MCQ
A solid cylinder rolls down an inclined plane from a height $h$. At any moment the ratio of rotational kinetic energy to the total kinetic energy would be
- A$1:2$
- ✓$1:3$
- C$2:3$
- D$1:1$
✓
Answer
Correct option: B.
$1:3$
b
The cylinder has rotational $KE = K _{1}=\frac{1}{2} Iw ^{2}$
The cylinder has rotational $KE = K _{1}=\frac{1}{2} Iw ^{2}$
where $I =\frac{ MR ^{2}}{2}$ about $COM$ and $w =\frac{ V }{ R }$
$\Rightarrow K _{1}=\frac{1}{2}\left(\frac{ M R ^{2}}{2}\right)\left(\frac{ V }{ R }\right)^{2}=\frac{1}{4} MV ^{2}$
If it has a velocity of $COM$, then translatory $KE$ is $K _{2}=\frac{1}{2} MV ^{2}$
$\therefore$ Total $KE = K = K _{1}+ K _{2}=\frac{3}{4} MV ^{2}$
Required ratio $\frac{ K _{1}}{ K }=\frac{1 / 4 MV ^{2}}{3 / 4 MV ^{2}}=\frac{1}{3}$
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