1. A 19-g paper clip is attached to the rim of a phonograph record with a diamet
ID: 1531260 • Letter: 1
Question
1. A 19-g paper clip is attached to the rim of a phonograph record with a diameter of 47 cm, spinning at 2.6 rad/s. What is the magnitude of its angular momentum (in kg m2/s)? Round your answer to the nearest ten-thousandth.
2. A hoop (I = MR2) of mass 3 kg and radius 1 m is rolling at a center-of-mass speed of 10 m/s. An external force does 810 J of work on the hoop. What is the new speed of the center of mass of the hoop (in m/s)? Round your answer to the nearest whole number
.
3. A uniform thin rod is hung vertically from one end and set into small amplitude oscillation. If the rod has a length of 2.7 m, this rod will have the same period as a simple pendulum of length ____cm.
Explanation / Answer
1)
angular momentum = I x omega
Inertia (I) = mr^2 = ((19/1000)x0.3^2) = 0.019 kg x 0.47^2 = 0.0041971 kg/m^2
omega = 2.6 rad/s
so, the magnitude of its angular momentum
=0.0041971 x 2.6 = 10.91 x 10^(-3) kg-m^2/s .......................................................ans
2)
Assuming that the hoop is not skidding then the kinetic energy of rotation is exactly equal to 1/2 m v^2
and so is the kinetic energy of translation.
ie HALF of the energy goes into accelerating the centre of mass.
1/2 * 3 * 10^2 +810 = 1/2 * 3 * v^2
or, 1.5x10^2 + 810= 1.5xv^2
or, v^2=640
or, v= 25.30 m/s..................................................................................................ans
3)
Formula for period of a simple pendulum:
T = 2*Pi*sqrt(Lsimple/g)
Formula for period of a physical pendulum:
T = 2*Pi*sqrt(I/(m*g*dcm)
For a rod about end:
Distance from center of mass to pivot: dcm = L/2
Rotational inertia about pivot: I = 1/3*m*L^2
Thus:
T = 2*Pi*sqrt(1/3*m*L^2/(m*g*L/2))
T = 2*Pi*sqrt(2*L/(3*g))
Compare:
2*Pi*sqrt(2*L/(3*g)) = 2*Pi*sqrt(Lsimple/g)
Solve for Lsimple:
Lsimple = 2*L/3
Thus, Lsimple = 180 cm .......................................................................ans
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