Question

suppose you start an antique car by exerting a force of 370 n on its crank for 0.330 s. what angular momentum (in kg m²/s) is given to the engine if the handle of the crank is 0.270 m from the pivot and the force is exerted to create maximum torque the entire time? kg m²/s. resources reading submit answer 8. -/12.5 points details my notes ask your teacher practice another when a diver gets into a tuck position by pulling in her arms and legs, she increases her angular speed. before she goes into the tuck position, her angular velocity is 5.5 rad/s, and she has a moment of inertia of 1.2 kg - m². once she gets into the tuck position, her angular speed is 11.5 rad/s. determine her moment of inertia, in kg - m², when she is in the tuck position. assume the net torque on her is zero. kg - m²

Explanation:

Step1: Recall the formula for angular momentum

The formula for angular momentum \(L\) is given by \(L = \tau t\), where \(\tau\) is the torque and \(t\) is the time. The torque \(\tau\) is given by \(\tau=rF\) (since the force is exerted to create maximum torque, \(\sin\theta = 1\) in \(\tau=rF\sin\theta\)).

Step2: Calculate the torque

Given \(F = 370\space N\) and \(r=0.270\space m\), then \(\tau=rF=(0.270\space m)\times(370\space N)=99.9\space N\cdot m\)

Step3: Calculate the angular momentum

Given \(t = 0.330\space s\) and \(\tau = 99.9\space N\cdot m\), then \(L=\tau t=(99.9\space N\cdot m)\times(0.330\space s)=32.967\space kg\cdot m^{2}/s\approx33.0\space kg\cdot m^{2}/s\)

Answer:

\(33.0\space kg\cdot m^{2}/s\)