A solid sphere of radius R makes a perfect rolling down on a plane which is inclined to the horizontal axis at an angle `theta`. If the radius of gyration is k, then its acceleration is

Options

A. `(g sin theta)/((1+(k^(2))/(R^(2))))`
B. `(g sin theta)/((R^(2)+k^(2)))`
C. `(g sin theta)/(2(R^(2)+k^(2)))`
D. `(g sin theta)/(2(1+(k^(2))/(R^(2))))`

Answer

A

Explanation

By forces balancing perpendicular to inclined plane,
`N="mg "cos theta`
By forces balancing parallel to inclined plane,
`"mg "sintheta=f_(r)=m(dv)/(dt)`
`implies" "(dv)/(dt)=(mg sin theta-f_(r))/(m)impliesv(t)=int g sin theta dt-(1)/(m)int f_(r) dt`
Now, torque, `tau=I(d omega)/(dt)=r xx F_(r)" and "I=mk^(2)`
`:." "mk^(2)(d omega)/(dt)=rF_(r)" or "(d omega)/(dt)=(rF_(r))/(mk^(2))`
`implies" "omega(t)=( R )/(mk^(2))int f_(r)dt`
or `(1)/(m)int f_(r)dt=(k^(2))/( R )/(omega(t)=(k^(2))/(R^(2))v (t)" "("using "omega=v//R)`
`implies" "v(t)=int g sin theta dt-(k^(2))/(R^(2))v(t)`
or `v(t)=(1)/((1+(k^(2))/(R^(2))))int g sin theta dt`
So, acceleration,
`a(t)=(dv(t))/(dt)=(g sin theta)/((1+(k^(2))/(R^(2))))`