In a circus performance, a large 3.0 kg hoop with a radius of 1.3 m rolls without slipping. If the hoop is given an angular speed of 6.8 rad/s while rolling on the horizontal and is allowed to roll up a ramp inclined at 24◦ with the horizontal, how far (measured along the incline) does the hoop roll? The acceleration of gravity is 9.81 m/s2 .

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-07-15T08:25:12+02:00

Answer:

The distance is 19.58 m.

Explanation:

Given that,

Mass = 3.0 kg

Radius = 1.3 m

Angular speed = 6.8 rad/s

Angle = 24°

Acceleration of gravity = 9.81 m/s²

We need to calculate the distance

Using formula of kinetic energy

Total Initial kinetic energy,

[tex]K.E = \dfrac{1}{2}(mv^2+I\omega^2)[/tex]

[tex]K.E = \dfrac{1}{2}(mv^2+mr^2\omega^2)[/tex]

[tex]K.E =\dfrac{(mr^2+mr^2)\omega^2}{2}[/tex]

[tex]K.E =mr^2\omega^2[/tex]....(I)

Now, Total potential energy

[tex]P.E=mgs\sin\theta[/tex]

[tex]P.E=mgs\sin24^{\circ}[/tex]....(II)

Equating equation (I) and (II)

[tex]mr^2\omega^2==mgs\sin24^{\circ}[/tex]

[tex](1.3)^2\times6.8^2=9.81\times s\times\sin24^{\circ}[/tex]

[tex]s = \dfrac{1.3^2\times6.8^2}{9.81\times\sin24^{\circ}}[/tex]

[tex]s=19.58\ m[/tex]

Hence, The distance is 19.58 m.