00:01
Hello student, to solve the given question, let us write the relation for torque that is equal to force into displacement, which is also given by moment of inertia multiplied by angular acceleration.
00:12
Now also here, moment of inertia in terms of radius of gyration is given by mk square.
00:19
So using these relations, let us find firstly for part a, angular acceleration.
00:25
Now here moment of inertia for g equals m is given.
00:30
X multiply by k is also given 19 to 10 to power minus 3 squared this will be calculated as 0 .0 486 kilogram meter square now the total moment of inertia will be given by i g plus m r square so here plug in the values we get 0 .0486 plus masses 6 multiply by r square that is 1 to 0 .0 0 into 10 to power minus 3 squared.
01:04
Hence, moment of inertia will be calculated as 0 .135 kilogram meter squared.
01:11
Now that we have calculated the moment of inertia, let us write the equation for torque, which will be p force multiplied by distance here will be taken as r equals moment of inertia multiplied by alpha.
01:27
So from here, alpha will be calculated as 20 multiply by 0 .06 divided by 0 .135.
01:37
Hence, from here we get the final answer for alpha, that is equal to 8 .89 radiance per second square...