Question

Suppose the particle of the previous problem has a mass m and a speed ν before the collision and it sticks to the rod after the collision. The rod has a mass M. (a) Find the velocity of the centre of mass C of the system constituting the rod plus the particle". (b) Find the velocity of the particle with respect to C before the collision. (c) Find the velocity of the rod with respect to C before the collision. (d) Find the angular momentum of the particle and of the rod about the centre of mass C before the collision. (e) Find the moment of inertia of the system about the vertical axis through the centre of mass C after the collision. (f) Find the velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision."

Answer 1

Answer:

serious grasp

Explanation:

gotten husky

Answer 2

The answer to each of the subparts has been provided chronologically as follows :-

(a) The question states that no external torque and force has been applied on the system.

Via the law of conservation of momentum, we have;

$mv$' = ($M$ + $m$) $v$'

⇒ $v$' = $\frac{mv}{M+m}$

(b) The velocity of the particle w.r.t. Centre Of Mass (COM) C before the collision = $v_c = v - v$'

⇒ $v_c = v - \frac{mv}{M + m} = \frac{Mv}{M+v}$

(c) The velocity of the particle w.r.t. COM C before the collision = $-\frac{Mv}{M+ m}$

(d) The distance of the COM from the particle = $x_c_m = \frac{m_1x_1 +m_2 x_2}{m_1+m_2}$

⇒ $r = \frac{M * \frac{L}{2}+ m * 0 }{M + m}$

⇒ $r = \frac{ML}{2(M+m)}$

∴ The angular momentum of the body about COM  = $M$ × $[\frac{mv}{(M+m)} ]$ × $\frac{1}{2} \frac{mL}{M +m}$ = $\frac{Mm^2vL}{2 (M+m)^2}$

(e) The moment of inertia about COM = $I = I_1 + I_2$

⇒ $I = m [\frac{ML}{2(m+M)}]^2 + \frac{ML^2}{12} + M [ \frac{mL}{2(m+M)} ]^2$

⇒ $I = \frac{mM^2L^2}{4(m+M)^2} + \frac{ML^2}{12} + \frac{M m^2L^2}{4(M+m)^2}$

⇒ $I = \frac{3mM^2L^2 + M(m+M)^2L^2 + 3Mm^2L^2 }{12(m+M)^2}$

⇒ $I = \frac{M(M+4m)L^2}{12(M+m)}$

(f) About COM;

$V_c_m = \frac{mv}{M+m}$

∴ $I$ω = $mvr = mv$ × $\frac{ML}{2(M+m)}$

⇒ ω = $\frac{mvML}{2(M+m)}$ × $\frac{12 (M+m)}{M (M +4m)L^2}$ = $\frac{6mv}{(M+4m)L}$