Ii the pulley in figure 1165 consists of two disks of...

(II) The pulley in FIGURE 11.65 consists of two disks of different diameters attached to the same shaft. The rope attached to the block of mass m₁ = 1 kg passes over a smooth nail, while the block of mass m2 = 3 kg is suspended vertically from one of the disks. The moment of inertia of the pulley is equal to 0.2 kg.m^(2); r₁ = 5 cm; and r2 = 10 cm. (a) Determine the tensions in the ropes and the modulus of the accelerations of the blocks. (b) If the figure shows the blocks at the initial instant and h = 2 m, at what instant will the blocks be at the same height?

Transcript

00:01 Okay, so here is our pulley, which has a rotational inertia of 1 .8 kilogram meters squared.

00:06 And it has a mass m1 hanging off of the large section of the pulley.

00:11 So that mass is 3 .2 kilograms, and r1 is 0 .6 meters.

00:17 And then we have a second mass hanging off of a smaller section that is 2 .1 kilograms at a distance of 0 .3 meters.

00:28 The accelerations for these two masses will be different, the linear accelerations, but this whole system is going to have the same rotational acceleration.

00:39 So to evaluate this and find the tension in the strings, which the tension will be different as well, we need to look at the sum of the forces on these two masses and the sum of the torques on this pulley.

00:52 So the first mass, m1, its net force, m1a1, it's going to equal the weight force down, m1g, minus the tension force up, tension 1.

01:05 And the forces on m2, the net force is going to be m2a2.

01:12 And now, since this is going to be accelerating up, the tension in the string is going to be greater, so we'll have t2 up, and then m2g down, trying to slow it down.

01:23 And then finally, the sum of the torques on the pulley will be i times alpha, and that's going to be equal to this counterclockwise torque r1t1, minus this clockwise torque r2t2.

01:39 And so now we need to substitute into this equation.

01:43 And since we are looking for alpha to start, we're going to sub in for t1 and t2.

01:50 So let's solve these two equations for the tension forces.

01:54 So this first equation becomes t1 equals m1g minus m1a, and we can find that by adding t1 to both sides and subtracting this term from both sides.

02:05 And then this equation here becomes t2 equals m2a plus m2g, and we get that just by adding this to both sides.

02:15 Now these accelerations, this should be a1 and this should be a2, and we can substitute in for those accelerations in terms of the rotational inertia or the rotational acceleration.

02:28 So this a here is going to be alpha r1 and this a here is going to be alpha r2.

02:35 And so we're going to plug those into this equation here...

Question

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