The small mass m sliding without friction along the looped track is to remain on the track at all times, even at the very top of the loop of radius r. (not unlike a roller coaster)

A)In terms of the given quantities, determine the minimum release height h. (i got this to be 2.5)

B)If the actual release height is 4h, calculate the normal force exerted by the track at the bottom of the loop.

C)If the actual release height is 3h, calculate the normal force exerted by the track at the top of the loop.

D)If the actual release height is 3h, calculate the normal force exerted by the track after the block exits the loop onto the flat section.

I'm more interested in the equations to use than receiving the answer.

A) To determine the minimum release height, we need to consider the energy of the system at the highest point of the loop. At the top of the loop, the only forces acting on the mass are its weight (mg) and the normal force (N) exerted by the track. Since the mass is not in contact with the track at the top, the normal force must be equal to zero. In this case, we can use the following equation:

N + mg = 0

Solving for N, we get:

N = -mg

Since the normal force cannot be negative, the mass should remain in contact with the track at the top. Therefore, we have:

N > 0

For the mass to remain on the track at all times, the normal force should be greater than zero. Therefore, the minimum release height h can be determined by finding the height at which the normal force becomes zero:

0 + mg = 0

So, there is no net force in the vertical direction at the top of the loop. By applying the centripetal force condition, we have:

mg = mv^2 / r

Where v is the speed of the mass at the top of the loop. Solving for v:

v = sqrt(rg)

Now, we can substitute this into the kinetic energy equation at the top of the loop:

mgh = (1/2)mv^2

mgh = (1/2)m(rg)

h = (1/2)g

Substituting the value of g (acceleration due to gravity), we have:

h = 5/2

So, the minimum release height is 2.5 units.

B) To calculate the normal force at the bottom of the loop when the actual release height is 4h, we need to consider the forces acting on the mass at that point. The forces acting on the mass are its weight (mg) and the normal force (N) exerted by the track. Since the mass is in contact with the track at the bottom, the normal force will be greater than zero. We can use the following equation:

N - mg = mv^2 / r

Substituting the value of v from the energy equation at the top of the loop (sqrt(rg)), we have:

N - mg = mg

Simplifying, we get:

N = 2mg

So, the normal force exerted by the track at the bottom of the loop is twice the weight of the mass.

C) To calculate the normal force at the top of the loop when the actual release height is 3h, we can use the same equation as in part A:

N + mg = mv^2 / r

Substituting the value of v from the energy equation at the top of the loop (sqrt(rg)), we have:

N + mg = mg

Simplifying, we get:

N = 0

So, the normal force exerted by the track at the top of the loop is zero. Remember, in order for the mass to remain on the track, the normal force must be greater than zero. Therefore, the mass will lose contact with the track at the top of the loop when released from a height of 3h.

D) To calculate the normal force exerted by the track after the block exits the loop onto the flat section, we need to consider the forces acting on the mass in that region. The forces acting on the mass are its weight (mg) and the normal force (N) exerted by the track. Since the mass is in contact with the track on the flat section, the normal force will be greater than zero. We can use the same equation as in part B:

N - mg = 0

Simplifying, we get:

N = mg

So, the normal force exerted by the track after the block exits the loop onto the flat section is equal to the weight of the mass.

To approach this problem, we can make use of energy conservation and centripetal force concepts. Let's go step by step to find the answers.

A) To determine the minimum release height h, we need to ensure that the block doesn't lose contact with the track at the top of the loop of radius r. At the top point, the normal force should be equal to zero for an instant.

At the top point, the block is undergoing circular motion with a radius equal to r. The net force acting on the block is the gravitational force (mg) downward. To ensure the normal force is zero, we need to equate the centripetal force with the gravitational force:

mv^2/r = mg

where v is the velocity of the block at the top point. The velocity of the block can be found by using the conservation of energy. The conservation of energy equation states that the initial kinetic energy plus the initial potential energy equals the final kinetic energy plus the final potential energy.

Initial kinetic energy (at the top point) = 0
Initial potential energy (at the release height h) = mgh

Final kinetic energy (at the top point) = 1/2mv^2
Final potential energy (at the top point) = mgh + 2mgr (since it has fallen by 2r)

Setting up the conservation of energy:

mgh = 1/2mv^2 + mgh + 2mgr

Simplify and solve for v:

gh = 5/2v^2

v^2 = 2gh/5

Now, substitute the value of v^2 into the centripetal force equation:

2mv^2 / r = mg

2m(2gh/5) / r = mg

4gh / 5 = gr

h = 5r/4

So the minimum release height, h, in terms of the given quantities, is 5r/4.

B) If the actual release height is 4h, we can use the conservation of energy equation again to find the velocity v at the bottom of the loop. Then, we can calculate the normal force exerted by the track at the bottom.

mgh = 1/2mv^2 + mgh + 2mgr

4mgh = 1/2mv^2 + m(4h)g + 2mgr

4gh = v^2 + 4gh + 4gr

v^2 = -4gr

Since the velocity is negative, it means the block is moving in the opposite direction of the chosen positive direction. However, we only need the magnitude of the velocity. So, we can drop the negative sign:

v^2 = 4gr

Now, we can find the normal force using the centripetal force equation:

mg + N = mv^2/r

mg + N = m(4gr)/r

mg + N = 4mg

N = 3mg

So, the normal force exerted by the track at the bottom of the loop is 3 times the gravitational force, or 3mg.

C) If the actual release height is 3h, we can again use the conservation of energy equation to find the velocity v at the top of the loop. Then, we can calculate the normal force exerted by the track at the top.

mgh = 1/2mv^2 + mgh + 2mgr

3mgh = 1/2mv^2 + m(3h)g + 2mgr

3gh = v^2 + 3gh + 2gr

v^2 = -2gr

Again, we can drop the negative sign and find the magnitude of the velocity:

v^2 = 2gr

Now, we can find the normal force using the centripetal force equation:

mg - N = mv^2/r

mg - N = m(2gr)/r

mg - N = 2mg

N = mg

So, the normal force exerted by the track at the top of the loop is equal to the gravitational force, or mg.

D) After the block exits the loop onto the flat section, there is no curvature, and it continues to slide without any circular motion. So, the normal force exerted by the track will be equal to the gravitational force, or mg.

To solve each part of the given problem, we'll need to consider the conservation of energy and the forces acting on the mass at different points in the loop. Let's break down each part and determine the equations we can use.

A) Determining the minimum release height (h) at the given quantities:

To determine the minimum release height, we need to set the gravitational potential energy at the top of the loop equal to the kinetic energy at the bottom of the loop, ensuring that the mass doesn't lose contact with the track.

At the top of the loop:
- Gravitational potential energy: mgh
- Kinetic energy: 0 (at the top, the mass comes to rest momentarily)

At the bottom of the loop:
- Gravitational potential energy: mgh
- Kinetic energy: 1/2mv^2

Setting these equal to each other, we have:

mgh = 1/2mv^2

Canceling out the mass and rearranging the equation, we get:

2gh = v^2

Now, let's consider the relationship between velocity and radius at the bottom of the loop.
- At the bottom of the loop, the net force acting on the mass is the sum of gravitational force (mg) and the centripetal force (mv^2/r).

Setting these forces equal to each other, we have:

mg + mv^2/r = mv^2/r

Canceling out the mass and rearranging the equation, we get:

v^2 = gr

Now we can substitute this relation into our previous equation:

2gh = gr

Rearranging the equation to solve for h:

h = r/2

Substituting the given radius (r) yields the minimum release height.

B) Calculating the normal force at the bottom of the loop (4h):

At the bottom of the loop, the forces acting on the mass are the gravitational force (mg) and the normal force (N).

We can use Newton's second law to determine the net force acting on the mass at the bottom of the loop:

mg - N = mv^2/r

To solve for the normal force (N), rearrange the equation:

N = mg - mv^2/r

Substituting the given release height (4h) allows us to calculate the normal force.

C) Calculating the normal force at the top of the loop (3h):

At the top of the loop, the forces acting on the mass are the gravitational force (mg) and the normal force (N).

Using Newton's second law, we can determine the net force acting on the mass at the top of the loop:

N - mg = mv^2/r

Rearranging the equation to solve for the normal force (N):

N = mg + mv^2/r

By substituting the given release height (3h), we can calculate the normal force.

D) Calculating the normal force after the block exits the loop onto the flat section (3h):

After the block exits the loop onto the flat section, the only force acting on the mass is gravity (mg).

Therefore, the normal force exerted by the track on the mass will be equal to the gravitational force (mg).

Substituting the given release height (3h), we can calculate the normal force.

Remember to substitute the appropriate numerical values for each part of the problem to obtain the final answers.

i can't figure how to set this up. it starts at height h and there's a loop at at ground level with radius r.