Physics 1116 : Prelim #17:30 - 9:00 PM, October 5, 2010

Name :

• Don’t forget to write down your name

• Breathe deeply and relax. Abandon all hope ye who enter here.

• Calculators are permitted, but laptops, communication devices are not.

• There are 6 problems (+1 bonus) on the exam and 13 pages, not including

the equation sheets. Please check that you have all of them. Feel free to

remove the equation sheets from the back of the exam.

• Please show your reasoning clearly on all problems. Partial credit will not be

awarded if we do not understand what you are trying to do. Feel free to use

the back of pages for your working (but let us know that you are doing so).

• Don’t panic if you have difficulty with a problem, and don’t get hung up on

any one particular problem. This test is designed to be sufficiently challenging

that you do not have to be perfect in order to get a good score.

Problem Score

1-3 30 pts

4 20 pts

5 25 pts

6 25 pts

Bonus 3 pts

TOTAL / 100

DO NOT GO TO THE NEXT PAGE UNTIL INSTRUCTED TO DO SO

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Short Questions

Question 1 10 pts

Find the center of mass, RCM = (XCM , YCM), of this plate. The light grey part of the plate

has a density ρ0 (kg/m2), while the dark grey part has a density 3ρ0.

x (m)

y (

m)

1 2 3 4 5 6

1

2

3

4

5

6

(XCM, YCM) =

2

Question 2 10 pts

Consider a ring of radius R and mass M , with the z-axis passing through the axis of the ring

as shown below. The ring sits in the z = 0 plane. Calculate the gravitational field due to the ring

at the point (x = 0, y = 0, z = L).

(0,0,L)

x

y

z

�G(x = 0, y = 0, z = L) =

3

Question 3 10 pts

Consider a box of mass m on a 45◦

incline with a coefficient of friction µ. In the first case,

you are trying to prevent the box from sliding down the incline, and will need to exert at least a

minimum force FA to do so. In the second case, you are trying to push the box up the incline,

and need to exert a minimum force FB to do so. What is the difference in the magnitude of the

forces, |FB|− |FA|?

45o

m

|FB|− |FA| =

END OF SHORT QUESTIONS

4

Question 4 20 pts

Tony Stark’s autonomous Ice Rover has a mass of 100 kg and is sliding without friction along

a Siberian ice sheet with a velocity of 5i m/s the +x axis. At t = 0, the Rover is at (x = 0, y = 0).

At some unknown time T before t = 5 s, the Mandarin activates a hidden bomb onboard the

Rover, and it explodes into 4 quarters of equal mass.

At t = 5 s, the first piece is observed by a SHIELD spy satellite at (x1 = 19, y1 = 6) m, the

second piece at (x2 = 31, y2 = −8) m, and the third at (x3 = 17, y3 = −2) m. However, the

fourth piece cannot be seen. The Mandarin is flying overhead with a velocity vector of 5i m/s,

and manages to measure the velocity of the first piece relative to him as (vx1� = −3, vy1� = +3)

m/s.

(a) Calculate the position of the center of mass of the four pieces at t = 5 s.

(xCM(t = 5), yCM(t = 5)) =(b) Determine the position vector of the unseen fourth piece at t = 5 s.

(x4(t = 5), y4(t = 5)) =

5

(c) At what time T did the Mandarin cause the Ice Rover to explode?

T =

(d) Determine the velocity vector of the fourth piece (relative to the ground).

(vx,4, vy,4) =

6

Question 5 25 pts

A block of mass m sits on a horizontal frictionless table and is attached to a post by a massless

rigid rod of length L. The block rotates around the post at a constant speed v.

(a) Draw a free body diagram of the mass, identifying all forces.

(b) Determine the tension in the rod in terms of m, v, and L.

T =

7

(c) Now, the single mass is replaced two masses (each has equal mass m to the mass in part

a) connected by two massless rods to the pivot. Each rod has a lengthL2 . Draw the free body

diagrams for m1 and m2.

(d) Find the tensions in the inner rod (T1) and the outer rod (T2) in terms of m, v, and L.

T1 =

T2 =

8

(e) Returning to the instance of a single mass in parts (a) and (b), let’s now consider what

happens if the the mass sliding on the table now has a coefficient of friction µ. At t = 0, the speed

of the block is v0. Calculate the tension in the rod as a function of time, T (t).

T (t) =

9

Question 6 25 pts

Consider two identical masses which are free to slide without friction on two rods oriented at

90◦, such that m1 slides horizontally and m2 slides vertically. The rods are fixed in space and

unable to move, and gravity is directed along the −y direction. Let the corner connecting the rods

be the origin, (x = 0, y = 0). The two masses are attached to three springs. Two of the springs

are identical (spring constant k, and relaxed length L) and have one end attached to the origin

and the other end attached to either mass. The third spring is attached between m1 and m2 and

has a spring constant ofk2 and a relaxed length of zero.

(a) Draw and label the free body diagrams for the two masses, m1, m2, properly labeling all

forces.

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(b) Write down the equations of motion for m1 and m2 along the x and y directions, respectively.

Note : you will need to include the force of spring # 3 in the equations of motion, Fsp3, but

for this part of the problem you do not need to solve for Fsp3

mx1 =

my2 =(d) Find the positions of m1 and m2 when the system is in equilibrium (i.e. at rest).

Hint : The “zero relaxed length” for spring 3 was picked to make the algebra slightly simpler.

You may want to think about an easy way to express the different components of the force from

spring 3.

(x1,eq, y1,eq) =

(x2,eq, y2,eq) =

11

(c) The system is initially in equilibrium, but m1 is given a small push and both masses start

oscillating about their equilibrium positions. Find the frequency of oscillations of m1 and m2.

ωm1 :

ωm2 :

12

Bonus Problem 3 pts

The “Casimir Effect” is a force that is exerted between separate objects due to quantum

mechanical fluctuations of the electromagnetic field. This effect was first predicted by physicists

Hendrik Casimir and Dirk Polder in 1948, and finally verified experimentally in 1997.

The force per unit area due to the Casimir effect (F/A) between the plates depends only on

3 physical quantities : Planck’s constant, h, the speed of light, c, and the distance between the

plates, d. The units of h are kg · m2· s

−1.

Determine how F/A depends on h, c, and d.

F/A =

13

Equation Sheet for Prelim # 1

VECTORS

�r(t) = x i+ y j+ z k �a · �b = axbx + ayby + azbz = |�a||�b| cos θ�a× �b = (aybz − azby )i+ (azbx − axbz )j+ (axby − aybx)k = |�a||�b| sin θ c

1-D MOTION

Constant acceleration : ∆x = v0∆t+ 1

2a(∆t)2 v2f = v2

0+ 2a(∆x)

v(t) = v0 + a(∆t)

POLAR COORDINATES

�r(t) = r(t) r drdt = θθ dθ

dt = −θr

�v(t) = rr+ rθθ �a(t) = (r − rθ2)r+ (rθ + 2rθ)θ

For Uniform Circular Motion : |�v(t)| = rθ = rω |�a(t)| = v2/r

FORCES

Newton’s Second Law : �F = m�a

Frictional Forces : |�Ffric| ≤ µFN

Air Drag : �Fdrag = −cv2 v Viscous Drag : |�Fdrag| = −bv v

Spring Forces : �Fspring = −k(�r− �r0) ω0 =�

km T = 2π

ω0

Gravitation : �Fab = −Gmambr2ba

rba �ga = −Gmar2 ra G = 6.67× 10−11 m3 kg−1 s−2

Gravity close to the earth’s surface : �Fgrav = −mg z g ≈ 10 m/s2

MOMENTUM�F = d�p

dt�p = m�v

Center of Mass : �RCM =1

MTOT

�

i

mi�ri �RCM =1

MTOT

�

vol

ρ(�r) �r dV

Momentum & Force : �Psys =�

i

�pi =�

i

mi�vid�Psys

dt = �Fext,sys

MATH & STUFF

2π = 360◦ sin2 θ + cos2 θ = 1

cos(0◦) = 1 cos(30◦) =√3/2 ≈ 0.866 cos(45◦) = 1/

√2 ≈ 0.707 cos(60◦) = 1/2

sin(0◦) = 0 sin(30◦) = 1/2 sin(45◦) = 1/√2 ≈ 0.707 sin(60◦) =

√3/2 ≈ 0.866

Quadratic equation : ax2 + bx+ c = 0 x = (−b±√b2 − 4ac)/2a

Taylor Series : f(x−x0) = f(x0)+dfdx |x0(x−x0)+

1

2!

d2fdx2 |x0(x−x0)2+

1

3!

d3fdx3 |x0(x−x0)3+ ....

SOLUTIONS TO SOME DIFFERENTIAL EQUATIONS

mx+ kx+ C = 0 k > 0 x(t) = A sin((�k/m)t+ φ) + B

mx+ Cx+D = 0 x(t) = Ae−Ct/m +B

mx− b2x+ C = 0 x(t) = Aebt/√m +Be−bt/

√m +D