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`        A small hamster is put into a circular wheel cage, which has a frictionless center pivot. A horizontal platform is fixed to the wheel bellow the pivot. Initialy the hamster is at rest at one end of the platform. When the platform is released, the hamster starts running but because of the motion of the hamster, the platform and wheel remains stationary. Find the magnitude of acceleration of hamster when it is at a distance x from the center of the horizontal platform. (See attached image for pic)`
2 years ago

Arun
23506 Points
```							DearLet the midpoint of the platform be a distance h below the pivotBecause of gravity the hamster exerts a torque mgx about the pivot of the wheel-cage. On the other hand, as the hamster moves it accelerates using friction with the platform. When its acceleration is a this produces a reaction force of ma on the platform, directed away from its midpoint. The torque due to this force is mah. The wheel-cage (and the platform) remains in static equilibrium if these two torques are equal, i.e.mgx = mah.After making due allowance for its direction, the acceleration can thus be written as a = -(g/h)x. This shows that the required motion of the hamster is simple harmonic motion with an angular frequency = root(g/h).RegardsArun (askIITians forum expert)
```
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