Object Sliding Down a Hemisphere

A Classical Mechanics Analysis

Let the hemisphere have radius r. The particle starts from rest at the very top and slides down the smooth surface. Let \theta be the angle between the vertical axis (through the centre of the hemisphere) and the line from the centre to the particle. So \theta = 0 at the top and increases as it slides down.

1 Forces acting on the particle

At any position on the sphere, the forces are:

  • Weight mg acting vertically downward
  • Normal contact force N acting radially outward from the surface

While the object is sliding along the surface of the sphere, we can initially think of its motion as due to a centripetal force where the radial direction is toward the centre of the sphere. The particle requires a centripetal acceleration

a_r = \frac{v^2}{r}

directed towards the centre.

However, this centrifugal force, and resulting acceleration, is due to two forces acting on the object: the weight always vertically down and the normal contact force 'pushing' on the object. The maths is easier if we consider the direction we resolve the forces to be along a line from the centre of the base of the hemisphere to the point of contact with the object. As a result the component of the weight along this axis varies with \theta as the object slides. The Normal contact force is always in line with this radial axis but its size is varying as the object slides, see below.

Resolving forces in the radial (towards centre) direction:

The component of weight towards the centre is:

mg\cos\theta

The normal force N acts away from the centre, so the radial equation of motion is:

mg\cos\theta - N = \frac{mv^2}{r}
θ mg mg cos θ N mg sin θ Centre Horizontal diameter (base)

2 Condition for leaving the surface of sphere

The particle leaves the surface when the contact force becomes zero:

N = 0

Subtly important: the surface cannot "pull" the particle. It can only push. Once N=0, the sphere loses contact and the particle follows projectile motion.

So at the leaving point:

mg\cos\theta = \frac{mv^2}{r}
\Rightarrow v^2 = gr\cos\theta \quad \text{(1)}

3 Finding the speed using energy

Because the surface is smooth, mechanical energy is conserved.

Loss in gravitational potential energy = gain in kinetic energy.

Vertical drop from the top to angle \theta is:

h = r(1 - \cos\theta)

So,

\frac{1}{2}mv^2 = mg r(1 - \cos\theta)
\Rightarrow v^2 = 2gr(1 - \cos\theta) \quad \text{(2)}

4 Determining the angle where the particle leaves the sphere

Now combine the force result (1) and the energy result (2):

gr\cos\theta = 2gr(1 - \cos\theta)

Cancel gr:

\cos\theta = 2(1 - \cos\theta)
\cos\theta = 2 - 2\cos\theta
3\cos\theta = 2
\cos\theta = \frac{2}{3}
\boxed{\theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ}

This is the angle (from the vertical centre line) at which the object loses contact with the hemisphere.

5 Speed as the particle leaves the surface

(a) Using the force argument

At separation, N=0, so:

mg\cos\theta = \frac{mv^2}{r}

Substitute \cos\theta = \frac{2}{3}:

v^2 = gr\left(\frac{2}{3}\right)
\boxed{v = \sqrt{\frac{2gr}{3}}}

(b) Using the energy argument

From energy:

v^2 = 2gr(1 - \cos\theta)

Substitute \cos\theta = \frac{2}{3}:

v^2 = 2gr\left(1 - \frac{2}{3}\right) = 2gr\left(\frac{1}{3}\right)
v^2 = \frac{2gr}{3}
\boxed{v = \sqrt{\frac{2gr}{3}}}

6 Key physical interpretation (exam insight)

  • • While the particle is in contact, the normal force provides part of the centripetal force.
  • • As it slides down, speed increases, so required centripetal force \frac{mv^2}{r} increases.
  • • Meanwhile, the radial component of weight mg\cos\theta decreases.
  • • The particle leaves the surface exactly when the sphere can no longer supply any normal reaction, i.e. N = 0.
  • • After this point, gravity alone cannot provide enough centripetal force to keep the particle on the curved surface, so it detaches and follows projectile motion.

Space reserved for animations and diagrams