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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. |
Answer» Solution :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)` |
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