Moderate -0.3 This is a straightforward application of conservation of angular momentum to a collision problem. Students need to set up L_initial = L_final with opposite signs for opposite rotations, then solve a single linear equation for the unknown moment of inertia. It's slightly easier than average because it's a direct one-step application of a standard principle with no geometric complications or multi-part reasoning required.
1 Two flywheels \(P\) and \(Q\) are rotating, in opposite directions, about the same fixed axis. The angular speed of \(P\) is \(25 \mathrm { rad } \mathrm { s } ^ { - 1 }\) and the angular speed of \(Q\) is \(30 \mathrm { rad } \mathrm { s } ^ { - 1 }\). The flywheels lock together, and after this they both rotate with angular speed \(10 \mathrm { rad } \mathrm { s } ^ { - 1 }\) in the direction in which \(P\) was originally rotating. The moment of inertia of \(P\) about the axis is \(0.64 \mathrm {~kg} \mathrm {~m} ^ { 2 }\). Find the moment of inertia of \(Q\) about the axis.
1 Two flywheels $P$ and $Q$ are rotating, in opposite directions, about the same fixed axis. The angular speed of $P$ is $25 \mathrm { rad } \mathrm { s } ^ { - 1 }$ and the angular speed of $Q$ is $30 \mathrm { rad } \mathrm { s } ^ { - 1 }$. The flywheels lock together, and after this they both rotate with angular speed $10 \mathrm { rad } \mathrm { s } ^ { - 1 }$ in the direction in which $P$ was originally rotating. The moment of inertia of $P$ about the axis is $0.64 \mathrm {~kg} \mathrm {~m} ^ { 2 }$. Find the moment of inertia of $Q$ about the axis.
\hfill \mbox{\textit{OCR M4 2004 Q1 [4]}}