A point mass block slides without friction on the loop-de-loop track of radius R as shown. From what height h must it be released from rest in order to make it around the loop without leaving the track?

Show the equation and then plug the numbers into that equation and show the answer at the end

Still need help!!

thu

To determine the height from which the point mass block must be released to make it around the loop without leaving the track, we can use the principle of conservation of mechanical energy.

At the highest point of the loop, all the potential energy is converted to kinetic energy. At the bottom of the loop, the kinetic energy is at its maximum.

Let's denote:

h = height from which the point mass block is released
R = radius of the loop
g = acceleration due to gravity (approximately 9.8 m/s^2)

The potential energy at the highest point is given by mgh, where m represents the mass of the block.

The kinetic energy at the bottom of the loop is given by (1/2)mv^2, where v represents the velocity of the block.

Since there is no friction and no other external forces acting on the block, the total energy is conserved, thus we can equate the potential energy at the highest point to the kinetic energy at the bottom of the loop:

mgh = (1/2)mv^2

The mass of the block cancels out, and we get:

gh = (1/2)v^2

To solve for h, we need to find an expression for v in terms of h and R. Since the block makes a complete loop, v can be determined from the centripetal acceleration as it goes around the loop.

At the top of the loop, the net force acting on the block is the gravitational force, which provides the centripetal force:

mg = mv^2/R

Simplifying this equation, we get:

v^2 = gR

Plugging this expression for v^2 back into the equation relating h and v, we have:

gh = (1/2)gR

Canceling the factor of g, we get:

h = (1/2)R

Substituting the values of R into this equation, we can find the height h:

h = (1/2)(R) = R/2

Therefore, the point mass block must be released from a height of h = R/2 to make it around the loop without leaving the track.

Assuming a specific radius R is given, you can plug in the value of R into the equation h = R/2 to find the exact value for the height.

To determine the height h from which the point mass block must be released in order to make it around the loop without leaving the track, we can use the principle of conservation of mechanical energy.

At the highest point on the loop, the point mass block only has potential energy, and no kinetic energy. As it slides down the loop, its potential energy is converted into kinetic energy.

Let's start by considering the initial and final states of the point mass block:

Initial state: The block is at height h above the bottom of the loop and is at rest. Its potential energy is given by mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height.
Final state: The block reaches the highest point of the loop, where it momentarily loses contact with the track. Its potential energy is zero, and it has maximum kinetic energy.

Using the principle of conservation of energy, we equate the initial potential energy to the final kinetic energy:

mgh = (1/2)mv^2

where v is the velocity of the block at the top of the loop.

To find v, we can use the concept of circular motion. At the top of the loop, the centripetal force required to keep the block moving in a circular path is provided by the gravitational force on the block. This means that:

mv^2 / R = mg

where R is the radius of the loop.

Now we can solve for v:

v^2 = gR

Substituting this value of v into the conservation of energy equation, we have:

mgh = (1/2)m(gR)

Simplifying, we obtain:

h = (1/2)R

Thus, the height from which the point mass block must be released is half the radius of the loop.

Let's plug in some numbers to see the answer for a specific case. For example, if R = 10 meters, then:

h = (1/2)(10) = 5 meters

Therefore, the point mass block must be released from a height of 5 meters to make it around the loop without leaving the track.