A rotating merry go round makes one complete revolution in 4.0s. a) What is the linear speed of a child seated 1.2m from the center? What is her acceleration(give components)?
2006-12-01 00:33:10 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
That's awfully fast. I hope child isn't too sick. I'd be puking.
The child travels 1.2m*2*pi meters in 4 sesconds, so her speed is about 7.5 meters in 4 seconds, or 1.885 m/s.
To get acceleration, you need to write a formula for her movement:
Her "x" position is 1.2*sin(t*2*pi/4)
her "y" position is 1.2*cos(t*2*pi/4)
her "x" acceleration is the second derivative of "x" position.
f(t) = 1.2*sin(t*2*pi/4)
f'(t) = 1.2*cos(t*pi/2)*pi/2
f''(t) = 1.2*pi/2*(-sin(t*pi/2))*pi/2
or -.3*pi^2*sin(t*pi/2)
similarly, her "y" acceleration is the second derivative of the "y" position:
f''(t) for this function is -.3*pi^2*cos(t*pi/2.)
The total acceleration is the square root of the sum of the squares of the x and y components, which thankfully simplifies because sin^2 anything + cos^2 anything is 1,
so the acceleration is
.3*pi^2 or 2.9 m/s/s
2006-12-01 00:50:14 · answer #1 · answered by firefly 6 · 1⤊ 0⤋
Assuming that the merry go round completes one revolution at uniform speed, then the angular speed is 2*pi/4 rad/s. Linear speed at 1.2 m from center = angular speed * 1.2
Only accl is centripetal accl, directed towards center. magnitude = v^2/r
lug in values to calculate
2006-12-01 08:47:45 · answer #2 · answered by astrokid 4 · 0⤊ 0⤋