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Chapter 1

Q. 1.7

The slender rod of mass m of Figure 1.20 is swinging through a vertical position with an angular velocity \omega _{1} when it is struck at A by a particle of mass m/4 moving with a speed v_{p}. Upon impact the particle sticks to the bar. Determine (a) the angular velocity of the bar and particle immediately after impact, (b) the maximum angle through which the bar and particle will swing after impact, and (c) the angular acceleration of the bar and particle when they reach the maximum angle.

 

Step-by-Step

Verified Solution

(a) Let t_{1} occur immediately before impact and t_{2} occur immediately after impact. Consider the bar and the particle as a system. During the time of impact, the only external impulses are due to gravity and the reactions at the pin support. The principle of impulse and momentum is used in the following form:

\binom{External  angular  impulses  }{ about  \omicron  between  t_{1}  and  t_{2}} = \binom{Angular  momentum }{ about  \omicron  at   t_{2}} – \binom {Angular  momentum }{ about  \omicron  at  t_{1}}

Using the momentum diagrams of Figure 1.20(b), this becomes

0=\left(m\frac{L}{2} \omega _{2} \right) \left(\frac{L}{2}\right) +\left(\frac{m}{4}a\omega _{2}  \right) \left(a\right)+ \frac{1}{12}mL^{2}   \omega _{2}

-\left[\left(m\frac{L}{2}\omega _{1}  \right)\left(\frac{L}{2} \right)- \left(\frac{m}{4}v_{ p} \right)\left(a\right)  +\frac{1}{12}m L^{2}\omega _{1}   \right]          (a)

 

which is solved to yield

\omega _{2} =\frac{4L^{2}\omega _{1}-3v_{p} a  }{4L^{2} +3a^{2} }        (b)

 

(b) Let t_{3} be the time when the bar and particle assembly attains its maximum angle. Gravity forces are the only external forces that do work; hence conservation of energy applies between t_{2} and t_{3}. Thus, from Equation (1.45),

T_{2} +V_{2} =T_{3} +V_{3}

T_{A} +V_{ A } =T_{B} +V_{B}         (1.45)

 

The potential energy of a gravity force is the magnitude of the force times the distance its point of application is above a horizontal datum plane. Choosing the datum as the horizontal plane through the support, using Equation (1.38) for the kinetic energy of a rigid body, and noting T_{3} = 0 yields

\frac{1}{2} m\left(\frac{L}{2}\omega_{2}  \right)^{2}+\frac{1}{2}\frac{1}{12}mL^{2}\omega ^{2}_{2}+ \frac{1}{2}\frac{m}{4} \left(a\omega _{2} \right) ^{2} -mg\frac{L}{2}-\frac{mg}{4}a

=-mg\frac{L}{2}\cos \theta _{\max } -\frac{m}{4}ga\cos \theta _{\max }          (d)

which is solved to yield

 \theta _{\max } =\cos^{-1}\left[1-\frac{\left(4L^{2}+3a^{2} \right) \omega ^{2}_{2}}{g\left(12L+6a \right) } \right]          (e)

(c) The bar attains its maximum angle at t_{3}, \omega _{3} = 0. Summing moments about \pmb{\omicron} using the free-body diagrams of Figure 1.20(c) assuming moments and positive clockwise gives

\left(\sum{M_{\omicron } } \right)_{ext} = \left(\sum{M_{\omicron } } \right)_{eff}            (f)

– \left(mg\right) \left(\frac{L}{2}\sin \theta _{\max } \right) -\left(\frac{mg}{4}\left(a\sin \theta _{\max } \right)  \right)

=\left(m\frac{L}{2} \alpha \right) \left(\frac{L}{2}\right) +\left(\frac{mg}{4}a\alpha  \right)\left(a\right) +\frac{1}{12}mL^{2}\alpha             (g)

which is solved to yield

\alpha =-\frac{\left(6L+3a\right)g\sin \theta _{\max }  }{4L^{2}+3a^{2} }           (h)

T=\frac{1}{2}m\overline{v} ^{2} +\frac{1}{2}\overline{I} \omega ^{2}           (1.38)