Question

. An electric sander consisting of a rotating disk of mass 0.7 kg and radius 10 cm rotates at 15 rev/s. When applied to a rough wooden wall the rotation rate decreases by 20%. (a) What is the final rotational kinetic energy of the rotating disk

Answer 1

The final rotational kinetic energy of the rotating disk is 9.95 J

Explanation:

Given data,

mass 0.7 kg radius 10 cm rotates at the speed of 15 rev/sec

We have the formula,

Kf= 1/2 I ω f²

I=1/2 Mr²2

ωf= (0.8) ωo

substitute in the formula we get

Kf= 1/2 (1/2 MR²2) (0.8) ² ( ωo)²

=(0.16) M R²2 ωo²2

=(0.16)(0.7)(0.10)²(15[2π)] ²

Kf=9.95 J

The final rotational kinetic energy of the rotating disk is 9.95 J