yaar someone solve it and show ..............i am getting
25mgh .......................... but answer is given
27mgh
in dc pandey the question is ....given
A solid sphere of mass m rolls down a inclined plane of height h . find the rotational kinetic energy of sphere ???
plzz m weak at this topixc so plzzz post full solution??
v^{2}= \frac{2gh}{1+\frac{k^{2}}{R^{2}}}
where k= radius gyration .
it's derivation is given in Hc verma i think?
yaar someone solve it and show ..............i am getting
25mgh .......................... but answer is given
27mgh
in dc pandey the question is ....given
Rotational KE is given by 12Iw2..
Bt find w first...
mgh=1/2mv2+1/2Iw2
substituting I=25mr2, and v=wr, we get w=10gh7r2
Now KErot=12Iw2=12.25mr2.10gh7r2=27mgh