Saved Bookmarks
| 1. |
A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity omega_(0) . When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platformwill vary with time t as |
|
Answer»
Let the tortoise move ALONG the chord PQ. When the tortoise MOVES from P to M its distance from axis point O decreases and so the moment of inertia decrease as `I=mr^(2)`, where r = distance of tortoise from O. When the tortoise moves from M to Q the distance r increases and so I also increases. `I_(P) = mR^(2) + (MR^(2))/2`..........(i) `I_(N) = mr^(2) + (MR^(2))/2`.......(ii)  By geometry, `r^(2) =d^(2) + [sqrt(R^(2) -d^(2))-vt]^(2)`........(iii) Angular momentum is conserved `I_(p)omega_(0) = I_(N)omega(t)` or `omega(t) =(I_(p)omega_(0))/I_(N)`.........(iv) `omega(t)` DEPENDS on `I_(N'), I_(N)` depends on r and r depends on time t. The function of t is non-linear. HENCE, `omega(t)` is a non-linear function of t. `omega` increases when tortoise from P to M and `omega` decreases when tortoise travels from M to Q. First increases and then decrease of `omega` is revealed in graph (b) and (d). But `omega(t)`is a non-linear function of t, hence graph (b) represents the variation of `omega`(t) w.r.t. time. |
|
