Loop Physics Question

Question

A small block of mass m = 1.50 kg slides, without friction

along the loop-the-loop track shown. The block starts from

the point P a distance h = 55.0 m above the bottom of the

loop of radius R = 20.0 m. What is the kinetic energy of the

mass at the point A on the loop?

I found the KE to be 221 J and the downward acceleration at the top of the loop to be 14.7 m/s^2

I need help finding this question.

What is the minimum height h for which the block will reach?

point A on the loop without leaving the track?

Answer 1

To determine the minimum height h for which the block will reach point A on the loop without leaving the track, we need to use the conservation of mechanical energy.

At point P, the block has only potential energy, which is given by:

U_P = mgh

where m is the mass of the block, g is the acceleration due to gravity, and h is the height of point P above the bottom of the loop.

At point A, the block has both potential energy and kinetic energy, which are given by:

U_A = mg(2R) K = (1/2)mv^2

where R is the radius of the loop, and v is the speed of the block at point A.

Since there is no friction, the mechanical energy is conserved, so we have:

U_P = U_A + K

Substituting the expressions for U_P, U_A, and K:

mgh = mg(2R) + (1/2)mv^2

Simplifying and canceling out the mass m:

gh = 2R + (1/2)v^2

We can also use the fact that at the top of the loop, the centripetal force required to keep the block moving in a circle is provided by the normal force of the track:

mg = N + mv^2/R

where N is the normal force.

Solving for the normal force and substituting into the expression for kinetic energy:

N = mg - mv^2/R K = (1/2)mv^2 = (1/2)mg(2R - h)

Substituting the expression for N and simplifying:

K = (1/2)mg(2R - h - v^2/g)

At the minimum height h where the block just reaches point A, the speed of the block at point A is just enough to keep it moving in a circular path. This means that the block will lose contact with the track if it has any less speed at point A.

At the top of the loop (point A), the normal force of the track must be equal to zero, otherwise the block would have a normal force inwards towards the center of the loop that would prevent it from continuing on its path. This means that the centripetal force is only provided by gravity, and we have:

mg = mv^2/R

Solving for the speed v:

v = sqrt(gR)

Substituting this expression for v into the expression for kinetic energy:

K = (1/2)mg(2R - h - gR)

Setting K to zero (since this is the minimum height where the block can reach point A), and solving for h:

h = 5R/2

Substituting the given values of R:

h = 50.0 m

Therefore, the minimum height h for which the block will reach point A on the loop without leaving the track is 50.0 meters.