Assignment 1, solutions

Problem 1:

A highway curve with a radius of 750 m is banked properly for a car traveling 120 km/h.  If a 1590 kg Porsche rounds the curve at 230 km/h,
(a)  how much sideway force must the tires exert against the road if the car does not skid?
(b)  what must be the bank angle for the Porsche to turn if there is no friction force on the tires?

Solution:

bulletConcepts:
Uniform circular motion,  Fc = mv2/r
bulletReasoning:
In part (a) the frictional force must provide part of the centripetal acceleration, in part (b) the horizontal component of the normal force must provide all the centripetal acceleration.
bulletDetails of the calculation:
(a) A free-body diagram of the car on the track is shown below.

When the car is traveling at v = 120 km/h the frictional force f = 0 and 
Ncosθ = mg, Nsinθ = mv2/r, tanθ = v2/(gr), θ = 8.6o.
When the car is traveling at v' = 230 km/h we have 
Ncosθ - mg - fsinθ  = 0,  Nsinθ + fcosθ = mv'2/r.
f(tanθsinθ + cosθ) = mv'2/r - mgtanθ.
f = 6226 N.
The tires must exert a sideway force of 6226 N.
For part (b) the horizontal component of the normal force must equal mv'2/r = 8653.4 N.
The vertical component of the normal force is mg = 1590 kg(9.8 m/s2) = 15582 N.
tanθ = 8653.4/15582  θ  = 29o.  The bank angle must be 29o.

Problem 2:

The shortest way from the US to Australia is via a tunnel that goes thought the center of the Earth.  If one could build such a tunnel and make it friction free, then an object dropped at the US side with zero initial velocity would emerge after some time on the other side in Australia.  Assuming that density of the Earth is uniform (which is not correct), calculate how long it would take for an object to pass through such a tunnel. 

Solution:

bulletConcepts
Newton's law of gravitation, Gauss' law
bulletReasoning:
The gravitational acceleration a distance r from the center is in the –r direction and its magnitude is found using “Gauss' law”.
bulletDetails of the calculations:
Imagine a tunnel bored through the center of the earth.  For a test particle a distance r from the center we have

(1/m)∫closed areaFn dA = 4πGMinside,

F4πr2 = m4πG(4/3)πr3ρ,  F = (4/3)Gmπρr = ma, a = (4/3)Gπρr.
At the surface g = (4/3)GπρR, so a = gr/R, where R is the radius of the earth.

F = -(mg/R)r.
If the coordinate system is chosen so that the origin is at the center of the Earth and the tunnel lies along the y-axis, then F = -(mg/R)y.
y = Rcos(ωt), ω = (g/R)1/2 = 2π/T,  T/2 = (Rπ2/g)1/2 is the time it take to pass through the tunnel.  T/2 = 2535 s.

Problem 3:

A block is given a quick push along a horizontal table.  The coefficient of kinetic friction between the block and the table is μk.  It is known that during the time interval t (immediately after the push) the block covers a distance d.  Find the distance that may be covered by the block during the subsequent time interval t'.  Find all possible answers.

Solution:

bulletConcepts:
Newton's 2nd law, friction, kinematics
bulletReasoning:
The force of kinetic friction is responsible for the acceleration of the block.
bulletDetails of the calculation:
Let the block's mass be m, its momentum just after it received the impulse be p = mv0i.
dpx/dt = -μkmg until the block comes to rest.
Therefore px decreases at a steady rate until the block comes to rest, ax = μkg.
The block comes to rest at time trest = v0/(μkg).
The block can cover a maximum distance dmax = v0trest - ½μkgtrest2 = ½v02/(μkg).
Case (a)  t ≥ trest, d = dmax, then the distance covered during the subsequent time interval t' is zero.
Case (b)  t + t' ≥ trest, then the distance covered during the subsequent time interval t' is dmax - d.
Case (c)  t + t' < trest, then the distance covered during the subsequent time interval t' is
d' = v0(t + t') - ½ μkg(t + t')2 – d = v0t' - ½ μkg t'2 - μkgtt'.

Problem 4:

Two wedges are placed mirror symmetrically so that the tip of one touches the tip of another as shown in the figure below.  The surface of each wedge is at angle θ = 30 degrees relative to the ground.  A small elastic ball is dropped from height h = 1 m with zero initial velocity.  How far from the tips of the two wedges (x) must the small ball be dropped, so that after bouncing from the two wedges it will reach the same height from where it was dropped?  Neglect any friction from air and consider the bouncing of the ball to be completely elastic.

Solution:

bulletConcepts:
Energy conservation, kinematics
bulletReasoning:
The ball's trajectory must be symmetric about the midpoint between the wedges.  It has to cross the midpoint with its velocity having no vertical component. 
bulletDetails of the calculation:
Just after the ball bounces off the plane its velocity makes an angle of 30o with the normal to the plane and therefore also an angle of θ' = θ = 30o with respect to the horizontal.  Assume it has speed v.  Its velocity components are vx = vcosθ,  vy = vsinθ.  It executes projectile motion and reaches its maximum height at tmax_height =  vsinθ/g.  During this time it travels a distance
d = vcosθ*vsinθ/g = v2sin(2θ)/(2g).  If the distance to the middle is d, it crosses the midpoint with its velocity having only a horizontal component.  The second half of its trajectory then mirrors the first half.  We need v2 = 2gd/sin(2θ). 
But ½mv2 = mg(h - d tanθ),  (h - d tanθ) = d/sin(2θ),  d (1/sin(2θ) + tanθ) = h.
d = h sin(2θ)/(1 + 2sin2θ) = 0.577 m.

Problem 5:

A brave physics student (an undergraduate, of course) climbs aboard a high powered merry-go-round and goes to the center, at r = 0.  At time t = 0, the platform starts from rest (Ω = 0) and begins to spin about its vertical axis with constant angular acceleration α.  Also at time t = 0, the student begins to crawl radially outward at constant speed v, relative to the platform.
Assuming the student does not slip, find the student's acceleration in the inertial frame of an outside observer.

Solution:

bulletConcepts:
Motion viewed from an inertial and a non-inertial frame
bulletReasoning:
In non-inertial frames fictitious forces appear. 
Consider a particle moving with velocity v in a reference frame K which moves with velocity V(t) relative to the inertial frame K0 and rotates with angular velocity Ω(t). 
The equations of motion are in the non-inertial frame are
mdv/dt = Finertial - mdV/dt  + mr × dΩ/dt - 2mΩ × v - mΩ × (Ω × r).
Here -mdV/dt = fictitious force due to acceleration of frame.
mr × dΩ/dt = fictitious force due to non-uniform rotation of frame.
-2mΩ × v = Coriolis force.
-mΩ × (Ω × r) = Centifugal force.
In the inertial frame we have
Finertial = mdv/dt + mdV/dt  - mr × dΩ/dt + 2mΩ × v + mΩ × (Ω × r).
bulletDetails of the calculation:
For our student dv/dt = dV/dt = 0.
Finertial = - mr × dΩ/dt + 2mΩ × v + mΩ × (Ω × r).
In the inertial frame
Ftangential = (mrα + 2mvαt),  Frad = -m(αt)2r,
Ftangential = 3mrα,  Frad = -m(αt)2r.
In the inertial frame the acceleration is therefore given by
atangential = 3rα,  arad = -(αt)2r.
Assume that the direction of Ω is the z-direction. 
Then atangential = 3rα(ф/ф) and  arad = -(αt)2r.
Transforming from polar to Cartesian coordinates we have
(r/r) = cosф i + sinф j,  (ф/ф) = -sinф i + cosф j.
Therefore ax = -(αt)2cosф - 3rα sinф,  ay = -(αt)2sinф + 3rα cosф.
r = vt, ф = ½αt2.
ax = -(αt)2cos(½αt2) - 3vtα sin(½αt2),  ay = -(αt)2sin(½αt2) + 3vtα cosф(½αt2).