Problem 3. A constant force F[N] with angle θ from the...

Problem 3. A constant force F[N] with angle $\theta$ from the horizontal axis is applied to wire wrapped around the axle with radius r[m] of yo-yo shaped cylinder which is initially stational at t=0 (see the attached figure). The mass and radius of the outer cylinder is M[kg] and R[m], respectively. The mass and thickness of wire are negligible. The cylinder has uniform density and the rotational inertia about the center is given by $MR^2/2$. Assume that the slip between the cylinder and the floor does not occur. (a) What is the magnitude of the acceleration of the center of mass? (b) Find the magnitude of frictional force between the cylinder and the floor. Answer with the direction of the frictional force. (c) Consider the case of r = R and $\theta$ = 0. Confirm that the acceleration obtained in (a) is larger than F/M which is the acceleration of an mass point with mass of M and applied force F. Then explain why the value can be larger than F/M. (d) Confirm that the work-kinetic energy theorem is satisfied between two times t=0 and t = t' (> 0). (Hint: Carefully consider the distance you have to pull the wire.)

Transcript

00:01 So here on this problem we are given this figure and this one is our cylinder this one is our pulley and this is the box now the radius of the cylinder r1 we are given us 50 centimeter and the mass of the pulley m2 we are given as 5k g and the radius r2 for the pulley it is given as 10 centimeter and the mass of the pulle and the mass of the for the box m3 we are given us 50 kgis now there are three parts in the question and here we are to assume that the cylinder rolls without slipping now in the first part we have to draw the free board diagram of the system so these are the free board diagram this is the free board diagram for the cylinder here you can see the tension force would be acting like this is a tension force this is the normal force offered by the surface the surface and the is the gravity force which is acting in the downward direction and the friction force fr now this one is the freeboard diagram for the pulley here you can see the tangent force t1 would be acting like this and the this side the other side the tension force we have taken it to be t2 now the normal force due to this surface would be acting like this is a normal force and this one is the gravity force now this is the free border diagram for the box we have only two forces here the first one is the gravity force the next one is the tension force which would be acting in the vertically upward direction so these are the free -board diagram for the system now next we'll be solving the b part now in the b part we have to calculate the mass of this cylinder m1 we have to calculate and we are given that the acceleration for the system is 5 meter per second squared so this is the acceleration given for the system now to calculate the mass of this body we have to this tension force 3 1 and to calculate this tension force t2 we can refer to this diagram this tension force t2 can be calculated if we know the tension force t2 now to calculate the tension force t2 we can refer to this diagram so from here we can calculate this tension force t2 then we'll go into the in the reverse order so for this we can apply newton's second law of motion here that is net force acting on the system is equals to mass times acceleration so the net force we have m3g minus t2 is equals to a mass we have to take for this body which is m3 times the acceleration so from here we can calculate the tension force t2 now m3 we are given in the equation as 50 times 9 .81 minus t2 we have to calculate and it is equals to m3 we have 50 times acceleration we have 5 so from here the tension force t2 it comes out to be as 240 .5 newton now next we're going to refer this diagram this one so here we're going to apply the talk equation that net torque about the center let this point be c is equals to i times alpha and we have to take the moment of inertia for this pulley about the center i see now alpha is the angular acceleration and in this case angular acceleration alpha would be equals to a over the radius is r2 and the i see that is the moment of inertia for this pulley is m r square over two now net talk acting on this body about this point will be t2 minus t1 times r2 so we have all the expression this is a net torque we can plug it here so net talk we have a t2 minus t1 times r2 is equals to i see we have now i see we take the mass for this pulley which is m2 so m2 times r2 square divide by 2 times alpha we have a over r2 now here we see that this r2 and r2 will cancel out and this are one more r2 will cancel out by this one now we can put the values to calculate t 1 so t2 we have calculated as 240 .5 minus t 1 so t 1 we have to calculate as equals to m 2 so m 2 we are given the equation as 5 times acceleration is 5 divided by 2 now from here the tension t 1 comes out to be as 2 to 8 newton so we have calculated the tension t 1 and t 2 2 2 2 therefore this diagram now to calculate the mass m1 of for this cylinder we can apply the nutrients second law of motion which says that net force acting on the system is equals to mass times acceleration now the net force we have a t1 minus f r that is a friction force is equals to m1 times the acceleration now here we don't have the value of fr also so for that we can use another equation that is a torque equation net torque is equal to i times alpha and we'll be taking torque about this point so we have to take i see here and i see here tc here and i see here so tc would be equals to f r1 this is the radius for the cylinder now i see for the cylinder we have a 1 over 2 m1 r1 square times alpha now f r this r1 and r1 will cancel out here we can see this will cancel it out so we are left with this one is f r as it is equals to 1 over 2 m1 r1 and alpha can be written as this is angular acceleration so it would be equal to linear acceleration divide by the radius r1 so here rver and r1 will cancel out so we have obtained the expression for f r to be 1 over 2 m1 times acceleration now we can put this expression here.

07:14 So t1 minus this is 1 over 2 m1 times acceleration is equals to this would be m1 times a as it is now you can put the values so t1 we have calculated as 2 to 8 newton minus 1 over 2 mass we have to calculate times acceleration is 5 is equals to 5 times m1 so from here we get 7 .5 times m1 and equals to 2 to 8 newton so from here m1 comes out to be as 30 .4 kgues so this is the answer for the b part now next we'll be solving the c part now in the c part we have to calculate the coefficient of kinetic friction so for that we can use this expression which is f r is equals to 1 over 2 m1 times acceleration now so, from here, we know that fr is equal to mu times the normal reaction which is n.

08:22 So in this case, if we balance the forces in the y direction, we can say that n would be equals to m1g.

08:29 So it would be m1g...

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