Want to verify textbook ans on angular velocity

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SUMMARY

The discussion centers on calculating the distance a 5kg hoop rolls up a 20-degree incline after being given an angular speed of 3 rad/s. The user initially calculated the distance using only rotational kinetic energy, resulting in a distance of 12.1m. However, the correct approach requires including both rotational and translational kinetic energy, leading to a textbook answer of 24m. The oversight was in neglecting the translational kinetic energy in the energy conservation equation.

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I got the following qn:

In a circus performance, a large 5kg hoop of radius 3m rolls without slipping. If the hoop is given an angular speed of 3 rad/s while rolling on the horizontal and allowed to roll up a ramp inclined at 20 degrees with the horizontal, how far(measured along the incline) does the hoop roll?

My solution is as follows:
Total initial energy=0.5 * I * (angular velocity)^2
=0.5 * 45 * 3^2
=202.5
Final energy = mgh
= 5gh, where h is the vertical height
Equating the 2,
202.5=5gh
h=4.128.

Since we want to find the distance along the slope, d, d=h/sin20 = 12.1m

This is the ans I got, but it seems that the back of the textbook ans is 24m . Is the textbook ans correct or wrong?
 
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You included the rotational kinetic energy but forgot the translational kinetic energy.
 
Thanks! How could i have forgotten that?
 

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