for your question. It looks like you may be using the wrong formula for this problem. The formula you are using, v = √(2gh), is used to calculate the final velocity of an object that has been dropped from a certain height (h) and is falling straight down due to gravity. However, in this problem, the hoop is not falling straight down, it is rolling down an inclined plane. This means that it has both translational and rotational motion, and therefore, the formula for its final velocity will be different.
To solve this problem, we can use the conservation of energy principle. The initial potential energy (mgh) of the hoop at the top of the inclined plane is converted into both translational and rotational kinetic energy as it rolls down. The equation for this is:
mgh = (1/2)mv^2 + (1/2)Iω^2
Where m is the mass of the hoop, h is the vertical distance it is dropped, v is the final velocity, I is the moment of inertia (for a hoop, it is equal to mR^2 where R is the radius), and ω is the angular velocity.
Since the hoop is released from rest, its initial velocity (v0) and initial angular velocity (ω0) are both zero. Therefore, the equation becomes:
mgh = (1/2)mv^2 + (1/2)mR^2ω^2
Rearranging for v, we get:
v = √(2gh - R^2ω^2)
To find the final angular velocity, we can use the formula ω = v/R, since the hoop is rolling without slipping. Substituting this into the equation above, we get:
v = √(2gh - v^2)
Solving for v, we get v = √(4gh/3) = 7.7m/s. This is the correct answer.
I hope this helps clarify the problem and the correct formula to use. Good luck with your online test!
