Question

Question #10c64

Answer 1

Consider the following image of the scenario

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.