The algebraic expression for the ice cube's speed at any angle θ is: v(θ) = v_b * sqrt(1 - 2 * (R * cos(θ)) / R) .
The ice cube's speed at angle θ can be found using conservation of energy. Here's how:
1. Define variables:
m: mass of the ice cube (assumed constant)
g: acceleration due to gravity
R: radius of the circle (half of the diameter)
v_b: speed of the ice cube at the bottom (3.0 m/s)
v(θ): speed of the ice cube at angle θ
h: height of the ice cube above the bottom at angle θ
2. Apply conservation of energy:
At the bottom (θ = 0°), the ice cube has maximum kinetic energy (1/2 mv_b^2) and minimum potential energy (0).
At any other angle (θ), the ice cube has kinetic energy (1/2 mv(θ)^2) and potential energy (mgh).
These two energies must be equal due to conservation.
Therefore:
1/2 mv_b^2 = 1/2 mv(θ)^2 + mgh
3. Express height (h) in terms of angle (θ):
The height h can be related to the angle θ using trigonometry.
Consider a right triangle formed by the radius R, the height h, and the portion of the circle subtended by angle θ.
We know R and can calculate h using:
h = R - R * cos(θ)
4. Substitute and solve for v(θ):
Substituting the expression for h into the conservation of energy equation and rearranging for v(θ), we get:
v(θ) = v_b * sqrt(1 - 2 * (R * cos(θ)) / R)
5. Verify for θ = 0° and θ = 180°:
For θ = 0° (bottom), the expression simplifies to v(0°) = v_b * sqrt(1 - 0) = v_b, which is 3.0 m/s as expected.
For θ = 180° (top), the expression simplifies to v(180°) = v_b * sqrt(1 - 4) = 0, which means the ice cube loses all its kinetic energy and stops at the top.
Therefore, the algebraic expression for the ice cube's speed at any angle θ is: v(θ) = v_b * sqrt(1 - 2 * (R * cos(θ)) / R)
This expression satisfies the conditions of having v_b at θ = 0° and 0 at θ = 180°.