(a) A uniform disk of mass 21 kg, thickness 0.5 m, and radius 0.6 m is located at the origin, oriented with its axis along the axis. It rotates clockwise around its axis when viewed from above (that is, you stand at a point on the axis and look toward the origin at the disk). The disk makes one complete rotation every 0.7 s. What is the rotational angular momentum of the disk? What is the rotational kinetic energy of the disk? (Express your answer for rotational angular momentum in vector form.)

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24-07-2019
Physics

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(a) A uniform disk of mass 21 kg, thickness 0.5 m, and radius 0.6 m is located at the origin, oriented with its axis along the axis. It rotates clockwise around its axis when viewed from above (that is, you stand at a point on the axis and look toward the origin at the disk). The disk makes one complete rotation every 0.7 s. What is the rotational angular momentum of the disk? What is the rotational kinetic energy of the disk? (Express your answer for rotational angular momentum in vector form.)

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nuuk nuuk · Answer 1 · 2019-07-29T10:20:56+02:00

Answer:L=33.93

R.E.=152.30

Explanation:

Given

mass of disk[tex]\left ( m\right )[/tex]=21kg

thickness[tex]\left ( t\right )[/tex]=0.5m

radius[tex]\left ( r\right )[/tex]=0.6m

t=0.7 sec for every complete rotation

therefore angular velocity[tex]\left ( \omega \right ) =\frac{\Delta \theta }{\Delta t}[/tex]

[tex]\left ( \omega \right )=\frac{ 2\pi }{0.7}=8.977[/tex] rad/s

Rotational angular momentum is given by

[tex]L=I\omega [/tex]

[tex]I[/tex]=[tex]\frac{mr^2}{2}[/tex]

I=3.78 [tex]kg-m^2[/tex]

[tex]L=3.78\times 8.977[/tex]

L=33.93 [tex]\hat{j}[/tex]

considering disk is rotating in z-x plane

Rotational kinetic energy=[tex]\frac{I\omega ^{2}}{2}[/tex]=152.30 J

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-01T13:19:31+02:00

The rotational angular momentum of the disk and the rotational kinetic energy of the disk is mathematically given as

L=33.93kg-m2/sec

R.K.E=152.30 J

What is the rotational angular momentum of the disk and the rotational kinetic energy of the disk ?

Question Parameter(s):

A uniform disk of mass 21 kg,

thickness 0.5 m,

and radius 0.6 m

Generally, the equation for the angular velocity is mathematically given as

w =d∅ /dt

Therefore

w=2\pi/0.7

w=8.977

For the Rotational angular momentum

I=mr^2/2

I=3.78 kg-m^2

Hence

L=3.78*8.977

L=33.93kg-m2/sec

In conclusion, when disk is rotating in z-x plane

R K.E=w^2/2

R.K.E=152.30 J