Question

At time t = 0 s, a wheel has an angular displacement of zero radians and an angular velocity of +26 rad/s. The wheel has a constant acceleration of -0.43 rad/s2. In this situation, the time t (after t = 0 s), at which the kinetic energy of the wheel is twice the initial value, is closest to...
150s How do you solve for this?? I keep getting a negative time value

Answer 1

Answer:

t= 25.04 s

Explanation:

Given  that

Initial angular speed ,ωi= 26 rad/s

Angular acceleration ,α=0.43 rad/s²

We know that kinetic energy of the wheel given as

$KE=\dfrac{1}{2}I\omega^2$

I=Mass monent of inertia

ω=Angular speed

If kinetic energy is two time then the final angular speed

$\omega_f=\sqrt{2}\ \omega_i$

We know that

$\omega_f = \omega_i + \alpha\ t$

$\sqrt{2}\ \omega_i= \omega_i +0.43\times t$

t= 25.04 s

After 25.04 s then kinetic energy will become two times of the initial kinetic energy.