Question #10c64

1 Answer
Dec 12, 2015

Consider the following image of the scenario
enter image source here

We know from second law of motion the acceleration of the COM is given by-
mgsinx-f=ma_{cm}
Where a_{cm} is the acceleration of center of mass

Calculating moments about COM, we get
f r= I\alpha
Where r is the distance from the center where the friction is acting, I is moment of inertia and \alpha is the angular scceleration about the COM.

Here, we use the fact that the object is rolling without slipping, therefore:
a_{cm}=\alpha r

Substituting this in the equation above we get
f r= I \frac{a_{cm}}{r}
f = I \frac{a_{cm}}{r^2}

Substituting a_{cm} from first equation into the equation above
a_{cm}= \frac{mgsinx-f}{m}
f= \frac{I}{mr^2}(mgsinx-f)
f(1+\frac{I}{mr^2})=mgsinx\frac{I}{mr^2}
f= \frac{mgsinx\frac{I}{mr^2}} {1+\frac{I}{mr^2}}

Dividing numerator and denominator by \frac{I}{mr^2} we get
f= \frac{mgsinx} {1+\frac{mr^2}{I}}
Hence proved.