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`        A system contains 2 identical small bobs of mass 2kg each connected to the ends of 1m long light rod. The system is rotating about a fixed axis through centre of the rod and perpendicular to it at an angular speed of 9 rad/s. An impulsive force of average magnitude 10 N acts on one of the masses in the direction of its velocity for 0.2s. calculate new angular velocity of system`
2 years ago

Sanju
106 Points
```							The impulse is provided by the force F = 10N in the direction of velocity which is perpendicular to the rod as the mass is moving in circular motion.Therefore, the torque, T = rF = 0.5m * 10N = 5 NmMoment of inertia of the system abt the line passing thru its center & perpendicular to it is given by:I = 2 * 2kg * (0.5m)2 = 1 kg m2 We have, T = I * αwhere α is the angular accelerationTherefore, α = T / I = 5/1 = 5 rad/s2 So, the new angular velocity, w = w0 +αt   w = w0 +αt = 9 + (5*0.2) = 10 rad/s
```
2 years ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions