Question

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 .

Answer 1

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,

$K.E = \dfrac{1}{2}(mv^2+I\omega^2)$

$K.E = \dfrac{1}{2}(mv^2+mr^2\omega^2)$

$K.E =\dfrac{(mr^2+mr^2)\omega^2}{2}$

$K.E =mr^2\omega^2$....(I)

Now, Total potential energy

$P.E=mgs\sin\theta$

$P.E=mgs\sin24^{\circ}$....(II)

Equating equation (I) and (II)

$mr^2\omega^2==mgs\sin24^{\circ}$

$(1.3)^2\times6.8^2=9.81\times s\times\sin24^{\circ}$

$s = \dfrac{1.3^2\times6.8^2}{9.81\times\sin24^{\circ}}$

$s=19.58\ m$

Hence, The distance is 19.58 m.