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:

Concepts: Uniform circular motion, F _{c} = mv^{2}/r | |

Reasoning: 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. | |

Details 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 = mv ^{2}/r, tanθ = v^{2}/(gr),
θ = 8.6^{o}.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/s ^{2})
= 15582 N.tanθ = 8653.4/15582 θ = 29 ^{o}.
The bank angle must be 29^{o}. |

**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:

Concepts Newton's law of gravitation, Gauss' law | |

Reasoning: The gravitational acceleration a distance r from the center is in the –r
direction and its magnitude is found using “Gauss' law”. | |

Details 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)∫ F4πr 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 = (Rπ^{2}/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 μ_{k}. 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:

Concepts: Newton's 2 ^{nd} law, friction, kinematics | |

Reasoning: The force of kinetic friction is responsible for the acceleration of the block. | |

Details of the calculation: Let the block's mass be m, its momentum just after it received the impulse be p = mv_{0}i. dp _{x}/dt = -μ_{k}mg
until the block comes to rest.Therefore p _{x} decreases at a steady rate until the block comes to
rest, a_{x} = μ_{k}g.The block comes to rest at time t _{rest} = v_{0}/(μ_{k}g).The block can cover a maximum distance d _{max} = v_{0}t_{rest}
-_{ }½μ_{k}gt_{rest}^{2}
= ½v_{0}^{2}/(μ_{k}g).Case (a) t ≥ t _{rest}, d = d_{max}, then
the distance covered during the subsequent time interval t' is zero.Case (b) t + t' ≥ t _{rest}, then the distance covered
during the subsequent time interval t' is d_{max} - d.Case (c) t + t' < t _{rest}, then the distance covered during the
subsequent time interval t' is d' = v _{0}(t + t') -_{ }½ μ_{k}g(t
+ t')^{2} – d = v_{0}t' -_{ }½ μ_{k}g
t'^{2} - μ_{k}gtt'. |

**
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:

Concepts: Energy conservation, kinematics | |

Reasoning: 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. | |

Details of the calculation: Just after the ball bounces off the plane its velocity makes an angle of 30 ^{o}
with the normal to the plane and therefore also an angle of θ' = θ = 30^{o} with
respect to the horizontal. Assume it has speed v. Its velocity components are
v_{x} = vcosθ, v_{y} = vsinθ. It executes projectile motion
and reaches its maximum height at t_{max_height} = vsinθ/g. During
this time it travels a distance d = vcosθ*vsinθ/g = v ^{2}sin(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 v^{2} = 2gd/sin(2θ). But ½mv ^{2}
= mg(h - d tanθ), (h - d tanθ) = d/sin(2θ), d (1/sin(2θ) + tanθ) = h.d = h sin(2θ)/(1 + 2sin ^{2}θ) = 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:

Concepts: Motion viewed from an inertial and a non-inertial frame | |

Reasoning: 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 K_{0} and rotates with angular
velocity Ω(t). The equations of motion are in the non-inertial frame are md v/dt = F_{inertial} - mdV/dt + mr × dΩ/dt
- 2mΩ × v - mΩ × (Ω × r).Here -md V/dt = fictitious force due to acceleration of frame.m r × dΩ/dt = fictitious force due to non-uniform rotation of
frame.-2m Ω × v = Coriolis force.-m Ω × (Ω × r) = Centifugal force.In the inertial frame we have F_{inertial} = mdv/dt + mdV/dt - mr × dΩ/dt
+ 2mΩ × v + mΩ × (Ω × r). | |

Details of the calculation: For our student d v/dt = dV/dt = 0.F_{inertial }= - mr × dΩ/dt + 2mΩ × v
+ mΩ × (Ω × r).In the inertial frame F _{tangential} = (mrα + 2mvαt), F_{rad} = -m(αt)^{2}r,F _{tangential} = 3mrα, F_{rad} = -m(αt)^{2}r.In the inertial frame the acceleration is therefore given by a _{tangential} = 3rα, a_{rad} = -(αt)^{2}r.Assume that the direction of Ω is the z-direction.
Then a_{tangential}
= 3rα(ф/ф) and a_{rad}
= -(αt)^{2}r.Transforming from polar to Cartesian coordinates we have ( r/r) = cosф i + sinф j,
(ф/ф) = -sinф i + cosф
j.Therefore a _{x} = -(αt)^{2}cosф - 3rα sinф, a_{y} =
-(αt)^{2}sinф + 3rα cosф.r = vt, ф = ½αt ^{2}.a _{x} = -(αt)^{2}cos(½αt^{2}) - 3vtα sin(½αt^{2}),
a_{y} = -(αt)^{2}sin(½αt^{2}) + 3vtα cosф(½αt^{2}). |