Question

You need to design an industrial turntable that is 45.0cmà in diameter and has a kinetic energy of 0.270 Jà when turning at 50.0 rev/min.

A)Ã What must be the moment of inertia of the turntable about the rotation axis?(kg*m^2)
B)If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?(kg)

Answer 1

a) The moment of inertia (l) of the turntable about the rotation axis is approximately
`\(0.276 \, \text{kg m}^2\)`.

b) If the workshop makes this turntable in the shape of a uniform solid disk, its mass (m) should be approximately
`\(10.89 \, \text{kg}\)`.

a. Given:

Kinetic energy (KE) = 0.270 J

Angular velocity
`(\(\omega\)) = \(50.0\) rev/min`

First, convert the angular velocity from rev/min to rad/s:

`\[\omega = \frac{2\pi * \text{revolutions}}{60 \, \text{s}} * \text{rev/min}\]`

`\[\omega = (2\pi * 50.0)/(60) \, \text{rad/s}\]`

`\[\omega = (5\pi)/(6) \, \text{rad/s}\]`

Now, calculate the moment of inertia (I) using the formula:

`\[ I = (2KE)/(\omega^2)\]`

`\[ I = \frac{2 * 0.270 \, \text{J}}{\left((5\pi)/(6)\right)^2}\]`

`\[ I = \frac{2 * 0.270 \, \text{J}}{\left((25\pi^2)/(36)\right)}\]`

`\[ I = (0.540)/(25\pi^2/36)\]`

`\[ I \approx 0.276 \, \text{kg m}^2`

This is the moment of inertia (I) of the turntable.

Now, for the mass (m) of the uniform solid disk:

b. Given:

Diameter (d) = 45.0 cm

Radius (r)
`= \(d/2 = 22.5\) cm = \(0.225\) m`

The formula for the moment of inertia (I) of a uniform solid disk is

Rearrange the formula to solve for the mass (\(m\)):
`\(I = (1)/(2) m r^2\).`

`\[ m = (2I)/(r^2)\]`

`\[ m = \frac{2 * 0.276 \, \text{kg m}^2}{(0.225 \, \text{m})^2}\]`

`\[ m = (0.552)/(0.050625) \, \text{kg}\]`

`\[ m \approx 10.89 \, \text{kg}\]`

Therefore, the mass of the uniform solid disk that forms the turntable would be approximately 10.89 kg.

Answer 2

(a) The moment of inertia of the turntable about the rotation axis is 0.02 kgm².

(b) If the workshop makes this turntable in the shape of a uniform solid, the mass of the disk is 0.79 kg.

How to calculate the moment of inertia of the turn table?

(a) The moment of inertia of the turntable about the rotation axis is calculated as follows;

K.E = ¹/₂Iω²

where;

• I is the moment of inertia
• ω is the angular velocity

I = 2K.E / ω²

The angular speed, ω = 50 rpm

= 50 rev/min x 2π/rev x 1 min / 60 s

= 5.24 rad/s

I = (2 x 0.27 ) / (5.24²)

I = 0.02 kgm²

(b) If the workshop makes this turntable in the shape of a uniform solid, the mass of the disk is calculated as;

I = ¹/₂MR²

where;

• R is the radius of solid disk = 45 cm / 2 = 22.5 cm
• M is the mass

M = 2I / R²

M = ( 2 x 0.02 ) / ( 0.225²)

M = 0.79 kg