A merry go round makes 2 revolutions every 2 s. A child is sitting 1.80 m from the center. Calculate
a)The period of the merry go round
b)the speed of the child in his position
c) the centripetal acceleration of the child
d) the net horizontal force on the child if his mass is 24.00 kg
a) How long does it take to make one revolution if it takes two seconds for two revolutions? You can do that in your head.
b) V = 2 pi*R/(period)
c) a = V^2/R
d) F = M a = M V^2/R
To answer these questions, we need to understand the concepts of period, speed, centripetal acceleration, and net force.
a) The period of an object in circular motion is the time it takes to complete one full revolution. In this case, the merry-go-round makes 2 revolutions every 2 seconds. This means that the period is 2 seconds.
b) The speed of an object in circular motion can be calculated using the formula: speed = (2πr) / T, where r is the radius and T is the period. In this case, the child is sitting 1.80 m from the center, and the period is 2 seconds. Plugging in these values, we get: speed = (2π * 1.80) / 2. The speed of the child in his position is approximately 5.65 m/s.
c) The centripetal acceleration is the acceleration experienced by an object in circular motion, directed towards the center of the circle. It can be calculated using the formula: centripetal acceleration = (speed^2) / radius. In this case, we know the speed of the child is 5.65 m/s, and the radius is 1.80 m. Plugging in these values, we get: centripetal acceleration = (5.65^2) / 1.80. The centripetal acceleration of the child is approximately 17.7 m/s^2.
d) The net horizontal force on the child can be calculated using Newton's second law of motion: force = mass * acceleration. In this case, the mass of the child is given as 24.00 kg, and we can use the centripetal acceleration we calculated earlier as the acceleration. Plugging in these values, we get: force = 24.00 kg * 17.7 m/s^2. The net horizontal force on the child is approximately 425.0 N.