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(II) A uniform stick 1.00 $\mathrm{m}$ long with …

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Problem 82 Hard Difficulty

A baseball bat has a "sweet spot" where a ball can be hit with almost effortless transmission of energy. A careful analysis of baseball dynamics shows that this special spot is located at the point where an applied force would result in pure rotation of the bat about the handle grip. Determine the location of the sweet spot of the bat shown in Fig. $51 .$ The
linear mass density of the bat is given roughly by
$\left(0.61+3.3 x^{2}\right) \mathrm{kg} / \mathrm{m},$ where $x$ is in meters measured fromthe end of the handle. The entire bat is 0.84 $\mathrm{m}$ long. The
desired rotation point should be 5.0 $\mathrm{cm}$ from the end where
the bat is held. [Hint: Where is the cu of the bat?]

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 11

Angular Momentum; General Rotation

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Video Transcript

The desired motion is pure tradition about the handle grip. Since the grip is have any lenient motion on access through the grip, Ah can be choosing as on axis fixed in an inertial reference frame. Now the peel rotation condition can be the thing. Us. This is the acceleration off the center off mass disc embedded in US Alpha Matt Dying's the distance off descended off moss minus distance off the clip on Big Grip. Over here is 0.505 Mita from the sin from the end off the back, back to the So that's the distance. That's the grip. Now we applying Newton's second law for Goethe, the translational motion off the center off mus c m on the rotational motion about the handle group. So we have neck force to the equal Do Senate forces F ear, and this is equal to mass times sent it off Mass. According to Newton, slow and for dark talk should be equal toe. I, Elsa and I hear is moment of inner self grip times, angular acceleration and the next talk over here is the force applied times the perpendicular distance. Now we use this relation off off force on dhe user at this place and flying back whilst times acceleration off them that off mass times a day is equal to ay, grip, bang. Satisfied. So now we can plug this value this expression off a C M. Over here. So doing that we get, I saw to the equal Do so Alfa back is basically Alfa, So I'm just writing out for you So I'll for things they see and minus the grip on DDE at the sickle times they on their physical Do I grip Thanks. I thought from here began Sol for D Which concert to be I grip all right whilst times they cme minus decrypt and noticed that the alphas cancel out from both sides. So this is the final expression for data we are going we will be using. But before that we need to calculate mm on DDE the moment of inertia after a dip. So let's do that now said on Let's say this is a question one that we will be using here. Sonal loving was calculated the moment of inertia off the back about an access through the drip, the moss off the back. It is in here on the location off the center off Mars, which is D C M. No, I grip and Gilligan as this is basically mass Times square off the radius. And here isn't the mosque, if some changing. So we take a dining element off Mass on Well, to play it with its radius squared off the radius. And therefore, the total moment off initial should be integration over this. Now the M Here, let me just write it in green. The in here is Lambda Times The X The lambda is the senior mass density off. Huh? Lamb buys the linear mass density on out here is X minus the grip. So we take the square that on blood this expression off again on we're integrating form from cedar toe the length off B X, which is Siegel point 84 meters. Now let's substitute the values that we have. So we have a big group to the equal. Go Cedar Point Siegel! Fife, I'm not writing that. I'm not including the units. Now, after I'm done with all the calculation, I'll write the unit for the center off months. So this is that on Lambda is 0.61 this 2.3 times x skirt and then we have the X. Now we're going to expand this whole expression. So we have so far we expanded on Write it in, Oh, descending order off the orders. So, force, we have extra depart for the highest order. And we have Lena function of X and the constant and we multiply this and we integrate this over the ex. Now that we know what what the limits off X began simply do the integral and put these limits on the day Christian On doing that, they get I grip to the equal toe cedar point 33 6865 plays in with their split. So let me just give you a generalized formula for the integration that we'll be applying over here. So if you have to integrate, that's six to the pot and over the edge six and the limb itself from a to B, then the integration off. This is next to the part and plus one over and bless one. Then you apply the limits A to B and therefore this is equal dough one over and bless one. They will depart in plus one minus. I need the departed Bliss one. So we're substituting beyond a very will be have X now over here the initial m a crazy or the lower limit. Ezio, this means that a is Siegel, This is Eagle. Therefore, we have be to depart and plus lan over and placement on. This is the expression that we are applying now. Then we're doing each indication so here we have extra depart for so four so n here is equal to four. So we should have be to depart for over beauty by five over five and be here is Cedar 0.84 on. After doing that after the integration, you multi played with the constant or the component coefficient off that power coefficient corresponding to that power on DDE. We use the same formula for each of these storms. Ah, just at the value off in keeps on changing. So here in his three year. And it's true. And here in this one and here in this city on remember that this formula is only applicable men. And is it vintage so next. So now that it's clear for the integral part, we move on to find the moss. So months is basically integration off all those small element off masters on Dhe. As I drove over here that this D M is equal to land the times, the small element off lend taken with this t x on again, the limits are going to be saying, which is your 0.84 zero does it a 00.84 meters. So I'm not going to do the full integration over here because, wow, it's quite similar to this one. Re plug this value off line now over here and integrate a question so and then applied the limits. So doing that the u get and to be equal toe one point 1644 gauges last. We have refined the position off the center off Mars, which is ex cm or basically BC, and that we have years. So this is equal. Do the boat must, which is signs some mission. And here we have integration over the multiplication off all ex ambience, all beings, all the small element off my cells times the distance again. We do the same. We have Eckstein's lamb dug lines, the X notice that Lambda is a constant so you can't take it out off the integral. So you have to write fishin off. Landau used the expression off lambda that we have. Which is this one? So little interest rate. This is Lambda, and you have to use it over here s mint, and then you'll have a polynomial on. Limits are going, Toby. See you expand the polynomial on Dhe used this integration drawing and solve it. You get the center off, Moss miscalculated. Toby Cedar point 53757 on. Remember to use this value for Mars over here. Now we have all the values that we need. The Solvay question one this box to question here. So we simply put them with those values in the data question. So we have the equal door administrated again. I agree. All life loss times and not in Mr Total. Maas Times position off the center of mass minus big rip. So the gravy? No. So this is basically zero point 05 meter, as I told over here on this stuff, the values which just calculator some on the line in them. So these three have the values on DDE. Simply plug them into this a question to find the on DDE comes out to be a zero point five line. Jeannie are nearly equal to zero point 59 team Doc. So this means the distance from the end off the bag to the so called sweet spot. Then the equal do be less day grip it 0.5 me, Dutch. And if you use this value over here, you've got this value, Toby. 0.643 meter. So this is the distance from the end off the back to the street sport.

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Problem

(II) A uniform stick 1.00 $\mathrm{m}$ long with …

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Physics for Scientists and Engineers with Modern Physics

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Watch More Solved Questions in Chapter 11

Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Meghan Miholics

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Meghan Miholics

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Meghan Miholics

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics