Dynamics: Work energy applications

Recall that centripetal force is provided by another force.

1. Sled sliding on frictionless surface

Let's consider a sled starting from from rest (v₁ = 0) at the top of an icy hill. The path from 1 to 2 is circular with radius R, that we assume uniform. We neglect frictions. We want to determine the speeds of the sled and the normal forces exerted by the ice on it at the points 2, 3, and 4.

We recall that centripetal force is provided by amother force. Here it provides the resultant N + W, that is the normal force exerted by the ice on the sled + the weight W = mg of the sled. Further, we apply the work-energy theorem. (1/2)m(vf² - vi²) = - mg (xf - xi)

1. At the point 1:

v₁ = 0

1. At the point 2:

(1/2) m (v₂² - v₁²) = - mg (0 - h)
(1/2) m v₂² = mgh
v₂ = [2gh]^1/2

v₂ = [2gh]^1/2

The second Newton's law is written as:
W + N provides centripetal force = Fc
This force is radial, then oriented toward the center of the circle.
Therefore:
The vector relationship is:
N + W = Fc
Their components along + y-axis gives:

+ N - mg = + Fc, that is: N - mg = mv₂²/R, Thus:
N = mg + mv₂²/R

N = mg + mv₂²/R

1. At the point 3:

At this point, the component of the weight is zero along the x-axis, so
N = Fc, that is:
N = mv₃²/R

N = mv₃²/R

Applying the work-energy theorem gives:
(1/2) m (v₃² - v₂²) = - mg (R - 0)
(1/2) m (v₃² - v₂²) = - mgR
v₃² = - 2gR + v₂²

Since v₂² = 2gh, we have:
v₃² = 2gh - 2gR

v₃² = 2gh - 2gR

1. At the point 4:

The vector relationship is:
N + W = Fc
Their components along + y-axis gives:
- N - mg = - Fc, that is N + mg = Fc, or
N + mg = mv₄²/R, Thus:
N = mv₄²/R - mg

N = mv₄²/R - mg

Applying the work-energy theorem (between the two points 4 and 2) gives:
(1/2) m (v₄² - v₂²) = - mg (2R - 0)
or
(v₄² - v₂²) = - 4Rg
v₄² = v₂² - 4Rg


Since v₂² = 2gh, we have:
v₄² = 2gh - 4Rg

v₄ = (2gh - 4Rg)^1/2

At this point, to still have contact between the sled and the circular support, we must have N > 0, that is:
mv₄²/R - mg > 0 or
v₄² > Rg


That is:
h > 5R/2