A conical pendulum is a weight or bob fixed on the end of a string suspended fro

Question

A conical pendulum is a weight or bob fixed on the end of a string suspended from a point. It moves in a horizontal circular path, as shown in the diagram below with the string at a constant angle theta to the vertical. The pendulum bob has a mass of 0.350 kg, the length of the pendulum is 0.690 m, and the angle that it swings at is 15.0 degree. What is the angular momentum of the mass about the vertical if it moves in the clockwise direction when viewed from above? We use the standard rectangular coordinate system with +x axis to the right, +y axis vertically up, and +z axis out toward you. Enter your answer using unit vector notation. (Take the angular momentum of the mass to be about the center of the circular path.) L = kg middot m^2/s

Explanation / Answer

angular momentum=cross product of r and p

where r =position vector

p=momentum vector

here momentum vector is tangential to the circular path

and position vector is from O to any given point on the circular path

let at any given moment,

mass m is at (x,0,z) coordinate.

O is at (0,l*cos(theta),0)=(0,0.6665,0) m

let tension in the string be T.

then balancing vertical forces:

T*cos(theta)=m*g

==>T=0.35*9.8/cos(15)=3.551 N

balancing forces in horizontal direction:

T*sin(theta)=mass*speed^2/radius

==>3.551*sin(15)=0.35*speed^2/(0.69*sin(15))

==>speed=0.6848 m/s

unit vector along the momentum vector =(-z,0,x)/sqrt(z^2+x^2)

momentum vector=0.35*0.6848*(-z,0,x)/sqrt(z^2+x^2)

but as z^2+x^2=radius of circle^2=(l*sin(theta))^2=0.031892

==>momentum vector=1.3421*(-z,0,x)

position vector =r=(x,0,z)-(0,0.6665,0)=(x,-0.6665,z)

cross product of r and p:

determinant of the below matrix:

i       j       k
x -0.6665       z
-1.3421*z 0 1.3421*x

=i*(-0.6665*1.3421*x)-j*(1.3421*x^2+1.3421*z^2) + k*(-1.3421*0.6665*z)

=(-0.8945*x) i - 0.0428 j -( 0.8945*z) k

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