Respuesta :
Answer:
The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
Explanation:
Given that,
Mass = 2 kg
Radius = 0.5 m
Angular speed = 3 rad/s
Force = 10 N
(I). We need to calculate the rotational kinetic energy
Using formula of kinetic energy
[tex]K.E =\dfrac{1}{2}\timesI\omega^2[/tex]
[tex]K.E=\dfrac{1}{2}\times mr^2\times\omega^2[/tex]
[tex]K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2[/tex]
[tex]K.E=2.25\ J[/tex]
(II). We need to calculate the instantaneous change rate of the kinetic energy
Using formula of kinetic energy
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
On differentiating
[tex]\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}[/tex]
[tex]\dfrac{K.T}{dt}=mva[/tex]....(I)
Using newton's second law
[tex]F = ma[/tex]
[tex]a= \dfrac{F}{m}[/tex]
[tex]a=\dfrac{10}{2}[/tex]
[tex]a=5 m/s^2[/tex]
Put the value of a in equation (I)
[tex]\dfrac{K.E}{dt}=mva[/tex]
[tex]\dfrac{K.E}{dt}=mr\omega a[/tex]
[tex]\dfrac{K.E}{dt}=2\times0.5\times3\times5[/tex]
[tex]\dfrac{K.E}{dt}=15\ J/s[/tex]
Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
