# Rotation/Energy

#### zawhein

An electric grindstone weighing 2kg and has a diameter of 20cm rotates counterclockwise
at 30 rpm.When power is turned off, it takes 60 seconds before the grindstone stops.

What will be grindstone its angular acceleration before it stops (assume constant
angular acceleration). positive direction - against the clock
.
How many turns will the grind take before it stops?

What is the momentum of tightness to the grindstone (massive cylinder)?

How much kinetic energy is stored in the grindstone in the moment it
turns off?

If we calculate this energy into gravitational energy, how high above the groundmust the grindstone shall be placed?

#### skeeter

An electric grindstone weighing 2kg and has a diameter of 20cm rotates counterclockwise
at 30 rpm.When power is turned off, it takes 60 seconds before the grindstone stops.

What will be grindstone its angular acceleration before it stops (assume constant
angular acceleration). positive direction - against the clock
$\alpha = \dfrac{\Delta \omega}{\Delta t} = \dfrac{0 - 30 \, rpm}{1 \, min} = -30 \, rev/min^2 = -\dfrac{\pi}{60} \, rad/sec^2$

How many turns will the grind take before it stops?
$\Delta \theta = \bar{\omega} \cdot \Delta t = \dfrac{(30 \, rpm + 0)}{2} \cdot (1 \, min) = 15 \, rev$

What is the momentum of tightness to the grindstone (massive cylinder)?
I've never heard the term "momentum of tightness" ... contextual clues make me think you mean "moment of inertia". If so,

$I = \dfrac{1}{2}MR^2$

How much kinetic energy is stored in the grindstone in the moment it
turns off?
$K_{rot} = \dfrac{1}{2}I \omega^2$

If we calculate this energy into gravitational energy, how high above the groundmust the grindstone shall be placed?
I'm guessing this means $K_{rot} = U_{grav} \implies \dfrac{1}{2}I \omega^2 = Mgh \implies h = \dfrac{I \omega^2}{Mg}$

• 1 person

Thanks