00:01
In this question the mass moment of inertia igd equals of 5 .5 .5 .4 square equals 0 .4 kilogram meter square.
00:21
Also vp equals omega 2 times root 5 .92 and vd 2 equals omega 2 times 2 .2 times 2.
00:40
So sum of ha at 1 equals sum of ha at 2.
00:49
If we substitute we will have mpvp1rp1 equals igd omega 2 plus mdvd 2 plus mbb2 rb2 rb2 so if we substitute by the values we will have 0 .01 times 800 multiply 2 .4 equals 0 .4 omega 2 plus 5 omega 2 times 2 square plus 0 .01 omega 2 root 5 .92 square.
02:00
So from this equation we can get omega 2 equals 0 .656 3dm per second.
02:10
Since the system is required to stop finally then t3 will be equal to 0 and t2 equals of igd omega 2 square plus of md vd vd d2 square plus half mbb to square.
02:44
If we substitute by the values we will have half times 0 .4 times 0 .6 .6 square plus half like 5 multiply 0 .6 .m .2 .m.
03:03
2 .4 and this bracket square plus half times 0 .01 multiply 0 .66 root 592 square so t2 will be equal 6 .2996 joules so for the potential energy datum is set as as as indicated, so h or here, phi equal 10 inverse 0 .4 over 2 .4, which is equal 9 .4623, heinz y1 equals 2 .4 cosine and yp equals root 5 .92 cosine ceta minus 9 .4623.
04:27
Thus the gravitational potential energy of desk and pollet with reference to datum is vgd equals mdgg, yd, equals 5 multiply 9 .81 multiply negative 2 .4 cosine theta equals negative 117 .72 cosine setta...