Question

Ch9 Review Assignment Exercise 9.36 < A wheel is turning about an axis through its center with constant anqular acceleration. Starting from rest, at t 0, the wheel turns through 8.50 revolutions in t = 11.5 s . At 11.5 s the kinetic energy of the wheel is 39,0 J Part A For an axis through its center, what is the moment of inertia of the wheel? Express your answer with the appropriate units. Value Units I = Request Answer Submit Provide Feedback

Answer 1

Answer is 0.905 Kg.m²

calculation:::

Let's we know the formula ,

$2\pi .(revolution)=\frac{1}{2}*a*t^{2}..............(1)$

t= time=11.5 sec

a=acceleration

revolution =The revolution is a unit of measurement of angle=8.5 unit

put the all value equation number 1 for finding the acceleration

  $2\pi .(8.5)=\frac{1}{2}*a*11.5^{2}$

$a=\frac{(2\pi )(2)(8.5)}{(11.5)^{2}}$

according to question ,kinetic energy=KE

and KE=39.0 J

and find the moment of inertia = I

and we know  

  $KE = {\frac{1}{2}}*I*\omega ^{2}.............................(2)$

and $\omega ^{2}$ =2.a.d

=2* 0.807* $2\pi$ * 8.5

=86.16

again from equation number (2)

  $I=\frac{2KE}{\omega ^{2}}$

$I=\frac{2*39.0}{86.16} Kg.m^{2}$

the answer is 0.905 kg.m²