Two children are playing on stools at a restaurant counter....

Two children are playing on stools at a restaurant counter. Their feet do not reach the footrests, and the tops of the stools are free to rotate without friction on pedestals fixed to the floor. One of the children catches a tossed ball, in a process described by the equation $\left(0.730 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(2.40 \mathrm{j} \mathrm{rad} / \mathrm{s})$ $+(0.120 \mathrm{~kg})(0.350 \mathrm{i} \mathrm{m}) \times(4.30 \mathrm{k} \mathrm{m} / \mathrm{s})$ $=\left[0.730 \mathrm{~kg} \cdot \mathrm{m}^{2}+(0.120 \mathrm{~kg})(0.850 \mathrm{~m})^{2}\right] \vec{\omega}$ (a) Solve the equation for the unknown $\vec{\omega}$. (b) Complete the statement of the problem to which this equation applies. Your statement must include the given numerical information and specification of the unknown to be determined (c) Could the equation equally well describe the other child throwing the ball? Explain your answer.

Two children are playing on stools at a restaurant counter. Their feet do not reach the footrests, and the tops
of the stools are free to rotate without friction on pedestals fixed to the floor. One of the children catches a tossed
ball, in a process described by the equation
$\left(0.730 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(2.40 \mathrm{j} \mathrm{rad} / \mathrm{s})$
$+(0.120 \mathrm{~kg})(0.350 \mathrm{i} \mathrm{m}) \times(4.30 \mathrm{k} \mathrm{m} / \mathrm{s})$
$=\left[0.730 \mathrm{~kg} \cdot \mathrm{m}^{2}+(0.120 \mathrm{~kg})(0.850 \mathrm{~m})^{2}\right] \vec{\omega}$
(a) Solve the equation for the unknown $\vec{\omega}$.
(b) Complete the statement of the problem to which this equation applies. Your statement must include the
given numerical information and specification of the unknown to be determined
(c) Could the equation equally well describe the other child throwing the ball? Explain your answer.

Key Concepts

Rotational Collision Dynamics

Rotational collisions involve interactions where bodies collide and exchange angular momentum. In these events, different parts of a system (such as a person and a stool being hit by a ball) interact in a way that conserves overall angular momentum but may redistribute it between rotational parts. Understanding this concept is essential when analyzing scenarios where the system's configuration changes, even momentarily, as momentum is transferred during interactions like catching or throwing.

Cross Product in Angular Momentum Calculation

The cross product is used in the calculation of angular momentum for point masses moving linearly. When a force or momentum is applied at a point offset from an axis, the angular momentum is given by the cross product of the position vector (lever arm) and the linear momentum vector. This mathematical operation captures both the magnitude and direction of the induced rotation, which is critical in solving problems where the path of the thrown object affects rotational motion.

Conservation of Angular Momentum

This principle states that in the absence of external torques, the total angular momentum of a system remains constant. In rotational dynamics, especially in collisions or interactions like a ball being caught, the angular momentum before the event must equal the angular momentum after. This key concept enables solving problems where rotating objects exchange momentum without external influences altering the system.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotational motion, analogous to mass in linear motion. In composite systems, such as a person on a stool, the overall moment of inertia is the sum of the individual contributions from each mass distributed at various distances from the axis of rotation. It plays a crucial role in determining how angular velocity changes when additional mass or forces are involved in the system.

Angular Momentum of a System

Angular momentum in a system is the combined rotational effect of objects and can have contributions from both rotational motion (described by the moment of inertia times angular velocity) and linear motion (through the cross product of the position vector and linear momentum). The equation in the problem accounts for these distinct contributions, enabling one to analyze how motions like catching a ball affect a rotating stool-child system.

Transcript

00:01 Okay, so we know that two kids are playing out of counter, the feet don't reach to the bottom, and the pool, the stools can rotate.

00:10 And so they're throwing a ball back and forth.

00:14 And this process is described by a certain equation.

00:18 It's kind of funny they give you the equation.

00:19 It's almost like reminds me of like, you know, like you're cheating, but then the other person didn't really write down enough information about what they did.

00:28 So you've got to kind of piece it together.

00:30 So that's kind of funny.

00:35 So i'm going to go ahead and write the equation out, and then i'll kind of approach the problem.

00:42 So 2 .4 in the j hat, and i guess this is in radiance per second.

00:51 0 .120 kilogram times 0 .350 meters in the ihatt direction.

01:01 Oh, shoot.

01:04 And then that was times 4 .3 kilometers per second.

01:10 And we are setting that equal to 0 .730 kilogram squared.

01:40 I guess that's got to be the moment of inertia of the stool probably.

01:45 All right, so we wanted to solve for omega.

02:10 Okay, is this all in the j -hat direction? oh, it's not kilometers per hour.

02:26 It's crossed in the k -hat direction, which does give you, so i to k gives you negative j.

02:39 So this is going to be a negative sign.

02:44 And then this is it's going to be in the j hat direction very subtle i was like going to be a little worried if this was in kilometers per second but now i can see that was a typo um and then all right so then omega and then omega is i guess going to also be in the j hat direction so we're going to take this whole left -hand side and divide it by the right hand side so let me go ahead and and pull up my calculator .730 times 2 .4 minus .120 times .35 times .4.

03:40 And then we, oh shoot, that's what we wanted to put in the denominator.

03:44 So in the numerator we want to do .730 plus .120 times .35 squared.

03:53 And then we want to divide by the thing i just wrote.

04:00 Okay, so i get omega is 0 .474 radiance per second in the j -hat direction.

04:22 And great.

04:24 So let me go back to the original problem and look up what we need to do next...

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