Dynamics: Work energy applications

Recall that the centripetal force is provided by another force.

1.Ball sliding a long another ball

At the top (position i) v1 = 0

At the position f, the projection of the weight along the tangent, that is the direction of the velocity v_f gives, according to Newton's second law:

mg sin θ = ma_t

a_t is the tangential acceleration of the object.

Along the line from the point f to the center of the circle, we have the vector relation:

N + mg cosθ = F_centripetal.
In magnitude:
+ N - mgcosθ = - F_centripetal

+ N - mgcosθ = - F_centripetal

The instant there is no contact, we have: N = 0 , therefore:
- mgcosθ = - F_centripetal , that is:
mgcosθ = m v_f²/R, or

v_f² = g R cosθ     (1)

Now, Let's choose the origin of the potential energy at the bottom, and apply the conservation of energy:

Mechanical energy at the point initial i = Mechanical energy at the point f

(1/2) m v_i² + mg (2R) = (1/2) m v_f² + mg(R + Rcosθ)
with v_i = 0, we obtain:
mg (2R) = (1/2) m v_f² + mg(R + Rcosθ) or

mgR - mgRcosθ = (1/2) m v_f²     (2)

Using the formal (1), we obtain:
gR - gRcosθ = (1/2) gR cosθ, that is:
1 = (3/2) cosθ
cos θ = 2/3 or θ = 48.2^o

Ball sliding: θ = 48.2^o

This result does not depend on any parameter.

2.Ball rolling a long another ball

Now, if the ball of radius r rolls. Its moment of inertia is I = (2/5)mr².

since v_f = ω r,
then
(1/2)Iω² = (1/2) (2/5) m r² v_f²/r² = (1/5) m v_f²

The relation (2) becomes:

gR - gRcosθ = (1/2) mv_f² + (1/5) m v_f² = (7/5) m v_f²
Using the formula (1), we obtain:
gR - gRcosθ = (7/10) gRcosθ
1 - cosθ = (7/10) cosθ
1 = (17/10) cosθ

cos θ = 10/17 or θ = 54.0^o

Ball rolling: θ = 54.0^o

This result depends on the spherical shape of the object.