = 0 (73)

Fnet,x = 0 (74)

Fnet,y = 0 (75)

Note that gravity acts on the center of mass, which is adistance l

4 from the corner.

In this problem, there are three unknowns: the x and y components of the corner force,(FCx, FCy), and the magnitude of the normal force of the wall on the stick FN,W . This meansthat we will need to use all three equations to solve for our unknowns. Since we do not knoweither components of the corner forces, we should place our “pivot” at the corner and calculatethe torque this point. The reason is that the torque by the corner force at this point is zerobecause the moment arm is zero. According to equation 73, we have

τnet = τN,W − τg = 0⇒ τN,W = τg ⇒ FN,W rW,⊥ = mgrg,⊥

FN,W3l

4sin θ = mg

l

4cos θ

FN,W =mg

3 tan θ(76)

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We can relate the two horizontal forces FCx and FN,W using equation 74,

Fnet,x = FCx − FN,W = 0⇒ FCx = FN,w =mg

3 tan θ(77)

We can solve for FCy using equation 75,

Fnet,y = FCy − Fg = 0⇒ FCy = Fg = mg (78)

This gives us ~FC =(

mg3 tan θ ,mg

)11. Answer: µs >

ba

The free-body diagram of the box is shown below. When the box is just about to tip, imagine

the left surface of the box is ε away from the incline, such that the normal force is only at thelower right edge of the box (i.e. at the green dot). Consequently, the static friction force acts atthe lower right edge of the box. Newton’s 2nd Law for the box along x′ and y′ direction reads,

Fx′,net = Fg sin θ − Fs = 0 (79)

Fx′,net = FN − Fg cos θ = 0 (80)

where ax′ = 0 and ay′ = 0 because the box is not slipping. In order for the box not to slip,

Fs = Fg sin θ ≤ µsFN = µsFg cos θ → µs > tan θ (81)

The net torque about the green dot is,

τnet =(Fg sin θ

)a2−(Fg cos θ

) b2

(82)

According to the rotational analog of Newton’s 2nd law, in order for the box to tip

τnet = Iα > 0→(Fg sin θ

)a2

>(Fg cos θ

) b2

tan θ >b

a(83)

Putting equations 81 and 83 gives us,

µs >b

a(84)

Since µs < 1, this relation tells us that the box will only tip if b < a (i.e. the box is taller thanit is wide).

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12. Answer µw > 1 and µs > 2

The free-body diagram for the disk and the stick (which has mass m) is shown below. In

this problem, there are 6 unknown variables: the normal force and static friction force from thewall, FN,W and FW , the normal force and static friction force with the stick, FN,S

8 and FS , andthe coefficient of static friction at each surface, µs and µW . The forces laws for friction providetwo constraints,

FS ≤ µSFN,S (85)

FW ≤ µWFN,W (86)

Both the stick and disk are in static equilibrium, and the following conditions hold for eachobject separately,

Fnet,x = 0 (87)

Fnet,y = 0 (88)

τnet = 0 (89)

We can directly solve for the normal force on the disk from the stick, FN,S using Newton’s 2ndLaw for the stick,

Fnet,y,stick = FN,S −mg = 0⇒ FN,S = mg (90)

By Newton’s 3rd Law, the normal force exerted by the disk on the stick is equal and oppositein direction to the normal force exerted on the disk from the stick. Since we now know all buta single vertical forces acting on the disk, We can solve for FW using Newton’s 2nd Law for thedisk in the y-direction,

Fnet,y,disk = FW − Fg − FN,S = 0⇒ FW = Fg + FN,S = 2mg (91)

We can now relate the static friction force from the stick FS and from the wall FW by calculatingthe net torque about the disk’s center of mass. At this point, only the torques due to FS andFW will be nonzero,

τnet = τS − τW = 0⇒ τS = τW ⇒ FSR = FWR⇒ FS = FW = 2mg (92)

8For each normal and static friction force on the disk, there is an equal in magnitude, but opposite in directionforce acting on the wall/stick.

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Since we know both FN,S and FS , we can find the conditions for µs (eqn. 85),

FS ≤ µsFN,S ⇒ 2mg ≤ µsmg ⇒ µS ≥ 2 (93)

We now know all forces in the horizontal direction except for FN,W . Therefore, We can solve forFN,W using Newton’s 2nd Law for the disk in the x-direction,

Fnet,x = FN,W − FS = 0⇒ FN,W = FS = 2mg (94)

Since we know both FN,W and FW , we can find the conditions for µW (eqn. 86),

FW ≤ µWFN,W ⇒ 2mg ≤ µW (2mg)⇒ µW ≥ 1 (95)

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