Answer to AP Physics C Free Response Practice Question Involving Angular Momentum

In the post dated 4^th March 2009, I had given a free response practice question involving the concepts of moment of inertia, angular momentum and elastic collision for the benefit of AP Physics C aspirants. As promised, I give below the answer along with the question:

An object of mass m collides elastically with the lower end B of a thin uniform rod AB of mass 3m and length L suspended vertically using a frictionless hinge at its upper end A (Fig.) so that it can rotate in a vertical plane. The only external force present is that of earth’s gravity. Just before collision the object was moving horizontally with speed v. Now answer the following questions:

(a) Given that the moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its middle and perpendicular to its length is ML²/12, determine the moment of inertia of the rod AB about the hinge.

(b) Calculate the horizontal speed of the object of mass m after the collision.

(c) Suppose the colliding object were of the same mass (3m) as that of the rod. Calculate the distance d from the hinge at which the colliding object (moving with the horizontal speed v) should hit the rod so that its kinetic energy is fully transferred to the rod.

(d) If the collision at the position obtained in part (c) is such that the colliding object gets attached to the rod, calculate (in terms of the given parameters) the angular velocity with which the rod starts moving after the collision

(a) By parallel axis theorem the moment of inertia I of the rod about the hinge is given by

I = ML²/12 + M(L/2)² since the distance of the parallel axis is L/2.

Here M = 3m so that I = (3mL²/12) + 3mL²/4 = 3mL²/3 = mL²

(b) Since the angular momentum as well as kinetic energy are conserved in elastic collisions we have, on equating the angular momenta,

mvL =mv₁L + I ω

where ω is the angular velocity of the rod and v₁ is the horizontal speed of the object of mass m immediately after the collision.

[There is external force (gravity) and hence we will be justified only if we equate angular momenta just before and just after the collision].

Since I = mL² the above equation becomes

mvL =mv₁L + mL²ω

Therefore, v = v₁ + Lω from which ω = (v – v₁)/L

Equating the kinetic energies immediately before and after the collision, we have

½ mv² = ½ mv₁² + ½ Iω²

Substituting for I and ω we have

½ mv² = ½ mv₁² + ½ mL² [(v – v₁)/L]²

Therefore, v² = v₁² +(v – v₁)²

This gives v₁² = vv₁.

Since v₁ cannot be equal to v the above equation is satisfied only if v₁ = 0

[So the colliding object will have its horizontal momentum killed by the collision and it will fall vertically downwards].

(c) The situation is shown in the adjoining figure. Since the colliding object transfers the entire kinetic energy to the rod, its velocity immediately after the collision is zero. The collision is again elastic. Let ω₁ be the angular velocity with which the rod starts rotating in this case.

Equating angular momenta just before and just after the collision, we have

(3m)vd = Iω₁ = mL²ω₁ so that ω₁ = 3vd/L²

Equating kinetic energies just before and just after the collision, we have

½ ×(3m)v² = ½ Iω₁² = ½ × mL² (3vd/L²)²

Or, 3d²/L² = 1 from which d = L/√3

(d) Since the colliding object gets attached to the rod, this is a case of inelastic collision in which case also angular momentum is conserved.

[Remember that kinetic energy is not conserved in inelastic collisions].

Equating angular momenta just before and just after the collision, we have

(3m)vL/√3 =[3m(L/√3)² + mL²]ω₂ where ω₂ is the angular velocity with which the rod starts rotating in this case.

[The quantity in the square bracket on the right hand side is the total moment of inertia of the system after the object gets attached to the rod].

Therefore, √3 v = 2Lω₂ from which ω₂ = √3 v/2L.