Experimental Physics Scripts

R U S T E M VA L I U L L I N

E X P E R I M E N TA L P H Y S I C SS C R I P T S

F E L I X B L O C H I N S T I T U T E F O R S O L I D S TAT E P H Y S I C S , A P -P L I E D M A G N E T I C R E S O N A N C E

DRAFT Experimental Physics 1 - Winter Term 2020/21 - ©Rustem Valiullin The PDF is compiled on Friday 8th January, 2021

Copyright © 2021 Rustem Valiullin

published by felix bloch institute for solid state physics, applied magnetic resonance

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Contents

I Mechanics 5

1 Kinematics 71.1 Motion along a line . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Constant velocity . . . . . . . . . . . . . . . . . . . 7

1.1.2 Varying velocity . . . . . . . . . . . . . . . . . . . 8

1.1.3 Constant acceleration . . . . . . . . . . . . . . . . 9

1.1.4 Complex motion . . . . . . . . . . . . . . . . . . . 9

1.1.5 Constant acceleration kinematic equations . . . . 9

1.1.6 Kinematic equations from calculus . . . . . . . . . 12

1.1.7 List of experiments . . . . . . . . . . . . . . . . . . 12

1.2 Scalars and vectors . . . . . . . . . . . . . . . . . . . . . . 12

1.2.1 Scalars and vectors . . . . . . . . . . . . . . . . . . 13

1.2.2 Vector components . . . . . . . . . . . . . . . . . . 13

1.2.3 Scalar and vector products . . . . . . . . . . . . . 13

1.2.4 List of experiments . . . . . . . . . . . . . . . . . . 13

1.3 Motion in 2D and 3D . . . . . . . . . . . . . . . . . . . . . 14

1.3.1 Average and instantaneous velocities . . . . . . . 14

1.3.2 Average and instantaneous accelerations . . . . . 15

1.3.3 Equations of motion in vector form . . . . . . . . 15

1.3.4 Circular motion in component form . . . . . . . . 17

2 Newton laws 19

3 Rotations 213.1 Rotational motion . . . . . . . . . . . . . . . . . . . . . . . 21

3.1.1 Rotational motion . . . . . . . . . . . . . . . . . . 21

3.1.2 Torque . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1.3 Rotational collisions . . . . . . . . . . . . . . . . . 23


Part I

Mechanics


1 Kinematics

This chapter covers basics of kinematics. It introduces the conceptsof motion without relating to physical mechanisms causing it. An-other simplification is that only particles are considered, i.e. motionof point-like objects are considered. This point may be associatedwith the center-of-mass of a bigger object. In this way, any position ofan object may easily be characterized by its coordinates.

Three fundamental concepts of kinematics are displacement, veloc-ity, and acceleration.

1.1 Motion along a line

In this lecture we make one more simplification and consideronly one-dimensional motion along a straight line. Thus, we will notneed the concept of vectors, which will be introduced in one of thesucceeding lectures. Nonetheless, we still need to distinguish twopossible directions one may move along a line, let say to the right andto the left.

By selecting an appropriate coordinate system, we may fix theorigin, initial and current positions, and positive and negative direc-tions.

Figure 1.1: Coordinate system1.1.1 Constant velocity

Let us first consider the simplest constant-velocity case. Let us selectarbitrarily a position x(t) at a time instant t. Given a time interval ∆tto move, the final position will be x(t + ∆t). The displacement ∆xduring ∆t is x(t + ∆t)− x(t). By definition, the average velocity v in v: By the over-bar symbol we typically

denote the time average, if not statedotherwise.

the time interval ∆t is

v =∆x∆t

=x(t + ∆t)− x(t)

∆t(1.1)

For the constant-velocity case, v is irrespective of the choice of ∆tor time t and the bar symbol may be omitted.


8 experimental physics scripts

• E X P E R I M E N T:Demonstration of the independency of v on ∆x

evaluation:

1.1.2 Varying velocity

If velocity is not constant, the average velocity becomes a functionof both t and ∆t. In this case, a useful concept is an instantaneousvelocity, which is obtained by taking the limit of ∆t→ 0:

Figure 1.2: Constant-velocity motion

v(t) = lim∆t→0

∆x∆t

=dxdt≡ x (1.2)

It turns out that in the limit of small ∆t, x approaches a constantvalue, which is not a function of ∆t anymore, but t only. This valuewe call instantaneous velocity. If velocity is changing along a trajec-tory, then, quite generally, v 6= v.

• E X P E R I M E N T:Demonstration of the dependency of v on ∆x

evaluation:

• E X A M P L E 1 . 1 . 1 :Calculate v and v for the displacement graph shown in Fig. 1.3.

solution:Let us denote the time when the velocity changes from v1 to v2 by t1.On the time interval t ≤ t1 v = v. For t > t1 we need to performsimple arithmetic average with the weights t1/t and (t− t1)/t:

v(t > t1) =t1

tv1 +

t− t1

tv2 (1.3)

Figure 1.3: Speeding up

• E X A M P L E 1 . 1 . 2 :Find v and s for an object moving for 1 h to the right with 60

km/h and then for one 1 h to the left with 40 km/h.

solution:Here simple arithmetic mean applies: v = 1

2v1 +

12 v2 = 10 km/h,

s = 12s1 +

12 s2 = 50 km/h.


kinematics 9

• E X A M P L E 1 . 1 . 3 :What would be the difference, if now not identical time intervals,but equidistant separations are considered. The average velocityv and the average speed s may not coincide. Find v and s for anobject moving from A to B with 60 km/h and back with 40 km/h.

solution:First, we find times t1 and t2 needed for paths AB and BA:t1 = L/|v1| = t1/s1 and t2 = L/|v2| = t2/s2. Thus,v = t1

t1+t2v1 +

t2t1+t2

v2, v = 1t1+t2

(L|v1|

v1 +L|v2|

v2

)= 0. The same result

could be obtained easier by noting that v = ∆x/∆t, and the total ∆xis zero. If we apply the same equation to find s, the result is1s = 1

2

(1s1+ 1

s2

). This is called harmonic mean. The result is then

s = 48 km/h.

1.1.3 Constant acceleration

Let us now consider not displacement vs. time, but velocity vs. timecoordinates. In analogue with Eq. 1.3 we may introduce a averageacceleration

a =∆v∆t

=v(t + ∆t)− v(t)

∆t(1.4)

If velocity varies linearly with time, a is irrespective of ∆t and tand is simply equal to a single-valued a = v = x ≡ d2x

dt2 .

Figure 1.4: Constant-accelerationmotion

1.1.4 Complex motion

Rally Paris-Dakar is one of the prominent examples of complex mo-tion with varying velocity and acceleration. Fig. 1.5 shows an exam-ple where the first derivative of the car position yields instantaneousvelocities and the second derivative (first derivative of velocity) in-stantaneous accelerations.

Figure 1.5: An example of complexmotion

• E X P E R I M E N T:Demonstration showing first and second derivatives of x(t) andcomparing the derived x with the directly measured one.

evaluation:

1.1.5 Constant acceleration kinematic equations

The following set of equations can be used to analyse different situa-tions with one parameter being unknown:



- Equation Missing1 v = v0 + at ∆x2 x = x0 + v0t + 1

2 at2 v3 v2 = v2

0 + 2a(x− x0) t4 x = x0 +

12 (v + v0)t a

5 x = x0 + vt− 12 at2 v0

Table 1.1: Basic kinematic equations

• E X E R C I S E 1.1. 1:Derive all equations in Table 1.1

• E X A M P L E 1 . 1 . 4 :Derivation of the equation in line 2. One may think that if onereplaces v in x = x0 + vt by v = v0+ at on gets the answer, but thisresults in x = x0 + v0t + at2. What causes the problem?

solution:The problem is that v is the instantaneous velocity, but in x = x0 + vtit has the meaning of the average velocity, i.e. we should writex = x0 + vt. For the constant-acceleration case we considering,v = v0 +

12 (v(t)− v0) =

12 (v + v0). Thus, x = x0 + v0/2 + v(t)/2.

With v(t) = v0 + at one gets the equation in line 2 of Table 1.1.

• E X A M P L E 1 . 1 . 5 :Derivation of the equation in line 3. A U-Bahn needs most quicklyapproach station B which is 2 km apart from station A. The max-imal acceleration tolerated by the passengers is 2 m/s2. Whatwould be the maximal speed attained by the U-Bahn?

solution:Obviously, the train need to move with the maximal accelerationhalf-time and maximal deceleration second half. The velocity alongthe acceleration path will be v = v0 + at. The path is

x = x0 + v0

(v−v0

a

)+ 1

2 a(

v−v0a

)2. By simplifying this equation one

gets the one on line 3. The result at half distance is v = 63.2 m/s,which is about 227.6 km/h.

• E X P E R I M E N T:A ball falls down from a height h with zero initial velocity. Aftereach collisions the speed changes by a factor of a < 1. Find thetotal time T that ball will stop bouncing. The coefficient a can bedetermined by measuring the speeds before and after collision.It may also be proven that a is relatively constant for differentspeeds.


kinematics 11

evaluation:Let us consider the problem in small steps just for the instructivepurpose.

• Time t0 to reach the ground is found from 0 = h− gt20/2. Thus

t0 =√

2h/g. Compile a plot h− t2 and determine g

• Just before meeting the ground velocity will bev1 = −gt0 = −

√2hg.

• Bouncing upwards will occur with velocity of av1.

• One needs now to establish the coefficient a. To find a one caneither experimentally measure to which height h1 will the balljump if it is released from a height h OR this can be done bydirectly measuring the velocities before and after collision.

• Time t1u to reach the upper part of the trajectory is found from0 = av1 − gt1u, i.e. t1u = av1/g.

• The height h1 the ball approaches is

h1 = 0 + av1t1u − 12 gt2

1u = 12

a2v21

g .

• The coefficient is found to be a =√

h1/h.

• The question is what is time t1d to reach the ground back. Would itbe equal to t1u or different? The fall time t1d is found from

0 = h1 − gt21d/2. One finally gets t1d =

√2h1

g = av1g . This shows the

symmetry between the rising and falling times. Experimentproving the equality of fall and rise times.

• Thus, the total time t1 during this bouncing period is t1 = 2av1g .

• It is easy to see that t2 = 2a2v1g and tk =

2akv1g = 2ak

√2hg .

• The total time T is found as T = t0 + ∑k=∞k=1 tk. The sum of

geometric progression ∑k=∞k=1 ak = ∑k=∞

k=0 ak − 1 = a1−a . Thus,

T =√

2h/g + 2a1−a√

2h/g =√

2h/g 1+a1−a .

The equivalence of the rising andfalling times is a consequence of thefact that the laws of motion in classicalmechanics exhibit time reversibility, i.e.the equations are invariant under achange in the sign of time.

• E X A M P L E 1 . 1 . 6 :The time reversibility may be seen by considering the same bounc-ing problem without energy loss.



solution:Indeed, the equation describing the upward trajectory (line 2 ofTable 1.1) is h = 0 + v0t− gt2/2. The downward trajectory (line 5 ofTable 1.1) is described by 0 = h− v f t + gt2/2. If we change t→ −t inthe last equation and rearrange it, the we arrive toh = 0− v f t− gt2/2 which is exactly the same as one for the upwardtrajectory.

1.1.6 Kinematic equations from calculus

Figure 1.6: Taking limit ∆x → dx

The total displacement or a complex motion can be found by divid-ing the path on parts of equidistant intervals in time, ∆ti, and assum-ing that within this interval of time the object moved with a constantaverage velocity vi. Then, the total displacement X = ∑N

i vi∆t. It iseasy see from Fig. 1.6 that shorter is the time interval ∆t (correspond-ingly larger N), closer it the result to the real displacement. Takinglimit ∆t → dt one finds dx = vdt, where v is now instantaneousvelocity. Summation is now replaced by integration and one gets

∫ X

x0

dx =∫ t

0vdt =

∫ t

0(v0 + at)dt (1.5)

which results in the already famous kinematic equation X − x0 =

v0t + at2/2. Similarly, one may find velocity as

∫ v f

v0

dx =∫ t

0adt (1.6)

1.1.7 List of experiments

1. Constant velocity case: v is irrespective ∆x

2. Varying velocity case: v depends on ∆x

3. Complex motion: x, v, a

4. Changing speed upon collision, determining the ratio

5. Proving the h vs. t2 law

6. Proving that rise and fall times are identical

7. Find time for a ball released from h to stop bouncing

1.2 Scalars and vectors

In this lecture we recall the basics of scalars and vectors.


kinematics 13

1.2.1 Scalars and vectors

A scalar quantity is specified by a single value with an appropriateunit and does not reflect any direction (mass, volume, temperature).A vector quantity has in contrast both magnitude and direction (ve-locity, flux, force).

Vectors might be useful in many instances, for example in de-scribing trajectories as shown in Fig. 1.7. They may be subdivided tomany small sections each captured by a vector. The total displace-ment then is the vector sum of individual ones.

Figure 1.7: Trajectory

Figure 1.8: Adding two vectors

Vectors can be added as geometrically shown in Fig. 1.8. In thisway a negative vector turns out to be one giving zero upon addingwith the original vector. Summation of vectors complies with thefollowing laws:

Commutative law ~a +~b =~b +~a

Associative law (~a +~b) +~c =~a + (~b +~c)

Substraction law ~a−~b =~a + (−~b)Cartesian coordinate system ia right-handed coordinate system. The indexfinger , the middle finger , and thethumb now give the alignments of thex-, y-, and z-axes, respectively.

1.2.2 Vector components

1.2.3 Scalar and vector products

• E X P E R I M E N T:A piece of mass is first pulled along a table horizontally. After-wards, it is pulled under certain angle. The work done is found asa scalar product of the force and displacement.

evaluation:The constancy of the force may be secured by hanging another masson one end of the chord.

• E X P E R I M E N T:A wire carrying electric current is placed in a magnetic field. Thewire is forced perpendicularly to both magnetic field and currentdirections.evaluation:Instead of wire one may use an electron beam.

1.2.4 List of experiments

1. Demonstration of scalar product (work)

2. Demonstration of vector product (Lorenz force)



1.3 Motion in 2D and 3D

In this lecture we extend motion to higher dimensions to con-sider curvilinear pathes. We us for that the concept of vector intro-duced in the preceding section.

1.3.1 Average and instantaneous velocities

Figure 1.9: Displacement

Fig. 1.9 shows the initial and final positions of a material point, thepath connecting these two points, and also two radius vectors in anarbitrarily selected coordinate system describing these points. Theaverage velocity ~vavg ≡ v (note that it is as well a vector quantity) isdefined as

~vavg =∆~r∆t

(1.7)

As any vector, it may represented by its vector components. The average velocity is not a functionof path connecting initial and finalpositions.

The instantaneous velocity ~v is defined as the limit of the averagevelocity at ∆t→ 0, i.e.

~v =d~rdt

(1.8)

Instantaneous velocity is always tangential to the path. This state-ment can most simply be proven by considering motion along acircular path. The radius vector of an object~r = Rr. Hence,

~v =d~rdt

= Rdrdt

+ rdRdt

(1.9)

The last term on right hand side of Eq. 1.9 is zero because R isconstant. Let us now check the direction of dr/dt. For this, we findthe dot product r · dr

dt :

d(r · r)dt

= 2r · drdt

= 0 (1.10)

Indeed, because r · r = const the derivative is zero. Thus, Eq. 1.10

proves that dr/dt is perpendicular to r.

• E X P E R I M E N T:Sharpening of a knife using rotating disk. All sparks are movingtangential to the disk.

evaluation:


kinematics 15

1.3.2 Average and instantaneous accelerations

By definition the average acceleration is

~aavg =∆~v∆t

(1.11)

and its direction coincides with that of the average velocity asshown in Fig. 1.10.

Figure 1.10: Average acceleration

Figure 1.11: Instantaneous acceleration

Let as consider the instantaneous acceleration, which is defined as

~a =d~vdt

=d2~rdt2 (1.12)

Quite generally,~a has both normal and tangential components tothe path. Indeed,

d~vdt

=d(vs)

dt= v

dsdt

+dvdt

s (1.13)

In the spirit of Eq. 1.10 it is easy to see that ds/dt is normal to sand may be denoted as n(⊥ s). Hence, Eq. 1.13 may be rewritten as

~a = at s + ann, (1.14)

where at and an are the tangential and normal components of theacceleration vector, respectively.

• E X A M P L E 1 . 3 . 1 :By considering circular motion establish the meaning of an.

solution:Let us for simplicity consider v being constant. The first term inEq. 1.13, vds/dt may be found in the following way.~v = d~r/dt = Rdr/dt, ω = dϕ/dt = v/R. Because ~v = vs, thecomparison yields dr/dt = ωs. Let us consider d(s · r)/dt. It is zeroon the one hand. On the other hand it yields rds/dt = −sdr/dt.Hence, ds/dt = −ωr. Finally, an = −vω.

1.3.3 Equations of motion in vector form

We may consider as example two most important equations,

~v = ~v0 +~at (1.15)

~r =~r0 +~vt +12~at2 (1.16)

These two equations can be rewritten in the component form.



• E X P E R I M E N T:To illustrate that the equations of motion for each component aredisentangled one may show the "schnips" experiment, where oneball is given a horizontal impulse, while the second one startssimultaneously free fall

evaluation:Both balls reach the ground at the same time instance.

The best example for this independency is a projectile. For hori-zontal motion one finds

x = x0 + v0xt = x0 + v0 cos(θ0)t (1.17)

For the vertical projection there are two equations

y = y0 + v0 sin(θ0)t− gt2/2vy = v0 sin(θ0)t− gt (1.18)

Figure 1.12: Projectile

We are interested to find the equation of path for the projectile, i.e.a function y = y(x). Expressing t from Eq. 1.19, the required equationresults as

y− y0 = tan(θ0)(x− x0)−g2

(g(x− x0)

v0 cos(θ0)

)2

(1.19)

With the equation of path one may find some quantities of inter-ests as shown by the following examples.

• E X A M P L E 1 . 3 . 2 :Find the height h at the highest point of the projectile.

solution:To find h one needs to differentiate y with respect to x and equationto zero. In this way one finds x = v2

0 sin(θ0) cos(θ0)/g as thex-coordinate at the extremum. Substituting this to find y results inh = (v0 sin(θ0))

2/2g

• E X A M P L E 1 . 3 . 3 :Find the angle θ0 such that the longest distance is provided.

solution:Because of the symmetry of the projectile and using the result of thepreceding example the distance is x = 2v2

0 sin(θ0) cos(θ0)/g. Bydifferentiating this equation with respect to θ0 one findscos2(θ0)− sin2(θ0) = 0. Hence at θ0 = 45◦ the distance will belongest. The later is also graphically seen from Fig. 1.13.

Figure 1.13: Projectile


kinematics 17

• E X P E R I M E N T:Shooting a monkey.

evaluation:Monkey is hit

• E X P E R I M E N T:Water stream.

evaluation:Proving the longest distance at θ0 = 45◦.

1.3.4 Circular motion in component form

The radius vector of a particle moving along a circle with a constantangular velocity ω = dφ/dt in the polar coordinates system is

~r = R cos(ωt)i + R sin(ωt) j (1.20)

Taking time derivative the particle velocity results as

~v = −Rω sin(ωt)i + Rω cos(ωt) j = ω(−R sin(ωt)i + R cos(ωt) j

)(1.21)

Figure 1.14: Figure demonstrating thatthe vectors given by Eqs. 1.20 and 1.21

are orthogonal.

By taking the time derivative once again, acceleration is obtainedas:

~a = −ω2 (R cos(ωt)i + R sin(ωt) j)

(1.22)

It is readily seen that the angular acceleration is antiparallel withthe radius vector. By finding its magnitude and with the directionestablished, the angular velocity for circular motion is

~a = −ω2Rr = −v2

Rr (1.23)


2 Newton laws


3 Rotations

3.1 Rotational motion

Rotational motion, such as of the Earth orbiting the Sun, canstill be descrined using the Cartesian coordinate system. This, how-ever, appears to be very inconvenient. It turns out that all laws wehave discussed so far can be reformulated using angular quantities.Moreover, as it will become evident, there will be close similaritiesbetween them.

3.1.1 Rotational motion

Consider a point like object moving on a circle with the radius r withan instantaneous velocity v. An infinitesimally short displacement dsalong the circle can, on the one hand, be found as ds = vdt and, onthe other hand, as ds = rdθ.

Figure 3.1: Angular quantities.

3.1.2 Torque

In order to bring a body out of translational equilibrium, i.e. a statewith either zero or constant velocity, an application of a force isneeded. What about rotations? To induce rotations, it is not enoughto apply a force. For example, if to push a ball lying on a frictionlesstable horizontally along the line passing through its center, the ballwill start sliding, but not rotating. If, however, the line is displacedout of the central point the ball rotation is induced in additions totranslational displacement. The same is with a wrench fixed at apoint about which it may rotate as shown in Figure . If the direc-tion of the force applied passes through the center of the bolt (radialforce F = Fr), the bolt cannot be rotated. For this, a tangential forcecomponent is needed. Another important point is that if the force isapplied close to the rotation point, a relatively strong force is neededto induce rotation. The distance between the point where force isapplied and the rotation axis (l′ in Figure 3.2) is called level arm. If toincrease the level arm, a weaker force is needed to obtain the same



result. Thus, it is a combination of the tangential component of theforce and of level arm, which is responsible for causing rotations. Therespective parameter is called torque and is defined as

~τ =~r× ~F (3.1)

where~r is the radius vector from the rotation axis to the pointwhere the force is applied.

Figure 3.2: A bolt/wrench system.

The tangential component of the force applied to an object causesit translational motion, namely acceleration: Ft = ma. Because theobject is forced to move along a circular path, a = rα, where α is theangular acceleration. Hence, Ft = mrα. If to multiply both sides by r,one obtains

rFt = αmr2 (3.2)

The LHS of this equation is, by definition, torque. The RHS may beseen as the product of the angular acceleration and moment of inertiaI. Thus, we arrive to the Newton’s second law modified for rotationalmotion, This is quite in spirit what we have

discussed before regarding replacementof the translational and rotational quan-tities: force-torque; mass-moment ofinertia;acceleration-angular acceleration.

τ = Iα. (3.3)

Figure 3.3: A disc unwrapping a thread.

• E X A M P L E 3 . 1 . 1 :Find the acceleration of the disc shown in Figure 3.3 and tension inthe thread.

solution:The gravity force is applied at the center of the disc, thecenter-of-mass point. It is displaced from the point where the disctouches the thread, the rotation point. Hence there will be a non-zerotorque τ = Rmg, which gives rise to the angular accelerationaccording to τ = Iα = Ia/R. The moment of inertia of the disc aboutthe rotation axis is found using the parallel axis theorem asI = ICM + mR2 = 1

2 mR2 + mR2 = 32 mR2. Now, the net force T −mg

applied to the disc causes its translational acceleration,T −mg = −ma. The minus sign on the RHS of this equation is usedto indicate the direction of this acceleration. The set of two equationscan be solved to yield a = 2

3 g and T = 13 mg.

Figure 3.4: A mass fixed attached to arotating disc.

• E X A M P L E 3 . 1 . 2 :Find acceleration of the mass M in Figure 3.4 if the mass of thedisc is m.


rotations 23

solution:The torque causing the disc rotation is τ = RT, where T is thetension. The governing equation is τ = Iα = 1

2 mR2α = 12 mRa. With

T −Mg = −Ma one has 12 ma−Mg = −Ma. Consequently,

a = gM/(M + 12 m). If m = M, we recover the solution from the

previous solution as one may intuitively expect. An alternative andshorter solution is an energetic one:Mgh = 1

2 Iω1 + 12 Mv2 = 1

4 mv2 + 12 Mv2. By differentiating on both

sides one readily finds the same solution.

• E X P E R I M E N T:Maxwell’s disc, acceleration and tension.

evaluation:

3.1.3 Rotational collisions

Let as consider an example of two billiard balls, with the one beingat rest and the second one rolling towards the first one withe velocityv0and colliding with it elastically. In the present context, an elasticcollision means that translational components obey typical features ofthe elastic collisions, and the rotational components of the collidingobjects remain not affected by the collision. Let as denote the rollingball with B1, and the ball at rest with B2.

Figure 3.5: Ball B1 just after a collision.

The state of B1 just after the collision is depicted in Figure 3.5.The frictional force Ff r causes (i) acceleration of the translationalvelocity of the ball and (ii) deceleration of the angular velocity. Asfar as along these processes a dynamic state in which ω = v/R isreached, the sliding mode of the ball cancels and the ball continuesto roll with a constant speed v1 f . The two equations governing theseprocesses are:

Ff r = mdvdt

(3.4)

RFf r = −Idω

dt(3.5)

By comparing these two equations one finds that

mRdv + Idω = 0 (3.6)

Integrating it from initial (just after the collision, index ’i’) to final(when the rolling mode sets in, index ’f’) conditions results in theidentity

mRv1 f + Iω1 f = mRv1i + Iω1i (3.7)



Quite generally, it turns out that, for sliding/rolling motion, aquantity mRv + Iω remains conserved:

mRv + Iω = const (3.8)

Later we will see that this is a consequence of a fundamental lawof conservation of the angular momentum. Let us use Eq. 3.8 toobtain the final velocity of B1. With

Iv0

R= mRv1 f + I

v1 f

R(3.9)

the final velocity is found as

v1 f = v0I

mR2 + I=

27

v0 (3.10)

where the last identity on the RHS of Eq. 3.10 was obtained byusing I = 2

5 mR2 for a solid sphere.

Figure 3.6: Ball B2 just after a collision.

The same derivations can be performed for the ball B2 with theinitial condition shown in Figure 3.6. One may start, however, di-rectly with Eq. 3.8, which is generally applicable. It yields for B2

mRv0 = mRv2 f + Iv2 f

R(3.11)

Consequently, the final velocity is found to be The fact that mv0 = mv1 f + mv2 fproves that the law of conservation ofthe linear momentum is fulfilled.

v2 f = v0mR2

mR2 + I=

57

v0 (3.12)


List of Figures

1.1 Coordinate system 7

1.2 Constant-velocity motion 8

1.3 Speeding up 8

1.4 Constant-acceleration motion 9

1.5 An example of complex motion 9

1.6 Taking limit ∆x → dx 12

1.7 Trajectory 13

1.8 Adding two vectors 13

1.9 Displacement 14

1.10 Average acceleration 15

1.11 Instantaneous acceleration 15

1.12 Projectile 16

1.13 Projectile 16

1.14 Figure demonstrating that the vectors given by Eqs. 1.20 and 1.21

are orthogonal. 17

3.1 Angular quantities. 21

3.2 A bolt/wrench system. 22

3.3 A disc unwrapping a thread. 22

3.4 A mass fixed attached to a rotating disc. 22

3.5 Ball B1 just after a collision. 23

3.6 Ball B2 just after a collision. 24


List of Tables

1.1 Basic kinematic equations 10


Experimental Physics Scripts

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Transcript of Experimental Physics Scripts