Trying to figure this out. I know it relates to the fact that an inscribed angle subtended by a diameter is a right angle, but I can't put the answer into appropriate mathematical language:
An objects slides down an inclined plane OP' starting from rest at O, which is the top of a circle of diameter OP, where P and P' lie in a horizontal line. Show that the point Q reached in the time, t, required to fall straight down from O to P lies on a circle with OP as diameter.
Thanks.
In the problem, it is not clear where the center is, only that OP passes through it. Hope that will help. This is my solution(?), please point out any problems or a simpler way withou using so much geometry if there is such an answer. Thanks.:
The time, t, it takes to fall from O to P is t = sqrt(d(OP))/4.
The distance along the inclined plane that the mass slides is given by the equation d(OQ) = 16(t^2)*Sin(A).
In time t, d(OQ) = d(OP)*Sin(A).
Drawing Q'P, where Q' is the intersection of OP' with the circle, we have a right triangle OPQ' because an inscribed angle subtended by a diameter is a right angle. Because angle OP'P and angle OPQ' are complementary to the same angle they are equal, that is OPQ' = A. Therefore d(OQ') = d(OP)*Sin(A) = d(OQ). Since OQ and OQ' are at the same angle from OP and they have the same length, Q and Q' coincide. That is, Q lies on the circle with diameter OP.