Solve Velocity Problem: Dropped from 3m, .5m Radius, .2kg Mass

[titusdna]

A hoop with a radius of .5m and mass of .2kg is released from rest and rolls down inclined plane. How fast is it moving after dropping from vertical distance of 3m? First I tried v=Iw/md but I had no info for angular velocity so I tried calculating using v=sq root 2xgxh and get 7.7m/s but my online test is saying it is wrong. Am I using the wrong formula?
Thanks

 

[OlderDan]

A hoop with a radius of .5m and mass of .2kg is released from rest and rolls down inclined plane. How fast is it moving after dropping from vertical distance of 3m? First I tried v=Iw/md but I had no info for angular velocity so I tried calculating using v=sq root 2xgxh and get 7.7m/s but my online test is saying it is wrong. Am I using the wrong formula?
Thanks

Energy will be conserved. You seem to have some idea what moment of inertia and angular velocity are, so you are not far away from the result. Kinetic energy can be broken into the kinetic energy of transaltion and the kinetic energy of rotation. The first depends on linear velocity of the center of mass. The second depends on angular velocity about the center of mass. Firgue out the connection between those velocities assuming the hoop is NOT SLIPPING.

 

[thepatientmental]

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!