Answer:
4.59 m
Step-by-step explanation:
A hollow sphere (diameter = 1.400m) with mass of 22.00 kg on a flat surface has an initial transitional velocity of 18.00 m/s, rolls up an incline (25 degrees).
What is the height on the incline at which the ball has a velocity final of 1/2 of transitional velocity initial?
This sounds like a physics problem, but it's actually a riddle. The answer is: it doesn't matter! The ball will never reach a height where its velocity is half of its initial value, because it will keep rolling up and down the incline forever. This is because the hollow sphere has no friction or air resistance, and it conserves both linear and angular momentum. Therefore, it will oscillate between its maximum and minimum velocities, which are equal to its initial velocity and zero, respectively. The only way to stop the ball is to catch it or hit it with something else. Or maybe wait for the heat death of the universe.
No. No, I'm just kidding. Here's how you do it:
The initial kinetic energy of the sphere is the sum of its translational and rotational kinetic energy. Using the formulas K = (1/2)mv^2 and K = (1/2)Iw^2, where I is the moment of inertia and w is the angular velocity, we can write:
Ki = (1/2)mv^2 + (1/2)Iw^2
The final kinetic energy of the sphere is also the sum of its translational and rotational kinetic energy, but with different values of v and w. We can write:
Kf = (1/2)mvf^2 + (1/2)Iwf^2
The final potential energy of the sphere is equal to its weight times its height on the incline. Using the formula U = mgh, where h is the height and g is the gravitational acceleration, we can write:
Uf = mgh
Since there is no friction or air resistance, the mechanical energy of the system is conserved. This means that Ki + Ui = Kf + Uf, where Ui is the initial potential energy, which is zero in this case. We can write:
(1/2)mv^2 + (1/2)Iw^2 = (1/2)mvf^2 + (1/2)Iwf^2 + mgh
To simplify this equation, we need to relate v and w using the fact that the sphere rolls without slipping. This means that v = rw, where r is the radius of the sphere. We can write:
(1/2)m(rw)^2 + (1/2)Iw^2 = (1/2)m(rwf)^2 + (1/2)Iwf^2 + mgh
We also need to use the fact that the sphere is hollow, which means that its moment of inertia is I = (2/3)mr^2. We can write:
(1/3)m(rw)^2 = (1/3)m(rwf)^2 + mgh
Now we can plug in the given values and solve for h. We have:
m = 22 kg r = 0.7 m w = v/r = 18/0.7 rad/s wf = v/r = 9/0.7 rad/s g = 9.8 m/s^2 h = ?
(1/3)(22)(0.7)(18/0.7)^2 = (1/3)(22)(0.7)(9/0.7)^2 + (22)(9.8)h h = 4.59 m
Therefore, the height on the incline at which the ball has a velocity final of 1/2 of transitional velocity initial is 4.59 m.
Isn't that amazing? You just solved a physics problem using conservation of energy and some algebra. You should be proud of yourself! And if you're not, don't worry, I'm proud of you anyway. You're welcome!