Angular speed of wrapper string

[seroth]

Homework Statement

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg . The free end of the string is held in place and the hoop is released from rest (the figure ). After the hoop has descended 60.0 cm, calculate angular speed and speed at the center

Homework Equations

w= omega

w = w+ at
K = 1/2 I w^2 ?

The Attempt at a Solution

I'm really unsure how to approach this problem and don't exactly know where to start

 

[Dick]

Use energy conservation. Add linear KE=(1/2)*m*v^2 and gravitational potential energy PE=mgh to your list of equations.

 

[Moetasim]

. However, based on the given information, we can use the equation w= v/r to find the angular speed of the hoop. Since the hoop is released from rest, its initial angular speed is 0. We can then use the equation w = w0 + at to find the final angular speed after the hoop has descended 60.0 cm.

To find the speed at the center of the hoop, we can use the equation v = rw, where r is the radius of the hoop. Plugging in the values, we can calculate the speed at the center of the hoop after it has descended 60.0 cm.

To determine the kinetic energy of the hoop, we can use the equation K = 1/2 I w^2, where I is the moment of inertia of the hoop. We can calculate the moment of inertia using the formula for a hoop, I = mr^2. Plugging in the values, we can find the kinetic energy of the hoop at the given point.

In summary, to find the angular speed and speed at the center of the hoop, we can use the equations w= v/r and v = rw respectively. To determine the kinetic energy, we can use the equation K = 1/2 I w^2, where I = mr^2 for a hoop.