Question

A Frisbee (160 g, 25 cm in diameter) spins a rate of 300 rpm with its center balanced on a fingertip. What is the rotational kinetic energy of the Frisbee if the disc has 70% of its mass on the outer edge (basically a thin ring 25-cm in diameter) and the remaining 30% is a nearly flat disk 25-cm in diameterthis is word for word what the question says.the final answer is 1.05 J ; 40.1cm -> i dont know how they came to that answer, and why there is an x value in cmAlthough I provided the answer, please show me how to come up with it

Answer 1

To solve this exercise it is necessary to take into account the concepts related to Moment of Inertia and Kinetic Energy.

The rotational kinetic energy is given by the function

$KE_r = \frac{1}{2} I\omega^2$

Where,

I = Inertia moment

$\omega =$Angular velocity

The moment of inertia must be divided by the mass fractions located outside and inside.

The general formula for the disk is,

$I = mR^2$

However the Moment of Inertia Total must be written as a sum between the Inertia outside and Inside,

$I_t = I_o+I_i$

$I_t = (0.70m R^2) + (\frac{0.30m R^2}{2})$

$I_t = 0.70 mR^2 + 0.15 mR^2$

$I = 0.85 MR^2$

Therefore replacing with our values we have that for a mass of 160g and radius of 25/2cm

$I = 0.85 * 0.160 * 0.125^2$

$I = 2.125*10^{-3}kgm^2$

At the same time we calculate the angular velocity given as

$\omega = 300 rpm = 300 * \frac{2pi}{60}$

$\omega = 31.42 rad/s$

Applying the equation for Rotational Kinetic Energy we can calculate the total value given as,

$KE_r = \frac{1}{2} I \omega^2$

$KE_r = \frac{1}{2}(2.125*10^{-3}) (31.42^2)$

$KE_r = 1.0489J$

The problem also say that the rotational kinetic energy is converted to gravitational potential energy with an efficiency of 60%, then

$PE = 60\% KE_r$

$mgh = 0.6*1.0489J$

Solving for h,

$h= \frac{0.6*1.0489}{0.160 *9.8}$

$h = 0.4013m \approx 40cm$