Question

A rider on a water slide goes through three different kinds of motion, as illustrated in the figure below. Use the data and the details from the figure to answer the following questions. Remember, when an object is in a circular motion it is accelerated toward the center of the circle with acceleration a = v2/r
a) At the end of the first section of the motion, riders are moving at approximately what speed? Show your work for partial credit.



A. 3 m/s B. 6 m/s C. 9 m/s D. 12 m/s



b) Suppose the acceleration during the second section of the motion is too large to be comfortable for the riders. What change could be made to decrease the acceleration during this section?



A. Reduce the radius of the circular segment

B. Increase the radius of the circular segment

C. Increase the angle of the ramp.

D. Increase the length f the ramp.



c) What is the vertical component of the velocity of the rider as he hits the water? Show your work for partial credit.



A. 2.4 m/s B. 3.4 m/s C. 5.2 m/s D. 9.1 m/s



d) Suppose the designers of the water slide want to adjust the height h above the water so that the riders land twice as far away from the bottom of the slide. What would be the necessary height above the water? Show your work for partial credit.



A. 1.2 m B. 1.8 m C. 2.4 m D. 3.0 m



e) During which section of the motion is the magnitude of the acceleration experienced by the rider the greatest?



A. The first B. The second C. The third D. It is the same in all sections

Answer 1

a) C. 9 m/s

First of all, let's calculate the difference in height between the starting point of the motion and the end point of the first section. It is given by:

$\Delta h = L sin \theta$

where

L = 6.0 m

$\theta=45^{\circ}$

Substituting,

$\Delta h = (6.0)(sin 45^{\circ})=4.2 m$

Assuming the rider starts from rest, its initial speed is zero. For the law of conservation of energy, the decrease in gravitational potential energy of the rider will be equal to its gain in kinetic energy, so we can write:

$mg\Delta h = \frac{1}{2}mv^2$

where m is the rider's mass, g = 9.8 m/s^2 is the acceleration of gravity, and v is the speed at the end of the first section. Solving for v, we find:

$v=\sqrt{2g\Delta h}=\sqrt{2(9.8)(4.2)}=9.1 m/s \sim 9 m/s$

b) B. Increase the radius of the circular segment

In fact, the acceleration during the second section of the motion (circular motion) is given by the formula for the centripetal acceleration:

$a=\frac{v^2}{r}$

where

v is the speed at the end of the first section

r is the radius of the circle

We notice that the acceleration is

- Proportional to the square of the speed

- Inversely proportional to the radius

So, we immediately see that if we increase the radius of the circle (choice B), the acceleration will decrease.

c) B. 3.4 m/s

When the rider exits the second section of the motion, he has a speed (completely horizontal) of 9.1 m/s (calculated in part (a); it didn't change, because the speed during the second section does not change).

The vertical component of his velocity is instead zero, since his motion is completely horizontal. Therefore, we can use the following SUVAT equation along the vertical direction:

$v_y^2 - u_y^2 = 2g\Delta h$

where

$v_y$ is the vertical component of the velocity as the rider hits the water

$u_y=0$ is the vertical component of the velocity as the rider starts the 3rd section

$\Delta h = 0.60 m$ is the difference in height

Solving for $v_y$,

$v_y = \sqrt{2g\Delta h}=\sqrt{2(9.8)(0.60)}=3.4 m/s$

d) C. 2.4 m

We want here the rider to land twice as far compared to before.

The horizontal distance travelled by the rider in section 3 is entirely determined by his horizontal motion. The horizontal component of the velocity, which is constant, is

$v_x = 9.1 m/s$

calculated at part (a) and remained unchanged during section 2. The horizontal distance travelled during section 3 is

$d=v_x t$ (1)

where t is the time of the fall. This can be rewritten as

$t=\frac{d}{v_x}$

We also know that the vertical displacement is:

$h=\frac{1}{2}gt^2$

Substituting t from (1) into this equation, we find

$h=\frac{gd^2}{2v_x^2}$

So we see that the height needed is proportional to the square of the distance: $h \propto d^2$. Therefore, in order to land at twice the previous distance, the height must be 4 times the previous one, so:

$h=4 (0.6 m)=2.4 m$

e) B. The second

We need to calculate the acceleration in each section.

In section 1 (motion along the slope), it is:

$a=g sin \theta = (9.8)(sin 45^{\circ})=6.9 m/s^2$

In section 2 (circular motion), the acceleration is the centripetal acceleration:

$a=\frac{v^2}{r}=\frac{9.1^2}{1.5}=55.2 m/s^2$

In section 3, the motion is free fall, so the acceleration is equal to the acceleration of gravity:

$a=g=9.8 m/s^2$

Therefore, the rider experiences the largest acceleration in section 2.