Question

How far does the stick's center of mass move during one rotation of the stick?

• A. $\frac{2}{3}\pi l$
• B. $\frac{5}{6}\pi l$
• C. $\frac{6}{5}\pi l$
• D. $\frac{5}{6}{\pi }^{2}l$

Answer 1

The correct option is B $\frac{5}{6}\pi l$

The new center of mass of the stick-puck system w.r.t. center of stick is

$r_{cm}=\frac{m_s\left(l/2\right)}{m_s+m_p}=\frac{m\left(d/2\right)}{m+m}=l/4$ (given- $m_s=m_p=m$)

velocity of center of mass of stick-puck system will be ,

$v_{cm}=\frac{m{v}_i}{m+m}=\frac{v_i}{2}$ .........................eq1

Before collision , the total angular momentum of the stick-puck system ,

$L_1=rp+0$ (as stick is stationary, hence its angular momentum is zero)

or $L_1=\left(l/4\right)\left(m{v}_i\right)$ ..............eq2

After collision, angular momentum of stick-puck system ,

$L_2=I_{cm}^{\prime }\omega$ ...............eq3

by law of conservation of angular momentum ,

$L_2=L_1$

or $I_{cm}^{\prime }\omega =\left(l/4\right)\left(m{v}_i\right)$

or $\omega =\frac{\left(l/4\right)\left(m{v}_i\right)}{I_{cm}^{\prime }}$ ........................eq4

where $I_{cm}^{\prime }=$ moment of inertia of stick-puck system about new center of mass,

$I_{cm}^{\prime }=I_s+I_p$ ,

now , $I_s=I_{cm}+m(d/4{)}^2$ (by parallel axis theorem)

and $I_p=m(l/4{)}^2$

hence $I_{cm}^{\prime }=I_{cm}+m(l/4{)}^2+m(l/4{)}^2=I_{cm}+m\left(l^2/8\right)$

but $I_{cm}=m\left(l^2/12\right)$

therefore $I_{cm}^{\prime }=m\left(l^2/12\right)+m\left(l^2/8\right)=m\left(5l^2/24\right)$ ................eq5

putting the value of $I_{cm}^{\prime }$ in eq4 , we get

$\omega =\frac{l/4\left(m{v}_i\right)}{m\left(5l^2/24\right)}=\frac{6v_i}{5l}$

therefore , distance travelled in one rotation by center of mass ,

$d=v_{cm}T$ ($T=2\pi /\omega$ time-period of rotation)

or $d=\frac{v_i\times 2\pi }{2\times \omega }=\frac{v_i\times \pi }{\left(6v_i/5l\right)}=5\pi l/6$