Physics 1116 Fall 2011 Prelim 0 Your Name:

1. Find the acceleration of m1 and the tension T1 in the upper string. Stringsand pulleys are massless, pulley rotation is frictionless.

Solution:

y1, y2, y3 point down from ceiling to m1,m2, and the lower pulley asshown. Two N2, two constraints (fixed string lengths), and one N3equation lead to:

m1g − T1 = m1y1

m2g − T2 = m2y2

y1 + y3 = 0

y2 − 2y3 = 0

T1 = 2T2

Note y2 = −2y1. Solve the system of equations.

y1 =m1 − 2m2

m1 + 4m2g

T1 = 6m1m2 g

m1 + 4m2.

2. A car of mass m travelling at speed v enters a turn of radius R, on aroad that is banked (tilted) at an angle θ as shown in the figure. Thecoefficient of friction between the wheels and the road is µ. Find themaximum and minimum values of v that will allow the car to completethe turn without skidding sideways.

Solution:

Draw free body diagram involving only N, f, and mg. This problemis most easily handled in the xy frame of horizontal and vertical axes,

Page 2

because the centripetal acceleration is horizontal. It can be done in theuv frame with axes parallel and perpendicular to the road surface, butthen the centripetal acceleration vector is partly in u and partly in v.The figure above gives all the options.

Generically, Newton’s 2nd for the car is given by

N v + f u−mgy = mr,

given the definitions of Fig. 2. You can work in your favorite coordinatesystem just by dotting the above equation with either u and v, or x andy.

Here I use the xy option. So after dotting the vector equation abovewith x and y, I get two equations:

Vertical balance: N cos θ + f sin θ −mg = my = 0

Horizontal balance: N sin θ − f cos θ = mx = mv2/R ≡ mA

Note that I have also imposed the requirement that the vertical accel-eration be zero (equivalent to requiring that the car not slide in u),and that the horizontal acceleration be consistent with uniform circularmotion at radius R.

Friction is a force that can point either way, but we have to start withsome choice of sign convention. Here I’ve chosen µN = f to be positivewhen it points along +u as shown above.

Now we solve for A = v2/R in two cases f = −µN and f = +µN . Thefirst case will correspond to the car sliding up (+u), indicating that weare probing a maximum velocity; while the second case addresses theminimum velocity.

Amin =v2

min

R=

(sin θ − µ cos θ

cos θ + µ sin θ

)g

Amax =v2

max

R=

(sin θ + µ cos θ

cos θ − µ sin θ

)g

so:

vmin =

(sin θ − µ cos θ

cos θ + µ sin θRg

)1/2

Page 3

vmax =

(sin θ + µ cos θ

cos θ − µ sin θRg

)1/2

.

Note if µ = 1 and θ = π/4 the min and max velocities are 0 and ∞.

Page 4

3. Here is a formula that describes a real problem in particle physics:

M =1

w

√(t2 − w2)(w2 − ν2)

The variables t, w, and ν are shorthand notations for the top quark,W boson, and neutrino masses. Assuming that the w and ν masses arewell-known constants, expand this function in a Taylor Series in x, wherex ≡ t− t0 and x� t0.

Solution:

Be careful: the small parameter is x.

M(x) =1

w

√((t0 + x)2 − w2)(w2 − ν2)

≈ M(0) + xdM

dx

∣∣∣∣x=0

+ . . .

≈ 1

w

√(t20 − w2)(w2 − ν2) +

t0w

(w2 − ν2

t20 − w2

)12x

Page 5