Answer :
Answer:
a) ω = 8.75 rad/s
b) I = 12.9 kgm^2
Explanation:
Given data:
Radius of the wheel r = 0.400 m
Speed of the suitcase v = 3.50 m/s
Height h = 4.00 m
Mass of the suitcase m = 15 kg
a) To determine angular velocity of the wheel when the suitcase reaches the ground is
we know that , ω = v / r
ω = (3.50 m/s) / (0.400 m )
ω = 8.75 rad/s
b) From the conservation of energy
we have KEi + Ui = KEf + Uf
where KEi , KEf are initial and final kinetic energies and Ui and Uf are initial and final potentail energies.
We know that KEi = 0 and Uf = 0
Ui = KEf
mgh = (1/2)mv^2 + (1/2) Iω^2
I = moment of inertia of wheel to determined
15×9.8×4 = (1/2) (15)(3.5)^2 + (1/2) I (8.75)^2
Calculating we get
I = 12.9 kgm^2
A force that acts on a body moving in a circular path and is directed toward the center around which the body is moving.
According to the question, The formula is used to find the omega is:-
[tex]w =\frac{v}{r}[/tex]
[tex]w= \frac{3.50}{0.400}[/tex]
Hence, the omega is ω = 8.75 rad/s
From the conservation of energy we have KEi + Ui = KEf + Uf
where Kei, KEf are initial and final kinetic energies and Ui and Uf are initial and final potential energies.
We know that KEi = 0 and Uf = 0
So, therefore the Ui = KEf
Hence [tex]mgh = \frac{1}{2}mv^{2} + \frac{1}{2} Iw^{2}[/tex]
After putting the value:-
[tex]15*9.8*4 = \frac{1}{2}*15*3.5^{2} + \frac{1}{2}*I*8.75^{2}[/tex]
Hence the value of I is[tex]12.9 kgm^2[/tex]
For more information, refer to the link:-
https://brainly.com/question/21338592