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# SirPl solve question no-23

Asked by shobhit 23rd June 2018, 12:03 AM
Moment of inertia I of laminar sheet passing through centre of mass and perpendicular to its length is given by,

I = M×L2/12 = 1.2 (given) ......................(1)

we get length L from eqn.(1) as L = 69 cm. Hence Centre of Mass of laminar sheet from AB is 34.5 cm

At first impact at P, speed vi of laminar plate at impact point  is given by, vi = ω×0.5 = 1×0.5 = 0.5 m/s

speed vf at impact point after impact at P is obtained from , M(vi - vf) = 6 ..............(2)

LHS of eqn.(2) is change in momentum , which is due to the impulse given to laminar sheet during impact.
we are given that impulse is 6N-s as given in RHS

solving for vf in eqn.(2) using the values of M and vi, we get vf = 0.3 m/s

Now, laminar sheet moving towards Q with initial speed 0.3 m/s at impact point and making second impact and getting bounce back.
speed after second impact at Q is obtained as we did above, because same impulse is given at Q also.
we get the speed at impact point after second impact at Q  as 0.1 m/s.
Hence when the laminar sheet is returning to P with angular speed ( ω = v/r) 0.1/0.5 = 0.2 rad/s

When third impact is happening at P, as calculated before,  laminar sheet will loose speed 0.2 m/s at impact point.
But before third impact the initial speed itself 0.1 m/s
hence laminar sheet will come to rest at third impact happening at P

Answered by Expert 24th June 2018, 5:45 PM
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