The given angular acceleration remains valid even if the...

The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let L = 5.3 ft and assume that the pendulum is placed in motion at 0 = 0°. What is the minimum angular velocity at this position for the pendulum to swing through a full circle? • Answer min = 4.930 rad/s

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 So here we are given a pendulum that consists of a 20 kg uniform disc.

00:09 So mass of this pendulum with the disc is having 20 kgs and there is this 3 kg uniform slender rod and it is released from the position as shown right.

00:24 This is a slender rod with the disc that is attached right and the moment is given to be equal to if you see here it's given to be 30 newton meter right.

00:52 30 newton meter.

00:54 So what you need to do is you need to calculate the moment of inertia of this system about the rotational point which is the a right and based on this you have to calculate the angular velocity when it has moved 90 degrees.

01:08 So for that what you need to do is you need to calculate the acceleration of the system right.

01:12 So first of all let's calculate the moment of inertia of the disc about the center and shift it to a right.

01:18 So moment of inertia of okay so what i can do is moment of inertia about a will be equal to let's consider the moment of inertia of this disc which is half mass of the disc m r square right.

01:37 R is the radius of the disc.

01:39 Here we have got the radius to be given equal to 0 .4 meters right.

01:46 So the radius of the disc is 0 .4 meters and with that what you need to do is you need to shift this to the point a itself right.

01:56 So this is this shifting distance will be the length l which is given to be 2 meters 2 plus r radius of it which is 0 .4.

02:09 So 2 .4 meters you have to shift right so that it will the moment of inertia will be calculated at the point a...

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