00:01
In this question, the moment of inertia of this circular plate or circular disk, that is cylinder, is given as, i is given as 0 .01, kg meter square.
00:18
Now, from this question, we can write torque equals to force into radius, that is, force is here 0 .1 newton and radius of rotation is where force is applied is 1 cm that is 0 .01 meter.
00:43
So we can write torque equals to 0 .001 that is 0 .001 or 1 by 1000 newton meter.
00:57
So this is the value of torque.
01:00
Now we also know that torque is equals to i alpha where i is the moment of the torque.
01:07
Inertia and alpha is the angular acceleration.
01:11
So we can write here alpha equals to torque by i.
01:16
So it will be 0 .001 divided by i that is moment of inertia that is 0 .01.
01:24
So on solving this we'll get angular acceleration will be equal to 0 .1 that is angular acceleration alpha equals to 0 .1 radian per second.
01:40
And square.
01:42
So this will be the angular acceleration.
01:45
Now in the part b of the question it is given that if the length of the string or the thread is 20 cm.
01:55
Now one rotation that is the circumference of this cylinder which is equals to one rotation.
02:07
So 2 pi radian that is one rotation is equals to circumference so circumference is 2 pi r that is 2 pi into 0 .01 meter so a string wrapped around this cylinder of length 2 pi into 0 .01 meter will have one rotation equals to 2 pi radian so one can write 1 meter length will be equals to 2 pi divided by 2 pi into r where r is 0 .01 meter.
02:55
So 20 meter length will cause rotation up to 2 pi by 2 pi r into 20 meter.
03:07
Sorry it is 20 centimeters so it will be 0 .2 meter 20 centimeter can be written as 0 .2 meter...