Circular Motion

[匿名]

2011-12-04 3:51 pm

A slide ( 滑梯) XY is in the shape of the 3rd quadrant of a circle with radius 4 meters ( that is X is 4 meters above ground). A ball with mass m is released from rest from point X, find the time taken to reach point Y. ( Assume no friction).

回答 (1)

Prof. Physics

2011-12-04 5:02 pm

✔ 最佳答案

Consider the moment when the ball is at a location where it makes an angle @ with vertical direction.

The forces acting on the ball is its weight (mg) and normal reaction (N)

Consider radial direction,

N - mgcos@ = mr d@/dt ... (1)

Consider transverse direction,

mgsin@ = -mr d^2@/dt^2 ... (2)

From (2):

gsin@ = -r (d@' / d@)(d@/dt) = -r (d@'/d@)@'

Integrating both sides,

-g cos@ = -r @'^2 / 2 + C

When @ = 90*, @' = 0 (starting at rest)

So, C = 0

2g cos@ = r @'^2

d@/dt = @' = sqrt [(2g cos@) / r]

Integrating both sides,

Time required, t = sqrt(r / 2g) int (兀/2 to 0) (1/ sqrt(cosx)) dx

From numerical integration, int (兀/2 to 0) (1/ sqrt(cosx)) dx = 3.1862

So, the time required, t = sqrt[(4) / 2(9.8)] (3.1862)

= 2.036 s

參考: Prof. Physics

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