IIT JEE , AIEEE Forum - Physics , Mechanics

neeraj_agarwal_1990

A conical pendulum .
(a) The horizontal component of angular mom. of mass about the point of supprt P is approximately 2.9 kg-m^2/s
(b) The vertical component of angular mom. of mass about about the point of support P is approx. 1.7
(c) Mag. of dL/dt (L=angular mom. vector of mass about P) is approx 10
(d) torque = dL(vector)/dt will not hold gud in this case.
choose correct options.(i think mass shud be given)

[ans:A,B,D]

neeraj_agarwal_1990

angle wid vertical is 30 and length is 1m

elessar_iitkgp

Let the tension in the string be T. Then,
Tcos

= mg
Tsin

= mv²/L
Dividing these two,
v² = gLsin

----------------(1)

You can see here that at each instant, the direction of angular momentum changes. Further, the angular momentum vector rotates about the point of suspension, tracing out a cone. This type of motion is called precession. For precession, the vertical component of angular momentum = troque.
ie,
dL_z/dt =

L_z = I

The other parts can be solved easily. Nudge me if you have any doubts in those too.

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