A 75.0 kg person is on a Ferris Wheel that has a radius R = 16 m. The tangential velocity of the Ferris Wheel is 8.25 m/s. Calcu
late the period of rotation of the Ferris Wheel. Enter your answer below and include units and round to 3 significant figures. Show your work on paper, scan it, and submit it in the last problem.
1 answer:
Answer:
The period of rotation is
T=8.025s
Explanation:
The person is undergoing simple harmonic motion on the wheel
Given data
mass of the person =75kg
Radius of wheel r=16m
Velocity =8.25m/s
The oscillating period of simple harmonic motion is given as
T=(2*pi)/2=2*pi √r/g
Assuming that g=9.81m/s
Substituting our data into the expression we have
T=2*3.142 √ 16/9.81
T=6.284*1.277
T=8.025s
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Explanation:
The given data is as follows.
mass = 0.20 kg
displacement = 2.6 cm
Kinetic energy = 1.4 J
Spring potential energy = 2.2 J
Now, we will calculate the total energy present present as follows.
Total energy = Kinetic energy + spring potential energy
= 1.4 J + 2.2 J
= 3.6 Joules
As maximum kinetic energy of the object will be equal to the total energy.
So, K.E = Total energy
= 3.6 J
Also, we know that
K.E =
or, v =
=
=
= 6 m/s
thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.