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Question
A heavy particle of mass m falls freely near the earth's surface. What is the torque acting on this particle about a point 50 cm east to the line of motion? Does this torque produce any angular acceleration in the particle?
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Solution
We know that
\[ \overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F} \]
Given:-
\[ \overrightarrow{r} = - 0 . 5 \hat i m\]
\[ \overrightarrow{F} = - mg \hat j \]
The torque becomes
\[ \overrightarrow{\tau} = 0 . 5\left( - \hat i \right) \times mg\left( - \hat j \right)\]
\[ \overrightarrow{\tau} = 0 . 5 \text{mg } \hat k \left[ \because \hat i \times \hat j = \hat k \right]\]
No, there will be no angular acceleration on the particle due to the torque.
Angular acceleration is given by \[\alpha = \frac{\tau}{I}\]. As the particle here moves in a straight line, the centre of rotation lies at a distance infinity \[(r = \infty );\] so, moment of inertia \[(I = m r^2 )\] of the particle is infinity.
\[ \therefore \alpha = 0\]
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