An asymmetrical dumbell consists or two balls of mass m1 = m and m2 = 3m connect

Question

An asymmetrical dumbell consists or two balls of mass m1 = m and m2 = 3m connected by a rod. The distance between the centers of the balls it epsilon. The dumbell is initially at rest on a frictionless surface At time (t = 0, two force. F1 = 2Fi + Fj and F2 +Fi +3Fj are applied and remain constant in magnitude and direction regardless of the dumbell's motion The quantities m and F are reference mass and force scales, respectively. We wish to determine the trajectory of the center of mass, as a function of , for the following cases. The rod's mass is negligible. The rod's mass is 1/2m. HINT: Treat the bar as a third particle located at x = 1/2 epsilon, y = 0.

Explanation / Answer

letter in bold indicated vectors .

( a ) if rod mass is negligible ,

r0 com( centre of mass ) = [ m1 r1 + m2 r2 ] / [ m1 + m2 ]

r0 =[ m1*0 + 3m * ( i + 0j ) ] / [ m + 3m ] = ( 3 / 4 ) i ;

a= F1 + F2 / ( m + 3m ) = ( 1 / 4 m ) [ 2F i + F j + Fi + 3F j ] ;

a = ( 0.25 / m ) [ 3F i + 4F j ] ;

dv / dt = ( 0.25F / m ) [ 3 i + 4 j ] ;

dv = ( 0.25F / m ) [ 3 i + 4 j ] dt ;

v = ( 0.25F / m ) [ 3t i + 4t j ] + C ;

v = 0 at t=0 ;

0r 0 = ( 0.25F / m ) [ 3*0 i + 4*0 j ] + C ;

or C = 0 ;

v = ( 0.25F / m ) [ 3t i + 4t j ] ;

dr / dt = ( 0.25F / m ) [ 3t i + 4t j ] ;

dr = ( 0.25F / m ) [ 3t i + 4t j ] dt   ;

r =  ( 0.25F / m ) [ ( 3/2 )t^2 i + 2 t^2 j ] + C ;----------------1

r = ( 3 / 4 ) i at t=0 ,

( 3 / 4 ) i = ( 0.25F / m ) [ ( 3/2 )* 0^2 i + 2 *0^2 j ] + C ;

or C = ( 3 /4 ) i ;

so r = ( 0.25F / m ) [ ( 3/2 )t^2 i + 2 t^2 j ] + ( 3 / 4 ) i   ;

or r = [ 0.375 ( F / m )  t^2 + 0.75 ] i + 0.5 t^2 j ;<---ans

ROD : in the rod case , everything remains same except r0 ;

in case of rod , r com = [ m /2 * ( /2 ) + 3m * ]   / [ m + m/2 + 3m ] = ( 13 /18 ) i ;

from 1

r =  ( 0.25F / m ) [ ( 3/2 )t^2 i + 2 t^2 j ] + C ;

now since r0 = ( 13 / 18 ) i ; at t =0

so

( 13 / 18 ) i = ( 0.25F / m ) [ ( 3/2 )* 0^2 i + 2 * 0 ^2 j ] + C ;

or C = ( 13 / 18 ) i

or r =  ( 0.25F / m ) [ ( 3/2 )t^2 i + 2 t^2 j ] + ( 13 /18 ) i ;

or r =

or r = [ 0.375 ( F / m )  t^2 + 0.722 ] i + 0.5 t^2 j ; < -- ans for rod

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.