IIT JEE , AIEEE Forum - Physics , Mechanics

computer001

done

using cos rule v find tht the length is L/2
w=2V/L
look @ fig:
so I=mL^2/3
Ke=1/2 * mL^2/3 * 4V^2/l^2
2/3 MV^2

im sure this time

karthik2007

doesn't match with the options given....

computer001

tht aint my fault...find a mistake in it then

karthik2007

The fault aint mine either....

I don't see any faults in the qn btw.

Tom_Bombadil

The answer is (5/3)mv², or Option D.

Consider the fact that the velocity of any point in a system having combined rotational and translational motion is the vector sum of the velocity of the center of mass and wr, where r is the distance of that point from the center of mass and w is the angular velocity.

Once the velocity of point A is found with respect to the axes, apply this idea to points A and B. Equations relating V₀ , V_CM and w are readily obtained.

Energy, finally, is the sum of linear and rotational kinetic energies.

karthik2007

buddy - please give a full solution?

ganesha1991

karthik bro it wud be great to know how u solved this problem and i still dont get why mine is wrong

karthik2007

buddy, I don't know who is right and who is wrong... so many replies. And I am not sure of what I have done. Looks like Tom is right... waiting for his reply

ganesha1991

where did u get this problem from

karthik2007

friend gave it from his JEE workbook

Tom_Bombadil

Velocity of point A = Sum of velocity of COM and wL/2, where wL/2 is in a direction perpendicular to the rod and pointing toward positive X axis.

Similarly, Velocity of point B = Sum of velocity of COM and wL/2, where wL/2 is in a direction perpendicular to the rod and pointing toward negative X axis.

Consider the components of V_COMalong X and Y (V_C,X and V_C,Y), and also consider that at this instant, the rod makes an angle pi -

with the X axis.

The equations become:

For point B:

V_C,X - w(L/2)sin(theta) = V₀

V_C,Y = w(L/2)cos(theta)

For point A, they become

V_C,X + w(L/2)sin(theta) = 2V₀

V_C,Y + w(L/2)cos(theta) = V₀sqrt(3)

Upon solving these:

V_C = V₀sqrt(3)

wL = 2V₀

Put these in the energy equation: (1/2)m(V₀)² + (1/2)Iw²

Tom_Bombadil

Just to avoid any possible confusion, the direction of wL/2 is really perpendicular to the rod, and it points in the "general" direction of +X for point A, and in the "general" direction of -X for point B

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