- multiplying the ratio between the two torques by the mass of the bicycle and rider
- adding the two torques together, and multiplying by the time for which both torques are applied
- multiplying the difference in the two torques by the time for which the new torque is applied
- multiplying both torques by the mass of the bicycle and rider

(c)

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View sample solutionThis is College Physics Answers with Shaun Dychko. We're going to find the change in the angular momentum of the bicycle wheel after the gears are changed. So we have <i>l prime</i> minus <i>l</i> is what we have to find. We have the moment of inertia of the bicycle wheel system multiplied by the final angular velocity minus the moment of inertia of the system times its initial angular velocity. Now the moment of inertia of the system doesn't change with changes in the position of the chain. So we can factor out this <i>i</i>, it's a common factor. It doesn't need a subscript and it doesn't need a prime here. So we have <i>i</i> times <i>omega prime</i> minus <i>omega</i>. So we should find an expression for <i>omega prime</i> in terms of torque then. We know that it equals the initial angular velocity plus the angular acceleration multiplied by time and we can substitute for <i>alpha</i> with the net torque divided by the moment of inertia. So we can then substitute this in for <i>omega prime</i> and we do that here. We see that the <i>omega initial</i> makes zero because it's <i>omega</i> minus <i>omega</i>, and then the term that's left over is this. The moment of inertia on the bottom cancels with this factor there and we're left with net torque multiplied by time. Net torque is the difference between the two torques and so the answer is C and we're multiplying it by the time during which the new torque is applied.