MHT CET · Physics · Rotational Motion
A mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is L. if the string is now halved, keeping angular velocity same, then the angular momentum will be
- A L
- B \(\frac{\mathrm{L}}{4}\)
- C \(2 \mathrm{~L}\)
- D \(\frac{\mathrm{L}}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{L}}{4}\)
Step-by-step Solution
Detailed explanation
The correct option is (B).
Concept: Angular momentum is given by \(\mathrm{L}=\mathrm{m} \omega \mathrm{r}^2\), where \(\mathrm{m}\) is the mass, \(\omega\) the angular velocity and \(\mathrm{r}\) is the radius of the circular motion (length of the string).
If the new length is half of the original string, then the new angular momentum would be,
\(\mathrm{L}_{\mathrm{n}}=\mathrm{m \omega}\left(\frac{\mathrm{r}}{2}\right)^2\)
Or
\(\mathrm{L}_{\mathrm{n}}=\left(\mathrm{M} \omega \mathrm{r}^2\right)=\frac{1}{4}=\frac{\mathrm{L}}{4}\)
Concept: Angular momentum is given by \(\mathrm{L}=\mathrm{m} \omega \mathrm{r}^2\), where \(\mathrm{m}\) is the mass, \(\omega\) the angular velocity and \(\mathrm{r}\) is the radius of the circular motion (length of the string).
If the new length is half of the original string, then the new angular momentum would be,
\(\mathrm{L}_{\mathrm{n}}=\mathrm{m \omega}\left(\frac{\mathrm{r}}{2}\right)^2\)
Or
\(\mathrm{L}_{\mathrm{n}}=\left(\mathrm{M} \omega \mathrm{r}^2\right)=\frac{1}{4}=\frac{\mathrm{L}}{4}\)
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