1. If the bus is moving to the right, the region has the shape of a < , with the point A at the point of the <. The path of the bus, and of the person who has caught the bus starting from the boundary of the region, are two sides of a right triangle of hypotenuse (bus) d, and one side (person) d/2. The angle opposite the d/2 side is therefore arcsin(½) = 30°.
2. First show that the distance between the cylinders stays constant: assume it is so, show it's consistent because the surface speed at the plate's points of contact will then be equal, so there is no slipping. Alternatively, assume the contrary, that one of the cylinders is rotating faster than it should if the distance were constant. To keep from slipping on one of the surfaces the cylinders would then have to separate, but to keep from slipping on the other surface they would have to approach each other.
Thus the angle α is constant. Next, find the path of the center of mass. Suppose the cylinders could remain in place and only rotate, by slipping at the horizontal surface. The CM would then move down at an angle α, say by a distance d. If the cylinders don't slip, they would in addition propel the system to the right by an equal distance d. Adding the two displacements, equal in magnitude and angle α between them, gives a displacement inclined at an angle α/2. Thus the center of mass moves as if on a frictionless ramp inclined by α/2.
The acceleration down such a ramp is g sin(α/2).
This problem was a bit confusing because the original website version (though not the problem book) specified "no friction". It should have said "no energy dissipation by friction". The force involved in rolling is in general not normal to the surface; its parallel component is due to static friction, which does no work.
3. (a) Separating the m dv/dt = F gives me-v/V dv = -F0 dt. Integrate from 0 to t on the left, vo to v on the right, solve for v:
v = -V ln
(b) This is zero when the argument of the log is 1, so t = (1 - e-vo/V)mV/F0.
(c) Use òlnx dx = x lnx - x
to integrate the above v over t. The result is a mess, when you substitute the t from (b) you get
1 - e-vo/V
4. (a) 4/3 (b) 17/15
(c) W = ò01 t6 (3t2 dt) + (2t5)(2t dt) = 19/21
5. From conservation of energy, T = 1/2mv2 = E - U = mgR(1 - cos θ).
When the puck leaves the surface, N = 0, the radial component of gravity is just enough to give the path curvature R: mg cos θ = mv2/R.
Solve for cos θ = 2/3, hence the puck has descended a distance R/3.
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version 3.67. On 04 Feb 2006, 16:38.