Question

A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is 5.35 rad/s at the bottom, what is the height of the inclined plane?

Answer 1

To solve this problem we will use the concepts related to energy conservation. Both potential energy, such as rotational and linear kinetic energy, must be conserved, and the gain in kinetic energy must be proportional to the loss in potential energy and vice versa. This is mathematically

$PE = KE_{lineal} + KE_{rotational}$

$mgh = \frac{1}{2}mv^2 +\frac{1}{2} I\omega^2$

Where,

m = mass

v = Tangential Velocity

$\omega$ = Angular velocity

I = Moment of Inertia

g = Gravity

Replacing the value of Inertia in a Disk and rearranging to find h, we have

$mgh = \frac{1}{2}mv^2 +\frac{1}{2} I\omega^2$

$mgh = \frac{1}{2}mr^2\omega^2 + \frac{1}{2}(\frac{1}{2}mr^2)\omega^2 )$

$h = \frac{3}{4} \frac{r^2\omega^2}{g}$

Replacing,

$h = \frac{3}{4} \frac{(1.6)^2 (5.35)^2}{9.8}$

$h = 5.607m$

Therefore the height of the inclined plane is 5.6m

Answer 2

The  height of the inclined plane when the uniform solid disk of radius 1.60 m and mass 2.30 kg rolls  should be 5.607 m.

Calculation of the height:

Since we know that

$mgh = 1/2mv^2 + 1/2 Iw^2$

Here

m = mass

v = Tangential Velocity

w = Angular velocity

I = Moment of Inertia

g = Gravity

Now

$mgh = 1/2mr^2w^2 + 1/2(1/2mr^2)w^2\\\\h = 3/4r^2w^2/g$

So,

$= 2/4(1.6)^2(5.35)^2 / 9.8$

= 5.607 m

hence, The  height of the inclined plane when the uniform solid disk of radius 1.60 m and mass 2.30 kg rolls  should be 5.607 m.

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