A horizontal disc is freely totating about a transverse axis passing through its centre at the rate of 100 revolutions per minute. A 20 gram blob of wax falls on the disc and sticks to the disc at a distance of 5 cm from its axis. Moment of inertia of the disc about its axis passing through its centre of mass is 2×10-4 kg m2. Calculate the new frequency of rotation of the disc.
Answers
The new frequency of rotation of the disc is 80 r±.
Explanation:
I1 = Moment of inertia of disc = 2×10−4 kgm^2
I2 = Moment of inertia of the disc+moment of inertia of the bob of wax on the disc.
I2 = 2 × 10^−4 + mr^2 = 2 × 10^−4 + 20 × 10^−3 (0.05)^2
I2 = 2 × 10^−4 + 0.5 × 10^−4 = 2.5 × 10^−4 kgm^2
By the principle of conservation of angular momentum
I1n1 = I2n2
⇒ 2 × 10^−4 × 100 = 2.5×10^−4 n2
n2 = 100 ×22.5 = 80 r±
Thus the new frequency of rotation of the disc is 80 r±.
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Particle is executing shm completes 300 oscillations in 2.5 min.It's maximum acceleration is 314 m/s^2.Calculate frequency,time period,and maximum velocity?
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Answer :-
The new frequency of rotation of the disc is 80 r±.
The new frequency of rotation of the disc is 80 r±.Explanation:
I1 = Moment of inertia of disc = 2×10−4 kgm^2
I2 = Moment of inertia of the disc+moment of inertia of the bob of wax on the disc.
I2 = 2 × 10^−4 + mr^2 = 2 × 10^−4 + 20 × 10^−3 (0.05)^2
I2 = 2 × 10^−4 + 0.5 × 10^−4 = 2.5 × 10^−4 kgm^2
By the principle of conservation of angular momentum
I1n1 = I2n2
⇒ 2 × 10^−4 × 100 = 2.5×10^−4 n2
n2 = 100 ×22.5 = 80 r±
Thus the new frequency of rotation of the disc is 80 r±.
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Particle is executing shm completes 300 oscillations in 2.5 min.It's maximum acceleration is 314 m/s^2.Calculate frequency,time period,and maximum velocity?
https://brainly.in/question/12745295