Constant acceleration equations:x(t) = x0 + v0t + at2/2 v(t) = v0 + at

Velocity and accelerationv = Δ x/Δ t a = Δ v/Δ t

Freely falling motion: g = 9.8 m/s2 ay = -g y(t) = y0 + v0yt - gt2/2

Friction: f = μ |n|Spring: f = -kxNewton’s law: Σ F = maComponents: Vx = vcos θ Vy = vsin θ

Kinetic Energy = 1/2 mv2 Gravitational Potential Energy = mgh (g constant)F(x) = -dU/dxWork = F·dr Spring Potential Energy = 1/2 kx2

Rotation:I = Σ mr2

τ = I αv = ω r, a = αr L = r x p K = 1/2 I ω2

Center of mass: xcm= Σ mr / Σ m

Gravitation:F(r) = Gmm/r2

U(r) = -Gmm/rG = 6.67 x 10-11 m3 kg-1 s-1

1) An astronaut inside a spacecraft is orbiting a black hole at a distance of 120km from its center. The black hole has mass of 1x1031 kg. If her 0.05kg ears are 10cm apart from each other, what is the tension between her ears? (The difference between the two forces).

The tension is the difference between the forces.

F(ear1) = G mear mBH / (R +d )2

F(ear2) = G mearmBH / (R -d ) 2

T = F(ear1) - F(ear2) = 7720 N

2) A spacecraft is in a circular orbit around the moon at 50km above the surface at RM = 1.74 x106 m. If a malfunction causes the thrusters to reduce the speed by 20.0 m/s, with what speed will it crash into the moon’s surface?

Find the orbital speed:

F = GmmE/r2 = mv2/rv2 = GmE/r where r = RM + hv = 1.655 x 103 m/s

If it decreases by 20 m/s, it becomes

v = 1.635 x 103 m/s

Now it’s just an energy problem to find the remaining kinetic energy when it reaches the surface.

Ki + Ui = Kf + Uf Ui = -G mMm/(Rm+h)Uf = -G mMm/(Rm)

-G mMm/(Rm+h) + 1/2 mvf2 = -G mMm/(Rm) + 1/2 mvf2

Solve for v2 and you find v2 = 1.682 x 103 m/s

3) A large 16kg roll of paper with radius R= 18cm rests against the wall and is held in place by a bracket attached to a rod through the center of the roll. The rod turns without friction in the bracket and the moment of inertia of thepaper and the rod about the axis is 0.26 kg m2. The other end of the bracket is attached by a frictionless hing to the wall such that the bracket makes an angle of 30 degrees to the wall. The coefficient of frction between the paper and the wall is 0.25 and a constant vertical force F=40N is applied to the paper which unrolls.

Identify the forcesw weight of rollF pulling forcef frictional forceFrod force of rod on papern normal force on paper

a) What is the magnitude of the force that the rod exerts on the paper?Horizontal: Frod sin θ = nVertical: Frod cos θ = f+w+F

Use f = μ nFrod cos θ = μ n +w+FFrod cos θ = μ (Frod sin θ) +w+FFrod = w+F / (cos θ -μ sin θ)f = μ Frod sin θb) What is the magnitude of the angular acceleration of the roll?τ = (F-f)R = (F - μ Frod sin θ) Rα = τ / I

4) A block of mass 5kg slides down an inclined surface with coefficient of friction of 0.25 . A string is attached to the block and wrapped around a flywheel fixed at axis O. The flywheel has mass 25kg and I = 0.5 kg m2 with respect to the axis O. The string pulls without slipping at a distance of 0.2m from the axis.

a) What is the acceleration of the block down the plane?b) What is the tension in the string?y componentn - mg cos θ = 0fk = μk mg cos θ

x componentmg sin θ - T - fk = mamg (sin θ - μk cos θ) - T = ma

Find tensionτ = Iα = Ia/R = RTT = Ia/R2

mg (sin θ - μk cos θ) - Ia/R2 = maa = 1.12 m/s2

T = 14N

5) For each pair of r and F vectors indicate whether the torque will be positive (out of the page), zero, or negative (into the page)vector problem

a) positive b) zeroc) negatived) negativee) zerof) positive

6) For each potential (U) or force (F) diagram, draw the corresponding force or potential. Remember F = -dU/dx

7) A 2kg shell is launched at 45 degrees above the horizon with velocity of 100 m/s.

a) Write the formulas for the horizontal and vertical distances travelled as a function of time: x(t) and y(t).

x componentx(t) = v0 cos(45)t y componenty(t) = v0 sin(45)t - 1/2 gt2

b) What is the time of flight of the shell? (When does it hit the ground?)y(t) = v0 sin(45)t - 1/2 gt2 =0v0 sin(45) - 1/2 gt =02v0 sin(45)/g = t

t = 2 100 sin(45) /(9.8m/s) = 14.4 s

c) What is the range of the shell? (How far away does it land?)

x(14.4) = 100 * cos(45) *14.4 = 1018.2 m

d) If after 10s, the shell explodes into two pieces of equal mass, with horizontal velocities of +2 m/s and -2 m/s relative to the center of mass of the shell. They have no vertical velocity relative to the center of mass. What is the range and time of flight of each piece?

y component for both is unchangedy(t) = v0 sin(45)t - 1/2 gt2

time of flight is unchangedx component for piece Ax(t) = v0 cos(45))t + 2.0*(t-10)

x(14.4) = (100*cos(45)) * 14.4 + 2*4.4= 1027.0mx component for piece Bx(t) = v0 cos(45))t - 2.0*(t-10)x(14.4) = (100*cos(45)) * 14.4 - 2*4.4= 1009.4m

8) A solid uniform sphere of mass M (I =2MR2/5 ) starts from rest and rolls down a 50m high hill. The top half has friction and the boulder rolls without slipping. The bottom half is covered with ice and has no friction. What is the linear (not angular) speed of the boulder when it reaches the bottom of the hill?

Break into two pieces, rolling then slipping.

For rollingK = 1/2 mv2 + 1/2I ω2, v = ωR, I =2MR2/5

mgh1 = 1/2 mv2 + 1/2 I ω2

v2 = 10gh1/7

For slipping, the rotational KE doesn’t change.mgh2 + 1/2 mv2 + Krot = 1/2 mv2final + Krot

v2B = 10gh1/7 + 2gh2 vB = 29.0 m/s

9) A fan blade rotates with angular velocity ω(t) = a - bt2.

a) Calculate the angular acceleration as a function of time.

α = d ω / dt = -2bt

b) Calculate the average angular acceleration over the period t=0 to t=3s

( ω(3) - ω(0) )/ 3