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A hollow vertical cylinder of radius R a...

A hollow vertical cylinder of radius R and height h has a smooth internal surface. A small particle is held in contact with the inner side of upper rim at a point P. It is given a horizontal speed `v_(0)` tangential to rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will be [Take acceleration due to gravity g]

A

`(h)/(2piR)`

B

`(v_(0))/(h)sqrt((h)/(2g))`

C

`(v_(0))/(2piR)sqrt((2h)/(g))`

D

`(v_(0)pi)/(g)sqrt((2gh)/(2R))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the number of revolutions made by a small particle as it travels from the upper rim of a hollow vertical cylinder to the lower rim after being given an initial horizontal speed \( v_0 \). ### Step-by-Step Solution: 1. **Understanding the Motion**: - The particle is released from point P at the top of the cylinder with a horizontal speed \( v_0 \). - It will fall vertically down to point Q at the bottom of the cylinder due to gravity. 2. **Vertical Motion Analysis**: - The height of the cylinder is \( h \). - The initial vertical velocity \( u_y = 0 \) (since it starts moving horizontally). - The vertical displacement \( y = h \). - Using the equation of motion: \[ y = u_y t + \frac{1}{2} g t^2 \] Substituting the known values: \[ h = 0 + \frac{1}{2} g t^2 \] Rearranging gives: \[ t^2 = \frac{2h}{g} \] Therefore, the time taken to fall is: \[ t = \sqrt{\frac{2h}{g}} \] 3. **Horizontal Motion Analysis**: - The horizontal distance traveled by the particle while it falls is determined by its horizontal speed \( v_0 \) and the time \( t \): \[ \text{Horizontal Distance} = v_0 \cdot t \] Substituting \( t \): \[ \text{Horizontal Distance} = v_0 \cdot \sqrt{\frac{2h}{g}} \] 4. **Calculating the Number of Revolutions**: - The circumference of the cylinder is \( 2\pi R \). - The number of revolutions \( n \) made by the particle can be calculated by dividing the horizontal distance by the circumference: \[ n = \frac{\text{Horizontal Distance}}{\text{Circumference}} = \frac{v_0 \cdot \sqrt{\frac{2h}{g}}}{2\pi R} \] - Simplifying gives: \[ n = \frac{v_0}{2\pi R} \cdot \sqrt{\frac{2h}{g}} \] ### Final Answer: The number of revolutions made by the particle is: \[ n = \frac{v_0}{2\pi R} \sqrt{\frac{2h}{g}} \]
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