A solid sphere of mass m and radius r rolls with out slipping along the track.it starts from rest with the lowest point of tthe sphere at aheight h above the bottom of the loop of radius R,much larger than r.(a) what is the minimum value of h (interms of R)such that the sphere completes the loop?(b)what are the force components on the sphere at the point if h=3R?

a) 27/10R

To find the minimum value of h for the sphere to complete the loop, we can start by analyzing the forces acting on the sphere at the highest point of the loop. At this position, the sphere is momentarily at rest before descending. The forces acting on the sphere at the highest point are the gravitational force (mg) directed downwards and the normal force (N) directed upwards.

(a) For the sphere to complete the loop, the net force towards the center of the loop (centripetal force) at the highest point must be greater than or equal to the gravitational force. The centripetal force is provided by the normal force (N). Therefore, we have the equation:

N ≥ mg

At the highest point of the loop, the normal force N can be calculated using the equation:

N = mg + mv²/R

Where v is the velocity of the sphere at the highest point and R is the radius of the loop.

Since the sphere starts from rest, its initial velocity (v₀) at the bottom of the loop can be calculated using the principle of conservation of energy. The potential energy at the highest point is fully converted to kinetic energy at the bottom of the loop, given by the equation:

mgh = (1/2)mv₀²

Simplifying this equation allows us to find the value of v₀:

v₀ = √(2gh)

Plugging this value of v₀ into the equation for N, we have:

N = mg + m(2gh)/R

To complete the loop, the net force towards the center of the loop must be greater than or equal to the gravitational force:

mg ≤ mg + m(2gh)/R

Simplifying this inequality:

0 ≤ m(2gh)/R

This equation tells us that for the sphere to complete the loop, h must be greater than or equal to zero. Therefore, there is no minimum value of h (in terms of R) for the sphere to complete the loop.

(b) For the point when h = 3R, we need to calculate the force components on the sphere. At this point, the forces acting on the sphere are:

1. Gravitational force (mg) acting vertically downwards.
2. Normal force (N) acting vertically upwards.
3. Frictional force (f) acting horizontally towards the center of the loop.

The force components at this point can be calculated as follows:

Vertical Force Component:
The vertical force component is the sum of the gravitational force and the normal force:

Vertical Force Component = mg + N

Since the sphere is in equilibrium at this point, the gravitational force is equal to the normal force:

Vertical Force Component = 2mg

Horizontal Force Component:
The horizontal force component is provided by the frictional force:

Horizontal Force Component = f

To find the frictional force, we need to determine the maximum value of static friction. The maximum value of static friction is given by the equation:

f(max) = μsN

Where μs is the coefficient of static friction. For a rolling sphere, the coefficient of static friction between the sphere and the track is given by:

μs = (2/7)(r/R)

Plugging in the value of μs, we have:

f(max) = (2/7)(r/R) * 2mg

Therefore, the horizontal force component is given by:

Horizontal Force Component = f(max) = (2/7)(r/R) * 2mg

So, the force components on the sphere at the point when h = 3R are:
Vertical Force Component = 2mg
Horizontal Force Component = (2/7)(r/R) * 2mg