# I just can't do the homework.

"Do not merely listen to the word,

and so deceive yourselves.

Do what it says."

-- James 1:22

## You simply cannot survive this course -- or any of the programs or careers associated with it -- unless you work the homework problems enough to understand them. Browsing at already completed homework solutions is not enough. You have to get your own hands dirty! You have to expend your own sweat! Homework problems are not an option! Homework problems are not "just for fun"! Homework problems are vital!

As you know, this is a "conventional Physics exam". It is not multiple choice. Show all your work. Hand in everything. Be sure to put your name on this exam. Hour Exam Four and the Final will be similar.

Possibly useful information:

 s = si + v i t + (1/2) a t2 ac = v 2 / r p = m v v = v i + a t W = F s cos PTot = p1 + p2 v 2 = vi 2 + 2 a (s - si) W = F s PTot,i = PTot,f sin = opp/hyp A B = A B cos  cos = adj/hyp A B = Ax B x + Ay B y + Az Bz tan = opp/adj dW = F x dx M = mi F = m a K = (1/2) m v2 F12 = - F21 Wnet = K fsmax = µs Fn P = W/ t fk = µk F n U = - W g = 9.8 m/s 2 Å 10 m/s 2 E = K + U Ei = Ef Fx = - [ dU / dx ] dU = - F x dx Fg = G Mm / r2 1. Momentum

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in a perfectly elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.00 m/s. After the collision, the orange disk moves along a direction that makes an angle of 37.0° with its initial direction of motion, and the velocity of the yellow disk is perpendicular to that of the orange disk (after the collision -- and this will always be the case for elastic collisions between equal masses). Determine the final speed of each disk.  This is problem 9.45. We can substitute 5.0 m/s = vo at the end.

Pfx = m vorg,x + m vyel,x

Pfx = m vorg cos 37o + m vyel cos 53o

Pfx = m vorg (0.8) + m vyel (0.6)

Pfx = m [ vorg (0.8) + vyel (0.6) ]

Pfx = m [ vorg (0.8) + vyel (0.6) ] = m vo = Pix

vorg (0.8) + vyel (0.6) = vo

Now, for the y-component,

Pfy = m vorg,y + m vyel,y

Pfy = m vorg sin 37o - m vyel sin 53o

Pfy = m vorg (0.6) - m vyel (0.8)

Pfy = m [ vorg (0.6) - vyel (0.6) ]

Pfy = m [ vorg (0.6) - vyel (0.8) ] = 0 = Piy

[ vorg (0.6) - vyel (0.8) ] = 0

Now we have two equations and two unknowns, vorg and vyel.

0.6 vorg = 0.8 vyel

vorg = 1.333 vyel

vorg (0.8) + vyel (0.6) = vo

(0.8)(1.33 vyel ) + 0.6 vyel = v o

1.667 vyel = v o

vyel = 0.6 v o

vyel = 0.6 (5.0 m/s) = 3.0 m/s

vorg = (1.333) (3.0 m/s)

vyel = 4.0 m/s

2. Rotation

A 15-kg mass and a 10-kg mass are suspended by a pulley that has a radius of 10 cm and a mass of 3.0 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 3.0 m apart. Treat the pulley as a uniform disk.

Idisk = (1/2) M R2

Detemine the speeds of the two masses as they pass each other. This is problem 10.32

Now, draw free-body diagrams for all three bodies, We will need the moment of inertia of the pulley,

I = (1/2) M r2 = (0.5) (3.0 kg) (0.10 m)2 = 0.015 kg m2

For mass m2, I have taken up to be positive; that means

Fnet,2 = T2 - m2 g = m2 a

For the pulley, net = 2 - 1 = I  1 = r T1 = (0.10 m) T1 2 = r T2 = (0.10 m) T2 net = 1 - 2 = (0.10 m) T1 - (0.10 m) T2 = I (0.10 m) [ T1 - T2 ] = I = (0.015 kg m2) The linear acceleration a and the angular acceleration are related by

a = r or = a / r

so we may write

(0.10 m) [ T1 - T2 ] = (0.015 kg m2) a / 0.10 m

(0.10 m) [ T1 - T2 ] = (0.15 kg m) a

T1 - T2 = (1.5 kg) a

And, for mass m1, where I have taken down as positive,

FNet,1 = m1 g - T1 = m1 a

(15.0 kg) (9.8 m/s2) - T1 = (15.0 kg) a

147 N - T1 = (15.0 kg) a

Now we have three equations -- from the three free-body diagrams -- and three unknowns.

T1 = 147 N - (15.0 kg) a

T1 - T2 = (1.5 kg) a

[ 147 N - (15.0 kg) a] - T2 = (1.5 kg) a

T2 = [ 147 N - (15.0 kg) a] - (1.5 kg) a

T2 = 147 N - (16.5 kg) a

T2 - m2 g = m2 a

[147 N - (16.5 kg) a] - (10.0 kg) (9.8 m/s2) = (10.0 kg) a

147 N - 98 N = (26.5 kg) a

a = 49 N / 26.5 kg

a = 1.85 m/s2

With this acceleration of 1.85 m/s2, how fast will the blocks be moving after they have moved a distance of 1.5 m?

v2 = vo2 + 2 a (y - yo)

v2 = 02 + 2 (1.85 m/s2) (1.5 m)

v2 = 5.55 m2 / s2

v = 2.36 m/s

Alternate method, using Energy Conservation

Ei = Ef

Ei = U1i = (15 kg) (9.8 m/s2) (3.0 m) = 441 J

This takes the reference position to be the "ground", where the 10 kg mass is sitting. Since everything starts at rest, there is no kinetic energy at all.

Ef = K1 + U1 + K2 + U2 + K3

K3 = Krot = (1/2) I 2

I = = (1/2) M R2 = (0.5) (3.0 kg) (0.10 m)2 = 0.015 kg m2

K3 = Krot = (1/2) I 2 = (0.5) (0.015 kg m2) 2 = (0.0075 kg m2) 2

K2 = (1/2) m2 v2 = (0.5) (10 kg) v2 = (5 kg) v2

K1 = (1/2) m1 v2 = (0.5) (15 kg) v2 = (7.5 kg) v2

These v's are, indeed, the same. And these v's -- this velocity -- are/is related to the angular velocity by

v = R = (0.10 m)  = v / R = v / 0.10 m

K3 = Krot = (0.0075 kg m2) ( v / 0.10 m )2 = 0.75 kg v2

U2 = m2 g h2 = (10 kg) (9.8 m/s2) (1.5 m) = 147 J

U1 = m1 g h1 = (15 kg) (9.8 m/s2) (1.5 m) = 220.5 J

Ef = K1 + U1 + K2 + U2 + K3

Ef = 7.5 kg v2 + 220.5 J + 5 kg v2 + 147 J + 0.75 kg v2

Ef = 13.25 kg v2 + 367.5 J

Ef = Ei

13.25 kg v2 + 367.5 J = 441 J

13.25 kg v2 = 73.5 J

v2 = 5.55 m2/s2

v = 2.36 m/s

This is an example of the "joy of mathematics" -- mathematics works because it really does describe the "real world". This is the same answer we got earlier -- by a totally different method! The ideas of Physics and mathematics are useful -- and, therefore, important and worthwhile -- because they explain and predict the "real world"!

3. Angular Momentum

A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg m2 and is rotating at 10 rev/min. A 25-kg child jumps onto the edge of the merry-go-round.

What is the new angular speed of the merry-go-round?

This is problem 11.28

Lf = Li

L i = I i i

Ii = 250 kg m2 i = 10 rev/min

L i = (250 kg m2) ( 10 rev/min) = 2 500 (kg m2 rev / min)

L f = I f f

If= IMgR + Ichild

Ichild = m r2 = (25 kg) (2.0 m)2 = 100 kg m2

I f = (250 + 100) kg m2 = 350 kg m2 f = L f / I f = L i / I f = [ 2 500 kg m2 rev / min ] / [ 350 kg m2 ] f = 7.14 rev / min

4. Calculus application

Show, by direct calculation, that the moment of inertia for a uniform rod, rotated about its Center of Mass, is (1/12) M L2. This one is "new". This is not a duplicate of a homework problem but is quite similar to ones we did in class. I = r2 dm = x2 dx = x2 dx

I =  x2 dx = (1/3) x3 I = (1/3) [ ( L / 2 )3 - ( L / 2 )3 ] = (1/3) [ L3 / 4 ] = M / L

I = [ M / L ] (1/3) [ L3 / 4 ]

I = (1/12) M L2