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A rod of mass m and length 2R can rotate about an axis passing through O in vertical plane. A disc of mass m and radius R is hinged to the other end P of the rod and can freely rotate about P. When disc is at lowest point both rod and disc has angular velocity w. If rod rotates by maximum angle 60o with downward vertical, then find w in terms of R and g. (all hinges are smooth) {Ans: (9g/16R)1/2}

Aakash , 10 Years ago
Grade 12th pass
anser 3 Answers
Ayan

Last Activity: 7 Years ago

In this case the angular momentum of the rod is 1/12*m*(2R) = 1/3*m*Rand the angular momentum of the disc is 1/2*m*R2. Now as the rod and the disc rotates with same angular momentum omega when the disc is at lowest point and the rod rotates maximum with angle 60° then the angular momentum is, omega = √9g/16R

Ayan

Last Activity: 7 Years ago

In this case the angular momentum of the rod is 1/12*m*(2R) = 1/3*m*Rand the angular momentum of the disc is 1/2*m*R2. Now as the rod and the disc rotates with same angular momentum omega when the disc is at lowest point and the rod rotates maximum with angle 60° then the angular momentum is, omega = √9g/16R

Sahil babbar

Last Activity: 7 Years ago

Use mechanical energy conservation note that rotational kinetic energy of the disc remains the same as no torque is there to change it but the translational kinetic energy when rod makes 60degrees is zero for the disc and for the rod the total ke(rotational as it is in pure rotation) is also zero apply it correctly as stated u will get the desired answer if u are not able to pls tell I`ll send the full solution

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