Mechanics 1, solutions

Problem 1:  (E)

Kinematics, motion with constant a

The horizontal component of the velocity is constant.

Problem 2:  (C)

Uniform circular motion, gravity

v02/R = GM/R2, T0 = ½mv02 = ½(GMm/R) = -½U(R).
E = 2T0 + U(R) = 0.

Problem 3:  (C)

Energy conservation

The change in potential energy of the CM is ΔU = mgΔh = LρgL/2 = (2*10*102/2) J.

Problem 4:  (E)

Nonuniform circular motion, kinematics

a = sqrt(at2 + ar2),  at = gcosθ,  ar = v2(θ)/L.
energy conservation:  ½mv2 = mgLsinθ.
Therefore ar = 2gsinθ.
a = g*sqrt(cos2θ + 4 sin2θ) = g*sqrt(1 + 3 sin2θ).

Problem 5:  (B)

Newton's 2nd law

We only have centripetal acceleration.  The horizontal force of the road on the tires must cancel the air resistance force and provide the centripetal acceleration.

Problem 6:  (A)

Momentum conservation

m3v0 = (m1 + m2 + m3)vCM.
vCM = m3v0/(m1 + m2 + m3) = v0/4.

Problem 7:  (C)

Rolling, a and v do not depend on M and R

For example:
mgsinθ - f = ma.
fr = Iα, αr = a.
a = mg/(m + I /r2),  a = 2gsinθ/3.

Problem 8:  (C)

Rolling, a and v do not depend on M and R

Problem 9:  (C)

Parallel axes theorem

Physical pendulum:  θ(t) = θmaxcos(ωt + φ), with ω2 = (mgd)/I.
Moment of inertia of the hoop about its center: I = mr2.
Moment of inertia of the hoop about the pivot point: I = 2mr2 (parallel axes theorem).
period = 2π/ω = 2π(2mr2/(mgr))½ = 2π(2r/g)½ = 6.28*(0.04)½ s = 6.28/5 s ~ 1.3 s.

Problem 10:  (C)

Hooke's law

k' = 2k + k/2 = 5k/2,  T =  2π (m/k')½ = 2π (2m/(5k))½.