Rubber ball Rotational Motion Question

[Theta]

Note: I refrained from using any of the latex references, so bear with me.

Homework Statement

A rubber ball with a radius of 3.2 cm rolls along the horizontal surface of a table with a constant linear speed v. When the ball rolls off the edge of the table, it falls 0.66 m to the floor below. If the ball completes 0.37 revolutions during its fall, what was its linear speed, v?

a) Variables extracted from the question

r = 3.2 cm = 0.032 m
h = 0.66 m
theta = (2pi)(0.37 revolutions) = (2pi x 0.37) rad

2. Homework Equations and physics principles/concepts

a) Equations

i. d = V_it + 0.5at²
ii. v = rw

b) Physics principles/concepts

i. Law of conservation of energy (?)
ii. Rotational kinetic energy and the moment of inertia (?)
iii. Kinematics (?)

The Attempt at a Solution

a) Energy is conserved, therefore:

K_i + U_i = K_f + U_f

b) There is no potential energy as the bottom of the table. Furthermore, we are analyzing the rotational motion of a rubber ball (ie. I = 0.4mr²). Therefore:

K_i + U_i = K_f
0.5mv_i² + 0.5(0.4mr²)(v_i/r)² = mgh + 0.5mv_f² + 0.5(0.4mr²)(v_f/r)²

c) We can clean up the equation by eliminating the mass and radii parameters, and multiplying each term by 2.

v_i² + 0.4v_i² = 2gh + v_f² + 0.4v_f²

d) Like terms can be collected and v_i can be solved for.

v_i = sqrt ((5/7)[(7/5)v_f² - 2gh])

e) We only need the final linear speed now. To determine that, we can use:

v_f = rw_f

i. As can be seen above, we need to determine w_f. To do this, regular kinematics can be applied to find the total time the ball spends in the air.

d = V_it + 0.5at²
d = 0.5at²
t = sqrt (2d/a)
t = sqrt [(2 x 0.66 m) / (9.81 m/s²)]
t = 0.37 s

ii. At the very beginning, the amount of revolutions (in radians) was determined. By dividing the revolutions by the time obtained above, we can obtain w_f, and ultimately v_f.

v_f = rw_f
v_f = (0.032 m)[(2pi x 0.37) / (0.37 s)]
v_f = 0.20 m/s

f) Finally, we can solve for initial linear speed.

v_i = sqrt ((5/7)[(7/5)v_f² - 2gh])
v_i = sqrt ((5/7)[(7/5)(0.20 m/s)² - 2(9.81 m/s²)(0.66 m)])
v_i = WHAT??!? AN IMAGINARY NUMBER?

So yeah, I do not know what I am doing wrong. Am I misinterpreting the question? Am I approaching the question in the wrong way? Does my logic make any sense?

 

[indr0008]

as the ball falls down, its rotational speed does not increase :)

can u imagine why?

and also, in the energy equation where should we put the potential energy, does the system lose or gain potential energy?

hope that helps :)

 

[Redbelly98]

ii. At the very beginning, the amount of revolutions (in radians) was determined. By dividing the revolutions by the time obtained above, we can obtain w_f, and ultimately v_f.

v_f = rw_f
v_f = (0.032 m)[(2pi x 0.37) / (0.37 s)]
v_f = 0.20 m/s

Since ω does not change, the ball has this same ω when it leaves the table. Can you use that information to find v_i=rω?

 

[Theta]

Oh, wow, I think I figured it out.

Thanks for the help you two.