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 A 2 m rod of mass 1 kg rotates at an angular speed of 15 rad/s about its ends. Then K.E. is
  • a)
    150 J
  • b)
    1500 J
  • c)
    100 J
  • d)
    1.5 J
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A 2 m rod of mass 1 kg rotates at an angular speed of 15 rad/s about i...
Given,
 l=2m
m=1kg
ω=15rd//s
K.E.=1/2I ω
=1/2x(ml2/3) ω2
=1/2x(1x4/3)(15)2
=150J
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Most Upvoted Answer
A 2 m rod of mass 1 kg rotates at an angular speed of 15 rad/s about i...
Given,
 l=2m
m=1kg
ω=15rd//s
K.E.=1/2I ω
=1/2x(ml2/3) ω2
=1/2x(1x4/3)(15)2
=150J
Free Test
Community Answer
A 2 m rod of mass 1 kg rotates at an angular speed of 15 rad/s about i...

Explanation:

Given data:
- Mass of the rod, m = 1 kg
- Length of the rod, L = 2 m
- Angular speed, ω = 15 rad/s

Formula:
The kinetic energy (KE) of a rotating object can be calculated using the formula:
KE = 0.5 * I * ω^2

where I is the moment of inertia of the object, which for a rod rotating about its end is given by:
I = (1/3) * m * L^2

Calculations:
- Moment of inertia, I = (1/3) * 1 kg * (2 m)^2 = 4/3 kg m^2
- KE = 0.5 * (4/3) kg m^2 * (15 rad/s)^2
- KE = 0.5 * (4/3) * 225 J
- KE = 150 J

Therefore, the kinetic energy of the rotating rod is 150 J, which corresponds to option 'A'.
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