 In this problem, we have a 3 meters long steel shaft that we need in order to transmit 0.75 megawatts of power at 120 revolutions per minute. We have the material properties, we have that the tau max is equal to 45 megapascals, g of the steel is equal to 79.6 gigapascals and the density of the material is 7.8 times 10 to the power of positive 3 kilograms per meter square. So in the first case, we have to find what is the minimum radius of this shaft if it is solid. In the second case, we have to find the minimum radius if the shaft is hollow and we have the relationship that r1 should be equal to r2 divided by 2. And finally, we have to compare the weight for both designs. So we have to decide which one is better. So first of all, we don't know how much is the torque that we need to apply. So this is the first thing that we need to calculate. What is the torque? We already know that power is equal to the torque times the speed of rotation. So from this equation, we can calculate what is the torque t that we need to transmit. Then from this equation, we have that the torque is equal to power divided by speed of rotation. The power that we need to transmit is equal to 0.75 megawatts. So 10 to the power of 6 watts divided by the speed of rotation. But in this case, in order to be consistent with the units, this must be given in radians per second. So in this case, we have 120 revolutions per minute. We multiply times 2 pi in order to have radians and we divide by 16 in order to have seconds. And from here, we have that the total torque is equal to 59.7 kNm. Now we have all the information that we need in order to calculate what is the radius of the solid shaft. So question A, solid shaft. We have that this is the free body diagram. This is the shaft. So we are applying a torque here like this and the section is solid. So we need basically to calculate where is our SR of the shaft. So this radius will be given by the maximum shear stress. Because we already know that the maximum shear stress occurs at the maximum radius. So then we can use the torsion formula which says that the torque divided by the polar moment of inertia is equal to the stress divided by r. So in this case, we have here the maximum stress. Of course, we have here the maximum radius, which is the radius of the shaft. Then we can solve here. We need to know first what is the polar moment of inertia. So we have that the polar moment of inertia is equal to pi-halves times the radius to the power of 4. So if we substitute here in this formula, we have that the torque, which is equal to 59.7. Sorry, to the power of 3 divided by the polar moment of inertia pi-halves times rs to the power of 4 is equal to the maximum shear stress that is given in the statement of the problem. 45 to the power of 6 pascals divided by the radius of the shaft. So from this formula, we are going to obtain three different solutions. So we take, of course, the positive one and from here we solve. We find that the radius is equal to 0.0945 meters. So this is the minimum radius if the shaft is solid. Now we can solve the problem in the case that the shaft is hollow. So now we have that the cross section is different. We have here our inner radius and we have rs, which is the outer radius of the shaft, which is the one that we have to calculate. Then again, we can use the torsion formula. We have that the torque divided by the polar moment of inertia is equal to the shear stress divided by the radius. Once again, we have that the maximum shear stress is found at the maximum radius. So this is rs. But in this case, this formula remains the same but the polar moment of inertia is different. So we need to calculate what is the polar moment of inertia for this section. And we have that the polar moment of inertia is equal to pi over 2 divided by 2 times rs to the power of 4 minus r inner to the power of 4. Then now we need to have this equation only as a function of rs. So we can now use here the relationship which says that the inner radius is equal to the radius of the shaft divided by 2. So if we apply this relationship here, we have that this is equal to pi half rs square minus rs divided by 2. So this is to the power of 4 to the power of 4. And if we rearrange this equation and we calculate we have that this is equal to 1.473 r2 or rs to the power of 4. So we can now substitute here this result and this is the equation that we have. We have again that the torque is equal to 59.7 times 10 to the power of positive 3 divided by this 1.473 times rs to the power of 4. And this is equal to 45 times 10 to the power of 6 divided by rs. So now as we did before we can solve this equation for s. We take the positive result and rs is equal to 0.0966 meters. So this is the outer radius if the shaft is hollow. And of course from this equation we can easily obtain that the inner radius is equal to the outer one divided by 2. So this is equal to 0.0483 meters. And now we can do the comparison of both designs. So we have here comparison we can divide the weight of the hollow shaft by the weight of the solid one. So this is equal to the density rho times the volume and the volume is L times the area of the hollow shaft. Same here rho times L times area of the solid shaft. We have same density we have same length so this is basically a ratio of areas. So in the first case this is equal to pi rs in the hollow case minus our inner to the power of 2 divided by pi rs in the solid case to the power of 2. So finally we have that this is equal to then this is the final result. And from here we can see that the solid shaft is always heavier than the hollow one. And this is something that we could expect because if you look at the torsion formula we have that the maximum shear stress is equal to the torque times the radius the outer one divided by the polar moment of inertia. So for a given constant torque this is a function of r and this is a function of rs to the power of 4 minus the inner radius to the power of 4. So basically what we have here and of course the solid shaft is a limit case in which the inner radius is equal to 0. So we can see here that the larger this value of r the smaller needs to be the difference between rs and the inner radius. So we have that for a larger rs in order to have the same maximum shear stress we have that rs squared minus the inner radius squared this amount decreases. And of course if the area decreases the final weight of the structure also decreases. And from here we have the nice result that when the radius tends to infinity the weight of the structure is equal to 0.