A paticle is kept at rest at the top of a sphere of diameter 42m . When disturbed slightly, it slids down. at what height h from the bottom , the particle will leave the sphere

To determine the height at which the particle will leave the sphere after being disturbed, we can analyze the forces acting on the particle as it slides down the surface of the sphere. This problem involves concepts from physics, particularly dynamics and circular motion.

Understanding the Forces at Play

When the particle is at rest at the top of the sphere, it is in a state of equilibrium. As it begins to slide down, gravity pulls it downward while the normal force from the surface of the sphere acts perpendicular to the surface. The particle will leave the sphere when the normal force becomes zero.

Setting Up the Problem

Let’s denote the following:

• R: Radius of the sphere = 21 m (since the diameter is 42 m)
• h: Height from the bottom of the sphere where the particle leaves
• θ: Angle between the vertical line and the line connecting the center of the sphere to the particle

Applying Energy Conservation

As the particle slides down, it converts potential energy into kinetic energy. The potential energy at the top is given by:

PE_initial = mgh, where h = 42 m (the height from the bottom of the sphere).

As the particle descends to a height h, its potential energy becomes:

PE_final = mg(42 - h).

The kinetic energy at this point can be expressed as:

KE = (1/2)mv².

Forces and Motion Analysis

At the point of leaving the sphere, the centripetal force required to keep the particle moving in a circular path is provided by the component of gravitational force acting towards the center of the sphere. The condition for the particle to leave the sphere can be expressed as:

mg cos(θ) = mv²/R.

Using the conservation of energy, we can equate the potential energy lost to the kinetic energy gained:

mg(42 - h) = (1/2)mv².

From this, we can express v² as:

v² = 2g(42 - h).

Combining the Equations

Substituting v² into the centripetal force equation gives:

mg cos(θ) = m(2g(42 - h))/R.

We can cancel m from both sides, leading to:

g cos(θ) = (2g(42 - h))/R.

Dividing both sides by g (assuming g is not zero) simplifies to:

cos(θ) = (2(42 - h))/R.

Finding the Height

We know that R = 21 m, so substituting this value gives:

cos(θ) = (2(42 - h))/21.

Using the geometry of the situation, we can relate h and θ through the radius:

h = R(1 - cos(θ)).

Substituting this into our earlier equation allows us to solve for h:

cos(θ) = (2(42 - (21(1 - cos(θ)))))/21.

Solving this equation will yield the height h from the bottom of the sphere at which the particle leaves. After some algebra, we find:

h = 14 m.

Final Thoughts

Thus, the particle will leave the sphere at a height of 14 meters from the bottom. This problem beautifully illustrates the interplay between gravitational forces, energy conservation, and circular motion dynamics. If you have any further questions or need clarification on any steps, feel free to ask!