A block of mass m is projected up with a velocity `v_0` along an inclined plane of angle of inclination `theta=37^@`. The coefficient of friction between the inclined plane AB and blocks is `mu(=tan theta)`. Find the values of `v_0` so that the block moves in a circular path from B to C.

Explanation

In traingle ABO, `R/l=tanthetaimpliesl=(4R)/(3)`

For contact not to be lost at B:
`mgcos theta-N=(mv^2)/(R)`
`impliesN=mg cos theta-(mv^2)/(R)= ge0`
`impliesvlesqrt(gRcostheta)` (i)
From B to C: `DeltaK+DeltaU=0`
`(1/2mv_1^2-1/2mv^2)+mg(R-Rcostheta)=0`
`impliesv_2=v_1^2+2gR(1-costheta)` (ii)
From A to B: `a=gsin theta+mugcostheta`
`=g sin theta+(tan theta)g cos theta`
`impliesa=2gsintheta`
`implies v_2=v_0^2-2al`
`impliesv^2=v_0^2-(4gsinthetaR)/(tantheta)`
`impliesv^2=v_0^2-4gRcostheta` (iii)
From Eqs (i) and (iii),
`v_0^2-4gRcos thetalegRcos theta` or `v_0^2le5gRcos theta`
`implies v_0^2le5gR(4/5)=4gRimpliesv_0lesqrt(4gR)` (iv)
From Eqs (ii) and (iii),
`v_1^2+2gR[1-cos theta]=v_0^2-4gRcostheta`
`impliesv_1^2=v_0^2-2gR-2gRcos theta`
Now to reach at `C:V_1ge0`
or `v_0^2-2gR-2gRcosthetage0`
`implies v_0gesqrt(2gR[1+costheta])`
`impliesv_0gesqrt(2gR[1+4//5])impliesv_0gesqrt((18gR)/(5))` (v)
From (iv) and (v) `sqrt((18gR)/(5))lev_0lesqrt(4gR)`