A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity ω. The coefficient of static friction between the bottom of the beaker and the surface of the disc is μ. The beaker will revolve with the disc if :
- \(R≤\frac{μg}{2ω^2}\)
- \(R≤\frac{μg}{ω^2}\)
- \(R≥\frac{μg}{2ω^2}\)
- \(R≥\frac{μg}{ω^2}\)
The Correct Option is A
Solution and Explanation
The correct option is(B): \(R≤\frac{μg}{ω^2}\)
To move together
ω2R ≤ μg
\(R≤\frac{μg}{ω^2}\)
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JEE Main Notification
Concepts Used:
Distance and Displacement
Distance:
The sum of the length of the path traveled by an object from one place to another is called distance. The path may or may not be directly from the initial point to the final point.
Distance is a scalar quantity and has only magnitude, also does not have any direction.
For example,
From the particular point, if a car travels to the east for 5 km and takes a turn to travel north for another 8 km, the total distance traveled by car shall be 13 km. The distance can never be zero or negative but should be always more than the displacement of the object. The distance of the object gives complete information about the path traveled by the object.
Read More: Difference between Distance and Displacement
Displacement:
The length of the shortest path from the initial point to the final point is called displacement. It is a vector quantity that consists of magnitude as well as direction.
For example,
Let's consider the same example given above, the total displacement of the object will be the length of the line joining the two positions. The displacement of an object is usually shorter or equal to the distance traveled by the object. The displacement of the object does not give the proper information about the path traveled by the object.