A uniform solid sphere of mass m and radius 'R' is imparted an initial velocity `v_(0)` and angular `omega_(0) = (2v_(0))/(R )` and then placed on a rough inclined plane of inclination 'theta' and coefficient of friction `mu = 2 tan theta` as shown. The time after which the sphere will start rolling without slipping is

Options

A. `(v_(0))/(2g sin theta)`
B. `(v_(0))/(4g sin theta)`
C. `(v_(0))/(6g sin theta)`
D. `(v_(0))/(8g sin theta)`

Answer

C

Explanation

`N = mg cos theta`
`f_(k) = muN = 2 tan thetatmgcostheta = 2mg sin theta`
`f_(k) - mg sin theta = ma`
`2mg sintheta - mg sin theta = ma`
` therefore a = g sin theta`
Now, `tau_(cm) = I_(cm) alpha`
`f_(k)R = (2)/(5)mR^(2)alpha`
`2mg sin theta = (2)/(5)mRalpha`
`alpha = (5g sin theta)/(R )`
Where the rolling without slipping starts
`v = omegaR`
`v_(0) + at = (omega_(0) - alphat)R`
`v_(0) + g sin thetat = omega_(0) R - 5g sin thetat`
`v_(0) + 6g sin thetat = 2v_(0)`
`therefore = (v_(0))/(6g sin theta)`