NAEST 2020 Oscillations of a Banana Ghor, Tension in the Str
By Question:
1: Look at the banana Ghor (the whole thing hanging from the rope) oscillating. Measure the time period of the Ghor using your mobile stop watch. The time period of oscillation is closest to
रस्सी से लटकते हुए केले के घोर के दोलन को देखें। इस दोलन का आवर्तकाल अपने मोबाइल के स्टाॅपवाच से निकालें। आवर्तकाल का मान इनमें से किसके सबसे नजदीक है
- 2.20 s
- 2.25 s
- 2.30 s
- 2.35 s
- 2.5 s
- $T$ remains the same in all parts of the motion
- $T$ is maximum when the Ghor is at its lowest position
- $T < Mg$ at the extreme position of the Ghor
- $T > Mg$ at the lowest position of the Ghor.
- पूरी गति के दौरान $T$ एक समान रहता है
- जब घोर अपने सबसे नीचे के स्थान पर रहता है तब $T$ सर्वाधिक होता है
- जब घोर अपने सबसे किनारे के स्थान पर रहता है तब $T < Mg$
- जब घोर अपने सबसे नीचे के स्थान पर रहता है तब $T > Mg$
- will increase
- will decrease
- will remain same
- will decrease in proportion to the weight of the Ghor
- बढ़ जायेगा
- घट जायेगा
- अपरिवर्तित रहेगा
- घोर के भार के अनुपात में घट जायेगा
Solution 1: Option (D) looks to be correct.
I measured time for 25 oscillations and it comes out to be 58.32 s. Thus, the average time period is 58.32/25 = 2.33 s.
The amplitude of the oscillations reduces as time passes. What about the variation in time period? Can somebody find the time periods using the initial 5 oscillations and the final 5 oscillations?
Solution 2: Options (B), (C) and (D) is correct.
The centre of mass of the Ghor moves in a circular arc. The radial component of its acceleration is equal to the centripetal acceleration $v^2/l$, where $v$ is the speed and $l$ is the distance of the centre of mass from the point of suspension. Note that the speed $v$ varies as the Ghor oscillates. The speed is zero at the two extreme ends and it is maximum at the lowest point. Thus, radial component of the acceleration is zero at the two ends and it is maximum at the lowest point (in the middle).
Let $\theta$ be the angle made by the rope with the vertical and T be tension in the rope. The force on the Ghor in the radial direction is $T-Mg\cos\theta$. Apply Newton's second law in the radial direction to get
\begin{align}
T - Mg \cos\theta = Mv^2/l.\nonumber
\end{align}
Hence, tension in the string is given by
\begin{align}
T = Mg \cos\theta + Mv^2/l.
\end{align}
Its value at the extreme positions is
\begin{align}
T = Mg \cos\theta < Mg.
\end{align}
And tension at the lowest point is
\begin{align}
T= Mg + M v_\text{max}^2/l.
\end{align}
Solution 3: Option (A) is correct.
The oscillating Ghor has a large number of green bananas and only a few yellow bananas. Hence, we can assume its centre of mass to lie below the yellow bananas. When all yellow bananas are plucked, the centre of mass further goes down. If we assume Ghor to be a simple pendulum then its time period $T=2\pi\sqrt{l/g}$ will increase due to an increase in $l$.
However, approximate Ghor to be a simple pendulum is a crude approximation. It is more close to a physical pendulum. The time period of a physical pendulum depends on the mass and its distribution (i.e., the moment of inertia $I$).
\begin{align}
T=2\pi\sqrt{\frac{I}{Mgl}}=2\pi\sqrt{\frac{k^2}{gl}},
\end{align}
where $k$ is the radius of Gyration $k=\sqrt{I/M}$. When yellow bananas are plucked, both $k$ and $l$ increases. The answer will depend on many factors.
A good apploximation is to assume Ghor to be solid cylinder of radius $R$ and length $L$. Its moment of inertia about the axis of rotation is
\begin{align}
I=Ml^2+ \frac{1}{12}ML^2+\frac{1}{4}MR^2
\end{align}
and time period is
\begin{align}
T=2\pi\sqrt{\frac{l}{g}+\frac{L^2}{12gl}+\frac{R^2}{4gl}}.
\end{align}
When all yellow bananas are plucked, $l$ increases and $L$ decreases. It is difficult to comment on the variation of $T$ unless we get the numerical values of these parameters.
The real problems are much more difficult than textbook problems.
More Details about Physical Pendulum
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