Question

A small bob is attached to a cord of length 1.2 m and is released from rest when A= 5º. Take d= 0.30 m. d 1.2 m Oce с B A 5. Required information Determine the amplitude 6c (Round the final answer to two decimal places.) The amplitude ec is

Answer 1

Solution:

Principle Used: Conservation of angular momentum

Analysis : Let the system of bob and string form a system. As the this system rotates about the fixed point in the wall the angular momentum of the system is conserved as no external torque acts on it. Now from the conservation of energy the angular velocity of bob just before collision

h = R(1-cos(theta)) = 1.2 (1-cos(5)) = 4.56E-3

PE = Angular KE

mgh = 0.5Iw²

Now the Mmoment of Inertia = ml² = m(1.2)² = 1.44m

w = 0.24925 rad/s

Now when the string sticks to the wall the angular moment remains conserved as the MOI changes to = m(l-d)² = 0.81m

I1w1 = I2w2

w2 = 0.44311 rad/s

Now from the principal of conservation of energy this angular KE will be converted into PE

0.5I2W2^{2 =} mgh'

h'=(l-d)(1-cos(theta')) = 8.106E-3

theta' = theta(c) = 7.695 degrees.

Result: Tha amplitude of theta_c is 7.7 degrees.

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