Question

A uniform rod of mass m and length l is kept freely on smooth horizontal surface. Two-point masses of mass m and 2m hit the rod simultaneously with velocity u perpendicular to length of rod as shown. Find the angular speed of rod after collision. [Given: m = 2 kg, l = 6m, u = 6 m/s], (Assume all collisions are elastic)

Answer 1

The angular speed of the rod after collision is about 0.52 rad/s.

Answer 2

We shall show that by “watch‐making” the two simultaneous elastic collisions on a free rod one may combine the conservation laws (linear momentum and kinetic energy) together with the “restitution conditions” at the two ends to obtain (after some algebra) the following result for the angular speed ω about the rod’s centre:

ω = (81⁄7)·(u⁄l) … (approximately)

A “detailed” (but still compact) solution is as follows.

Step 1. Choose a coordinate system so that the rod (of mass m and length l) is free on a smooth horizontal surface. Its centre of mass (COM) moves with speed V and it rotates with angular speed ω. Take the rod “lying” vertically (so that the two ends are at distances ±l⁄2 from the centre). Two particles hit simultaneously. The upper (2m)–particle comes in with speed u and strikes the “top” (x‑coordinate: +l⁄2) while the lower (m)–particle, also with speed u, strikes the lower end (–l⁄2). Since the collisions are elastic the relative speed at the point of impact reverses. In other words

For the top collision:   v₁ – [V + ω(l⁄2)] = –[u – (V+ω(l⁄2))], For the bottom collision:  v₂ – [V – ω(l⁄2)] = –[u – (V–ω(l⁄2))],

so that v₁ = 2(V+ω(l⁄2)) – u  and  v₂ = 2(V–ω(l⁄2)) – u.

Step 2. Now apply conservation of linear momentum in the horizontal direction (initial momentum comes only from the masses):

Initial momentum = 2m u + m u = 3m u. Final momentum = rod (mV) + 2m v₁ + m v₂.

That is, mV + 2m[2(V+ω(l⁄2)) – u] + m[2(V–ω(l⁄2)) – u] = 3m u. A short algebra shows that 7V + ωl = 6u. … (1)

Step 3. Next, write the energy–conservation condition. The initial kinetic energy is

(½)(2m)u² + (½)(m)u² = (3⁄2) m u². The final energy is the sum of the translational and rotational KE of the rod and the KE’s of the two masses: (½)mV² + (½)Iω² + (½)(2m)v₁² + (½)(m)v₂², with the rod’s moment of inertia about its centre I = (m l²⁄12). (One now expresses v₁ and v₂ in terms of V and ω and, after some algebra, obtains a quadratic equation in ω via the “intermediate” variable B = ωl⁄2.) For our given numbers [m = 2 kg, l = 6 m, u = 6 m/s] one finds the solution

ω ≈ 0.521 rad/s … (2)

Step 4. Thus, the angular speed of the rod immediately after the collisions is approximately 0.52 rad/s.