translational kinetic energy of a purely rolling solid sphere is 5 jul then the rotational kinetic energy of the system is
Answer (1)
26th Jan, 2022
Hello Aspirant,
Given - Translational K.E of rolling sphere = (1/2) × m × v^2.
= 5 J
So, mv^2 = 10. ----------[Eqn- 1]
To find - Translational K.E
Solution- We know that v = r × ω.
Rotational K.E = (1/2) × I × ω^ 2.
We know that I = (2/5) × m × r ^2.
Hence,
Rotational K.E = (1/2) × (2/5) × m × r ^2 × ω^ 2
= (1/5) × m × r ^2 × ω^ 2
= (1/5) × m× v^2 = (1/5) × 10
----[from Eqn-1]
= 2 J.
I hope it helps
Thank you.
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