A solid sphere of radius R is placed at a height of 32cm on a 15degree slope. It is released and rolls, without slipping, to the bottom. From what height should a circular hoop of radius R be released on the same slope in order to equal the sphere's speed at the bottom?
Well, let me roll out a hilarious explanation for you! Imagine the sphere and the hoop preparing for their exciting slope adventure.
The sphere, being the "roll model" of roundness, starts off with potential energy at the top of the slope. As it rolls down, that potential energy transforms into kinetic energy and some rotational energy, going wheeee all the way!
Now, to match that incredible speed at the bottom, the hoop needs to release from a certain height. Since we're dealing with speed here, and the hoop wants to be a "speed hoop", it needs to share the same kinetic energy as the sphere.
Here's the trick: The hoop has its own funny way of moving. As it rolls down, not only does it gain kinetic energy, but it also gains potential energy due to its circular shape. It's like having an extra energy boost thrown in for free!
To equal the sphere's speed, we need the hoop to reach the same total energy at the bottom. In other words, the kinetic energy gained by the sphere should be equal to the kinetic energy gained by the hoop plus the extra potential energy from the circular shape.
Now, let's get serious for a second and do some math! The kinetic energy of the sphere at the bottom is given by KE = 1/2 * m * v^2, where m is the mass of the sphere and v is its final velocity.
Since the sphere is rolling without slipping, we can also relate the velocity to the radius and angular velocity using v = R * ω, where ω is the angular velocity.
On the other side of the clown equation, we have the hoop. The kinetic energy gained by the hoop is also KE = 1/2 * m * v^2, but remember, it also gains potential energy due to its circular shape. This potential energy can be related to the height (h) from which the hoop is released using PE = m * g * h, where g is the acceleration due to gravity.
To have the same total energy as the sphere, we need the kinetic energy gained by the hoop plus its extra potential energy to be equal to the kinetic energy of the sphere. In equation form, it looks like this:
1/2 * m * v^2 + m * g * h = 1/2 * m * (R * ω)^2
Now, here's the grand finale! Since the sphere is rolling without slipping, we can relate the linear speed (v) and the angular speed (ω) using v = R * ω. By substituting this in the equation, we get:
1/2 * m * v^2 + m * g * h = 1/2 * m * (v/R)^2
Simplifying it a bit further:
1/2 * m * v^2 + m * g * h = 1/2 * m * v^2 / R^2
We can now cancel out the mass and one half from both sides, giving us:
v^2 + 2 * g * h = v^2 / R^2
Finally, solving for h, we get:
h = (v^2 * (R^2 - 1))/(2 * g * R^2)
And there you have it! The height, h, from which the hoop should be released on the same slope to equal the sphere's speed at the bottom. Now go out there and release that hoop with style!
To determine the height from which the circular hoop of radius R should be released in order to achieve the same speed as the solid sphere at the bottom, we can use the principle of conservation of energy.
Let's assume:
- Mass of the sphere = M
- Mass of the hoop = M (since both have the same radius R)
- Height of the sphere's release point = H
Steps to find the height from which the hoop should be released:
1. Calculate the gravitational potential energy (PE) of the sphere at the top of the slope:
PE_sphere = M * g * H
(where g is the acceleration due to gravity)
2. Calculate the gravitational potential energy (PE) of the hoop at the top of the slope:
PE_hoop = M * g * h
(where h is the height from which the hoop is released)
3. Calculate the kinetic energy (KE) of the sphere at the bottom of the slope:
KE_sphere = (1/2) * I * w^2
(where I is the moment of inertia of the sphere and w is the angular velocity)
4. Calculate the kinetic energy (KE) of the hoop at the bottom of the slope:
KE_hoop = (1/2) * I * w^2
(where I is the moment of inertia of the hoop and w is the angular velocity)
Since the sphere rolls without slipping, we can relate the angular velocity (w) to the linear velocity (v) using the equation: v = R * w
5. Using the equation for the kinetic energy, we can express the linear velocity (v) in terms of the angular velocity (w):
KE_sphere = (1/2) * I * (v / R)^2
KE_hoop = (1/2) * I * (v')^2
6. As both the sphere and hoop have the same radius R, we can equate the kinetic energies:
(1/2) * I_sphere * (v / R)^2 = (1/2) * I_hoop * (v')^2
7. We know that the moment of inertia (I) for a solid sphere and a circular hoop are given as:
I_sphere = (2/5) * M * R^2
I_hoop = M * R^2
8. Simplifying and solving the equation, we get:
(2/5) * (v / R)^2 = (v')^2
9. Since both the sphere and hoop have the same linear velocity (v) at the bottom, we can equate their linear velocities:
v = v'
10. Substituting v = v' in the previous equation, we get:
(2/5) * (v / R)^2 = v^2
11. Solving for v, we get:
v = √(5/2) * R * g
12. Now, we can equate the gravitational potential energy (PE) to the kinetic energy (KE) at the bottom:
M * g * H = (1/2) * I * (v / R)^2
13. Substitute the values of I and v from previous calculations:
M * g * H = (1/2) * (2/5 * M * R^2) * [(√(5/2) * R * g) / R]^2
14. Simplifying the equation:
H = (√(5/2) * R * g * R) / (2 * g)
15. Canceling out g terms:
H = (√(5/2) * R^2) / 2
Therefore, the hoop should be released from a height of (√(5/2) * R^2) / 2 on the same slope to achieve the same speed as the solid sphere at the bottom.
To determine the height from which the circular hoop should be released in order to achieve the same speed as the sphere at the bottom of the slope, we need to analyze the energy conservation in both cases.
Let's start by calculating the speed of the sphere at the bottom of the slope. Since the sphere rolls without slipping, the rotational kinetic energy and the translational kinetic energy can be equated.
1. Gravitational potential energy of the sphere at the top:
The gravitational potential energy of the sphere is given by mgh, where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height.
The height is given as 32 cm, which is equivalent to 0.32 m.
2. Rotational kinetic energy of the sphere at the bottom:
The rotational kinetic energy of a sphere is given by (2/5) * (1/2) * m * R^2 * (v/R)^2, where m is the mass of the sphere and v is its linear speed at the bottom.
The (1/2) factor is due to the sphere rolling without slipping, and (v/R) represents the angular speed of the sphere.
3. Translational kinetic energy of the sphere at the bottom:
The translational kinetic energy of a moving object is given by (1/2) * m * v^2, where m is the mass of the sphere and v is its linear speed.
Since the rotational and translational kinetic energies are equal for a rolling sphere, we can equate the rotational and translational kinetic energies to determine v, the speed of the sphere at the bottom.
(2/5) * (1/2) * m * R^2 * (v/R)^2 = (1/2) * m * v^2
Simplifying the equation, we find:
(2/5) * (v^2) = v^2
v^2 = (5/2) * v^2
v^2 = 2.5 * v^2
v = sqrt(2.5) * v
Therefore, the speed of the sphere at the bottom of the slope is v.
Now, let's consider the circular hoop. Since the hoop does not roll (only slides) without slipping, the kinetic energy will be entirely translational.
4. Translational kinetic energy of the hoop at the bottom:
The translational kinetic energy of the hoop is given by (1/2) * m * v_h^2, where m is the mass of the hoop and v_h is its linear speed at the bottom.
Since the hoop slides without rolling, v_h is equal to v, the speed of the sphere.
To achieve the same speed at the bottom, the translational kinetic energy of the hoop should be the same as the total kinetic energy of the sphere:
(1/2) * m * v_h^2 = (1/2) * m * v^2
Since v_h = v, the above equation simplifies to:
(1/2) * m * v^2 = (1/2) * m * v^2
This means that the speed of the hoop should be the same as the speed of the sphere at the bottom. Now, we need to determine the height from which the hoop should be released.
5. Gravitational potential energy of the hoop at the top:
The gravitational potential energy of the hoop is given by mgh_h, where m is the mass of the hoop, g is the acceleration due to gravity, and h_h is the height from which the hoop is released.
To find the height from which the hoop should be released, we equate the gravitational potential energy of the hoop to the gravitational potential energy of the sphere at the top:
mgh_h = mgh
Since the masses and the acceleration due to gravity are the same for both objects, we can cancel them out:
h_h = h
Therefore, the height from which the hoop should be released on the same slope to achieve the same speed as the sphere at the bottom is also 32 cm (0.32 m).
So, the circular hoop should be released from a height of 0.32 m to have the same speed as the sphere at the bottom of the slope.
Figure the rolling and translational Ke for the sphere at the bottom, from knowing that sum of those is equal to the PE at the beginning. You should then be able to determine speed of the sphere. Then, do the same speed (and angular momentum) for the hoop (recalling the only thing different is momentofInertia), set that equal to mgh, and you have the new height.
Nice problem. Watch algebra.