Question

A uniformly dense solid disk with a mass of 4 kg and a radius of 2 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 20 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 5 s

Answer 1

Answer:

$\theta=12.5\ rad$

Explanation:

Given:

• mass of disk, $m=4\ kg$

• radius of the disk, $r=2\ m$

• rate of increase in kinetic energy, $\dot {KE}=20\ J.s^{-1}$

• time of observation, $t=5\ s$

Now the moment of inertia of the disk:

$I=\frac{1}{2} m.r^2$

$I=\frac{1}{2}\times 4\times 2^2$

$I=8\ kg.m^2$

We know that Kinetic energy:

$KE=\frac{1}{2}\times I.\omega^2$

$\dot {KE}\times t=\frac{1}{2} \times 8\times \omega^2$

$20\times 5=0.5\times 8\times \omega^2$

$\omega =5\ rad.s^{-1}$ is the angular velocity after 5 seconds.

• Since the disk starts from the rest, its initial angular speed, $\omega_i=0\ rad.s^{-1}$

Using equation of motion:

$\omega=\omega_i+\alpha.t$

$5=0+\alpha\times 5$

$\alpha=1\ rad.s^{-2}$

Now the angle rotated in the give time:

$\theta=\omega_i.t+\frac{1}{2} .\alpha.t^2$

$\theta=0+0.5\times 1\times 5^2$

$\theta=12.5\ rad$

Answer 2

Explanation:

Formula to calculate energy change is as follows.

                        dE = $\tau \times \theta$

                 $\tau = \frac{dE}{\theta}$

               $\frac{dE}{\theta}$ = 20 J/rad

It is given that mass is 4 kg, and radius is 2 m. Hence, we will calculate the moment of inertia as follows.

                     I = $\frac{1}{2}mv^{2}$

                       = $\frac{1}{2}\times 4 kg \times (2)^{2}$

                        = 8 $kg m^{2}$

Also,       $\tau = I \times \alpha$

                20 = $8 \times \alpha$

or,              $\alpha = \frac{20}{8}$

                             = 2.5 $rad/s^{2}$

$w_{o}$ = 0 (rest),            t = 4 sec

Also,     $\theta = w_{o}t + \frac{1}{2} \alpha t^{2}$

                          = $0 + \frac{1}{2} \times \frac{5}{2} \times (5)^{2}$

                          = 15.625 rad

or,                      = $\frac{15.625}{2\pi}$ rev

                         = 2.48 rev

Therefore, we can conclude that angular displacement is 2.48 rev.