A 1.4 kg, 20-cm-diameter turntable rotates at 120 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?
Help I am stuck!!!
Help I am stuck!!!
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Apply conservation of angular momentum .. (M o I for turntable disc, I = ½.MR²)
• Initial ang. mom. I(ωi) = (½.MR²)(ωi) .. [½ x 1.40kg x (0.10m)² ] x (120 x 2π rad / 60s) = 0.088
• Final ang .mom. (treating 0.50kg blocks as point masses at 0.10m radius .. M.o I = mR²)
= I(table)(ωf) + [2 x I(point mass)(ωf)
= ½MR² x (ωf) + 2.mR².(ωf)
= (ωf) {(½ x 1.40kg x 0.10²) + (2 x 0.50kg x 0.10²)
= (ωf) {0.017}
Mom before = mom.after ..
0.088 = 0.017.(ωf) .. .. ωf = 0.088/ 0.017 = 5.176 rad/s ..
= (5.176 x 60 / 2π) rpm .. .. ►49.4 rpm
• Initial ang. mom. I(ωi) = (½.MR²)(ωi) .. [½ x 1.40kg x (0.10m)² ] x (120 x 2π rad / 60s) = 0.088
• Final ang .mom. (treating 0.50kg blocks as point masses at 0.10m radius .. M.o I = mR²)
= I(table)(ωf) + [2 x I(point mass)(ωf)
= ½MR² x (ωf) + 2.mR².(ωf)
= (ωf) {(½ x 1.40kg x 0.10²) + (2 x 0.50kg x 0.10²)
= (ωf) {0.017}
Mom before = mom.after ..
0.088 = 0.017.(ωf) .. .. ωf = 0.088/ 0.017 = 5.176 rad/s ..
= (5.176 x 60 / 2π) rpm .. .. ►49.4 rpm