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A particle of mass m is moving  along a trajectory given  by x= x0+ a cos , y = y0 +  b sin t The torque, acting on the particle about the origin, at t=0 is :

(a)

−m(xob )

(b)

m(−xob + yoa) )

(c)

+myoa

(d)

Zero


Question ID - 51322 | Toppr Answer

A particle of mass m is moving  along a trajectory given  by x= x0+ a cos , y = y0 +  b sin t The torque, acting on the particle about the origin, at t=0 is :

(a)

−m(xob )

(b)

m(−xob + yoa) )

(c)

+myoa

(d)

Zero

1 Answer - 5876 Votes

3537

Answer Key : (c) -

x= x0+ a cos

y = y0 +  b sin t

⇒  vx = −a sin( t),  vy =b cos( t),      

       ax =−a cos( t),  a=−b sin( t),        

At t=0,  x=x0 +a,          y=y0

                        ax =−a ,        a=0

=m(−a )x y0( )

      =+my0a  .



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