The axle of a uniform cylinder of mass m and radius R is connected to a spring of constant k. A horizontal board of mass m rests on top, with no slipping between cylinder-board or cylinder-ground. If angular frequency is p, find 11p²m/k.

Question

G1 Physics · Accepted Answer

Let cylinder centre speed be u. No slip on ground gives angular speed u/R; no slip with the board makes board speed 2u. Cylinder KE=(1/2) μ \mu μ ²+(1/2)(mR²/2)(u²/R²)=3mu²/4. Board KE=(1/2)m(2u)²=2mu². Total KE=(11/4) μ \mu μ ²=(1/2)(11m/2)u², so mₑ=11m/2. Hence p²=k/mₑ=2k/(11m), giving 11p²m/k=2.