Question

Starting from rest, a hoop of 15 cm radius rolls down a hill to a point 10.0 m
below the starting point. How fast is the hoop rotating at this point? ​

Answer 1

Hoop is rotating with an angular velocity of 66.67 rad/s

Explanation:

If the mass of hoop is m and radius r then Moment of inertia (I) of hoop

$I=mr^2$

When the hoop is at the top there will be potential energy

As the hoop rolls down the hill the potential energy will get converted into linear and rotational kinetic energy

if the linear velocity of the hoop is v and angular velocity ω

Then

$mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$\implies mgh=\frac{1}{2}(mv^2+mr^2(\frac{v}{r})^2$

$\implies mgh=\frac{1}{2}\times 2mv^2$

$\implies gh=v^2$

$\implies v=\sqrt{gh}$

$\implies v=\sqrt{10\times 10}$      (Taking g = 10 m/s²)

$\implies v=10$ m/s

Therefore, angular velocity of the hoop

$\omega=\frac{v}{r}$

$\implies \omega=\frac{10}{0.15}$

$\implies\omega=66.67$ rad/s

Hope this answer is helpful.

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