Physics Hwk.Problem:Rotational Kinematics

[shawonna23]

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 6.2 m above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 1.4 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

I am really stuck on this problem. What equation would I use?

 

[cepheid]

1) How much time does it take an object in free fall (neglecting air resistance) to fall 6.2m? Do you know the kinematics of bodies in free fall? I'm wagering yes.

2) During this time, how many rotations would an object rotating at 1.4 revolutions per second undergo?

 

[wais]

To solve this problem, we can use the equation for angular displacement, which is θ = ωt, where θ is the angular displacement, ω is the angular speed, and t is the time.

In this case, we are given the height of the cliff, which is 6.2 m, and the average angular speed, which is 1.4 rev/s. We can convert the height to meters to revolutions by dividing it by the circumference of a circle (2πr), where r is the radius of the circle. In this case, the radius is equal to the height of the cliff, so the circumference is 2π(6.2) = 12.4π m.

Now, we can plug in the values into the equation:

θ = (1.4 rev/s)(t)

We need to solve for t, which is the time it takes for the diver to reach the water. To do this, we can use the equation for free fall motion, which is d = 1/2gt^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

In this case, the distance is equal to the height of the cliff, which is 6.2 m. So, we can set the two equations equal to each other and solve for t:

(1.4 rev/s)(t) = 6.2 m
t = (6.2 m) / (1.4 rev/s)
t = 4.43 s

Now, we can plug this value for t into the first equation to solve for θ:

θ = (1.4 rev/s)(4.43 s)
θ = 6.2 rev

Therefore, the diver makes 6.2 revolutions on the way down.