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Chapter 3: Problem 168

A can filled with water is revolved in a vertical circle of radius \(4 \mathrm{~m}\) and the water just does not fall down. The time period of revolution will be (a) \(1 \mathrm{sec}\) (b) \(10 \mathrm{sec}\) (c) \(8 \mathrm{sec}\) (d) \(4 \mathrm{sec}\)

Short Answer

Expert verified
The time period of the revolution is 4 seconds, therefore, the correct answer is (d) \(4\; sec\)

Step by step solution

01

Calculate the minimum necessary velocity

The minimum necessary velocity (v) to keep the water in the can at the topmost point can be calculated using the formula \( v = \sqrt{gr} \) (where g represents the acceleration due to gravity and r is the radius). Using g = \(9.8\; m/s^2\) and r = 4 m, we obtain \( v = \sqrt{9.8 \times 4} = 6.26\; m/s\).
02

Calculate the time period

Next, we use the formula for the period (T) of revolution, which is \( T = \frac{2\pi r}{v} \). Substituting the radius r = 4 m and velocity v = 6.26 m/s, we get \( T = \frac{2\pi \times 4}{6.26} = 2\; sec\). However, in the problem, we are rotating the can two times (up and down) in one period, therefore, the actual time period will be twice the calculated, it will be 2*2 = 4 sec.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Minimum Velocity in a Vertical Circle
When an object is in circular motion, such as the can of water in our exercise, it needs a specific speed to remain in motion without falling. This necessary speed at the highest point of a vertical circle is called the **minimum velocity**. At this point, the gravitational force helps the water to stay in the can. If the speed is too low, the water would fall out due to gravity. For an object to maintain motion in a vertical circle, the minimum velocity at the topmost point is derived from the equation: \[ v = \sqrt{gr} \] Where:
  • **g** is the acceleration due to gravity, approximately \(9.8\, \text{m/s}^2\) on Earth.
  • **r** is the radius of the circle.
This formula helps determine the minimum speed an object needs to stay in the loop. In our scenario, with a circle radius of 4 meters, we calculated a minimum velocity of approximately 6.26 m/s. This speed ensures that the water stays inside the rotating can at the top of its path.
The Dynamics of a Vertical Circle
Understanding motion in a **vertical circle** involves recognizing how forces interact to keep an object in continuous circular motion. As an object moves along a vertical circle, forces like gravity and tension (or centrifugal force) govern the motion, influencing the speed required to maintain the circle. In the case of a can revolved in a vertical circle:
  • When at the top of the circle, gravity acts downward, pulling the object down, while the velocity we calculated ensures that the object completes the loop safely.
  • At the bottom of the circle, the velocity needed is higher as it has to counteract the gravitational force trying to pull it back to the center.
For water to not spill, it's crucial to maintain this balance between velocity and the forces at play, especially at the topmost point where the risk of the contents falling out due to gravity is highest. This balance ensures smooth rotation and avoids interruptions in the circular path.
Calculating the Time Period of Revolution
The **time period of revolution** is the time it takes for an object to make a complete circle path. To find this, we use the calculated velocity and other circle properties.The formula to calculate the time period \(T\) is: \[ T = \frac{2\pi r}{v} \] Where:
  • **2蟺r** represents the circumference of the circle, the total distance traveled in one complete revolution.
  • **v** is the velocity.
In this problem, we determined the initial calculated period to be 2 seconds using the radius of 4 m and minimum velocity of 6.26 m/s. However, remember, since the can travels twice through the vertical path (once going up and once coming down), the actual period is double this value, leading us to a total time period of 4 seconds. This crucial adjustment accounts for the full journey in both upward and downward arcs of the vertical rotation.

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Most popular questions from this chapter

A ball rolls off top of a staircase with a horizontal velocity \(u \mathrm{~m} / \mathrm{s}\). If the steps are \(h\) metre high and \(b\) mere wide, the ball will just hit the edge of nth step if \(n\) equals to (a) \(\frac{h u^{2}}{g b^{2}}\) (b) \(\frac{u^{2} 8}{g b^{2}}\) (c) \(\frac{2 h u^{2}}{g b^{2}}\) (d) \(\frac{2 u^{2} g}{h b^{2}}\)

If a cycle wheel of radius \(4 m\) completes one revolution in two seconds. Then acceleration of the cycle will be (a) \(\pi^{2} m / s^{2}\) (b) \(2 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}\) (c) \(4 \pi^{2} m / s^{2}\) (d) \(8 \pi \mathrm{m} / \mathrm{s}^{2}\)

A particle moves with a constant speed \(v\) along a circular path of radius \(r\) and completes the circle in time \(T\). What is the acceleration of the particle \(\quad\) [Orissa JEE 2002] (a) \(m g\) (b) \(\frac{2 \pi v}{T}\) (c) \(\frac{\pi r^{2}}{T}\) (d) \(\frac{\pi v^{2}}{T}\)

With what angular velocity should a \(20 \mathrm{~m}\) long cord be rotated such that tension in it, while reaching the highest point, is zero (a) \(0.5 \mathrm{rad} / \mathrm{sec}\) (b) \(0.2 \mathrm{rad} / \mathrm{sec}\) (c) \(7.5 \mathrm{rad} / \mathrm{sec}\) (d) \(0.7 \mathrm{rad} / \mathrm{sec}\)

A cyclist taking turn bends inwards while a car passenger taking same turn is thrown outwards. The reason is [NCERT 1972] (a) Car is heavier than cycle (b) Car has four wheels while cycle has only two (c) Difference in the speed of the two (d) Cyclist has to counteract the centrifugal force while in the case of car only the passenger is thrown by this force
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