Question

A particle is projected with velocity u horizontally from the top of a smooth sphere of radius 'a' so that it slides down the outside of the sphere. If the particle leaves the sphere when it has fallen a vertical distance \(\left(\frac{a}{4}\right)\), the value of \(u\) is : (1) \(\sqrt{a g}\) (2) \(\sqrt{2 a g}\) (3) \(\frac{\sqrt{a g}}{2}\) (4) \(\frac{3 \sqrt{a g}}{2}\)

Short answer

Option (4) \( \frac{3 \beta(gca)} ...\beta \) is correct.

Step-by-step solution

Analyze the vertical distance fallen

The particle falls a vertical distance of \(\frac{a}{4}\) from the top of the sphere.

Determine the height from the center of the sphere

If the particle falls \(\frac{a}{4}\) vertically, the vertical height from the top of the sphere to where the particle leaves is \(\frac{3a}{4}\).

Use the conservation of energy

Apply the principle of conservation of energy. The initial potential energy at the top of the sphere is converted into kinetic energy and potential energy at a height of \(\frac{3a}{4}\). Initially, the particle has potential energy \(mga\) and no kinetic energy.

Calculate the final energies at the height \( \frac{3a}{4} \)

At a height \(\frac{3a}{4}\), potential energy is \(\frac{3mga}{4}\) and kinetic energy is \( \frac{1}{2}mu^2\). Therefore: \[ mga = \frac{3mga}{4} + \frac{1}{2}mu^2 \] Simplify this equation to solve for \( u \).

Simplify the energy equation

Substitute the energies into the conservation of energy equation and solve for \( u \): \[ mga = \frac{3mga}{4} + \frac{1}{2}mu^2 \] \[ mga - \frac{3mga}{4} = \frac{1}{2}mu^2 \] \[ \frac{mga}{4} = \frac{1}{2}mu^2 \] Further simplify this to: \[ \frac{ga}{4} = \frac{1}{2}u^2 \]}

Solve for \( u \)

Isolate \( u \) to solve for its value: \[ u^2 = \frac{ga}{2} \] \[ u = \frac{\beta}{\beta(gca)} \]

Calculate the final value of \( u \) and compare with given options

Solve for \( u \): \[ u = \frac{\beta}{ \beta(a g)}\frac{2}{ oal} \]

Key concepts

Conservation of Energy

When a particle is projected from the top of a smooth sphere, the principle of conservation of energy becomes crucial to solve for the initial velocity. The principle states that the total energy in a system remains constant unless acted upon by an external force. For this problem, we consider both potential and kinetic energy.

Initially, the particle is at rest at the top of the sphere with only gravitational potential energy given by \(mga\), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(a\) is the radius of the sphere. As the particle slides down, some of this potential energy converts into kinetic energy. When the particle falls a vertical distance \(\left(\frac{a}{4}\right)\), it reaches a new height of \(\frac{3a}{4}\) from the center of the sphere.

The energy conservation equation at this height can be written as:

\[ mga = \frac{3mga}{4} + \frac{1}{2}mu^2 \]

Here, the left side represents the initial potential energy while the right side represents the sum of the remaining potential energy and the acquired kinetic energy. This equation allows us to solve for the particle's initial velocity \(u\).

Vertical Distance Calculation

Understanding how to calculate the vertical distance the particle has fallen is key to solving the problem. Initially, the particle is at the top of the sphere. We're given that it falls a vertical distance of \(\left(\frac{a}{4}\right)\).

To determine how far the particle has fallen relative to the center of the sphere, we subtract this distance from the initial height of the sphere's radius \(a\). This gives us:

\[ \text{Height from the center} = a - \frac{a}{4} = \frac{3a}{4} \]

This calculation is important because it helps us determine the new potential energy of the particle when it is at \(\frac{3a}{4}\) from the center. By knowing this height, we can correctly apply the principle of conservation of energy and solve for the initial velocity.

Kinetic and Potential Energy

Kinetic and potential energy are two fundamental concepts used to describe the energy transformations of the particle as it moves down the sphere.

Initially, at the top of the sphere, the particle has potential energy due to its height and no kinetic energy because it hasn't started moving yet. The potential energy at the top is given by:

\[ PE_0 = mga \]

As the particle falls, it loses some potential energy which transforms into kinetic energy. When the vertical distance fallen is \(\frac{a}{4}\), the height of the particle from the center of the sphere is \(\frac{3a}{4}\), giving it a new potential energy:

\[ PE_f = \frac{3mga}{4} \]

The energy that has been converted into kinetic energy can be found using:

\[ KE = \frac{1}{2}mu^2 \]

By the conservation of energy principle, the sum of the potential and kinetic energy at this point must equal the original potential energy:

\[ mga = \frac{3mga}{4} + \frac{1}{2}mu^2 \]

Rearranging and solving for the initial velocity \(u\) gives us:

\[ u = \sqrt{\frac{ga}{2}} \]

This equation shows how the height fallen affects both the potential and kinetic energy, and ultimately, the initial speed of the particle.