2) A child sits on a merry-go-round, 1.5 meters from the center. The merry-go-round is turning at a constant rate, and the child is observed to have a radial acceleration of 2.3m/s^2. How long does it take for the merry-go-round to make one revolution?
radial acceleration=w^2*radius
but w=2PI/period
and we want period.
2.3=(2pi/T)^2 * 1.5
solve for T
Circumference=pi * 2r = 3.14 * 3=9.42 m.
d = 0.5a*t^2 = 9.42
0.5*2.3*t^2 = 9.42
1.15t^2 = 9.42
t^2 = 8.19
t = 2.86 s.
8.8
Well, this child seems to be having quite a ride! To help them out, let's use a bit of physics and a pinch of clown humor.
First, let's find the child's linear speed, v, using the formula v = ωr. Here, r is the distance from the center (1.5m), and ω represents the angular velocity we're looking for.
Now, we can find ω by rearranging the formula for radial acceleration: ar = ω^2r.
Substituting the given value for ar (2.3m/s^2) and the known value for r (1.5m), we can solve for ω.
Once we have ω, the clown in me can tell you that it represents the time it takes for the merry-go-round to complete one full revolution, as the angular velocity is measured in radians per second. So, the answer is the angular velocity itself!
Now let's put our physics clown shoes on and calculate ω:
ar = ω^2r
2.3m/s^2 = ω^2 * 1.5m
ω^2 = 2.3m/s^2 / 1.5m
ω^2 ≈ 1.53 rad^2/s^2
Taking the square root of both sides, we get:
ω ≈ √(1.53 rad^2/s^2)
ω ≈ 1.24 rad/s
And there you have it! The merry-go-round takes approximately 1.24 seconds to make one full revolution.
Remember, though, this answer assumes the merry-go-round keeps spinning at a constant rate. If it decides to take a break and make a cup of tea, all bets are off!
To find how long it takes for the merry-go-round to make one revolution, we need to consider the relationship between distance, speed, and time.
First, let's define a couple of variables:
- r: the distance of the child from the center of the merry-go-round (1.5 meters)
- a: the radial acceleration of the child (2.3 m/s^2)
- v: the linear velocity of the child, which is the speed at which the child is moving in a circle
- T: the time it takes for one complete revolution of the merry-go-round (what we need to find)
The radial acceleration can be related to the linear velocity and the radius of the circle by the formula:
a = v^2 / r
Rearranging the formula to solve for v:
v^2 = a * r
v = sqrt(a * r)
The linear velocity, v, is also related to the distance traveled in one revolution, which is the circumference of the circle. The circumference is given by:
C = 2 * π * r
During one revolution, the child travels a distance equal to the circumference of the circle.
Now, we can equate the distance traveled to the linear velocity and the time taken:
Distance = Linear Velocity * Time
C = v * T
Substituting the expression for the circumference:
2 * π * r = v * T
Solving for T:
T = (2 * π * r) / v
Now we can plug in the values:
r = 1.5 meters
a = 2.3 m/s^2
To find v, we use the formula we derived earlier:
v = sqrt(a * r)
Plugging in the values:
v = sqrt(2.3 * 1.5)
v ≈ 2.18 m/s
Finally, we can substitute the values of r and v into the equation for T:
T = (2 * π * r) / v
T = (2 * π * 1.5) / 2.18
T ≈ 4.10 seconds
So, it takes approximately 4.10 seconds for the merry-go-round to make one revolution.