A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ba

Question

A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.7

A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.7 degree past the lowest point on its way up, its total acceleration is (?22.5 i hat bold + 20.2 j hat bold ) m/s2. For that instant, do the following. Sketch a vector diagram showing the components of its acceleration. (Do this on paper. Your instructor may ask you to turn in the sketch.) Determine the magnitude of its radial acceleration. m/s2 Determine the speed of the ball. m/s Determine the velocity of the ball.

Explanation / Answer

let t be time measure in seconds.
start the particle moving at point (2,6)
at time t the particle will be at
(2+2t,2*sqrt(3t+3))
let D(t) be the distance from the particle to
the origin then
D(t)=sqrt( (2+2t)^2 +4*(3t+3) )
D(t)=sqrt(4t^2+8t+4+12t+12)
D(t)=sqrt(4t^2+20t+16)
now to find the rate at which this distance is changing we
simply find the dirivative of D(t)
D'(t)=(1/2)*(8t+20)*(4t^2+20t+16)^(-1/...
D'(t)=(4t+10)*(4t^2+20t+16)^(-1/2)
now to find the rate at which it is changing at (2,6) we simply
take D'(0)
D'(0)=10/sqrt(16)=10/4=5/2=2.5
so the distance from the particle to the origin is changing at a rate
of 2.5 units per sec

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