Two metal disks one with radius r1 254 cm and mass m1 0850...

Two metal disks, one with radius R1 = 2.54 cm and mass M1 = 0.850 kg and the other with radius R2 = 5.04 cm and mass M2 = 1.58 kg , are welded together and mounted on a frictionless axis through their common center. (Figure 1). A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block, suspended from the free end of the string. If the block is released from rest at a distance of 2.10 m above the floor, what is its speed just before it strikes the floor?

Transcript

00:02 Hi, here in this given problem there are two solid metal disk, smaller one and a larger one.

00:14 The larger one, having a radius r2, smaller one, having a radius r1.

00:23 And these values are r1 is equal to 2 .54 centimeter and its mass is a 0 .850 kilogram.

00:38 Similarly for the second disk which is the larger one r2 is equal to 5 .04 centimeter and its mass is m2 is equal to 1 .58 kilogram.

00:57 Then there is a mass suspended with the help of a thread over the smaller disk like this and free to fall down under its weight mg creating a tension t in the thread this mass small m this is given as 1 .50 kilogram and at an instant it is at a height h which is given as 2 .10 meter above the ground.

01:46 Now first of all we find combined rotational inertia of both of the disk together combined rotational energy of the two disks that will be given by i1 plus i2 so using the expression for the rotational inertia for i 1 this is half m1 r1 square plus for i2, half m2 r2 square.

02:34 Taking this half as a common out, then for m1 this is 0 .850 kilogram.

02:40 R1 that is 2 .54 centimeter or 2 .54 into 10 to the power minus 2 meter.

02:48 Plus for m2 this is 1 .58 kilogram and here this is 5 .04 into 10 to the power minus 2.

02:58 Square and finally it comes out to be equal to 2 .28 into 10 dash the power minus 3 kilogram into meter is square now using free body diagram of the falling mass as it is moving down so net force acting on it will be given by m g minus t and this net force should be equal to m into a using newton's second law of motion.

03:45 We can make it equation number one.

03:48 Then torque acting over the smaller disk because of the string wound over it.

04:02 That will be given by tension in the string into radius of the disk...

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