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1. The problem statement, all variables and given/known data
In Figure 1130(see attatched), a constant horizontal force F_app of magnitude 12 N is applied to a wheel of mass 8 kg and radius 0.70 m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 0.75 m/s2.
(a) In unitvector notation, what is the frictional force on the wheel?
[? Ni]
(b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?
[3.92 kg*m^2 ]
2. Relevant equations
[tex]\tau=r\times F[/tex]
[tex]\tau=I\alpha[/tex]
[tex]I_{hoop}=mr^2[/tex]
[tex]\alpha=\frac{a_{com}}{r}[/tex]
3. The attempt at a solution
I took the axis of rotation to be perpindicular to the wheel's center of mass.
There are four forces acting on the wheel: mg, normal, applied force, and friction. Of those four forces, only friction does not pass through the axis of rotation, so friction alone (symbol is lowercase f) contributes to torque.
[tex]\tau=r\times f=I\alpha[/tex]
[tex]fr=(mr^2)(\frac{a_{com}}{r})[/tex]
[tex]f=ma_{com}(\frac{r^2}{r^2})[/tex]
[tex]f=(8kg)(.75m/s^2)=6N[/tex]
BZZZZZ. I lose.
Though on the bright side, my solution for the rotational inertia earned me a pretty green check mark...
In Figure 1130(see attatched), a constant horizontal force F_app of magnitude 12 N is applied to a wheel of mass 8 kg and radius 0.70 m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 0.75 m/s2.
(a) In unitvector notation, what is the frictional force on the wheel?
[? Ni]
(b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?
[3.92 kg*m^2 ]
2. Relevant equations
[tex]\tau=r\times F[/tex]
[tex]\tau=I\alpha[/tex]
[tex]I_{hoop}=mr^2[/tex]
[tex]\alpha=\frac{a_{com}}{r}[/tex]
3. The attempt at a solution
I took the axis of rotation to be perpindicular to the wheel's center of mass.
There are four forces acting on the wheel: mg, normal, applied force, and friction. Of those four forces, only friction does not pass through the axis of rotation, so friction alone (symbol is lowercase f) contributes to torque.
[tex]\tau=r\times f=I\alpha[/tex]
[tex]fr=(mr^2)(\frac{a_{com}}{r})[/tex]
[tex]f=ma_{com}(\frac{r^2}{r^2})[/tex]
[tex]f=(8kg)(.75m/s^2)=6N[/tex]
BZZZZZ. I lose.
Though on the bright side, my solution for the rotational inertia earned me a pretty green check mark...
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