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प्रश्न
A flywheel of mass 8 kg and radius 10 cm rotating with a uniform angular speed of 5 rad/sec about its axis of rotation, is subjected to an accelerating torque of 0.01 Nm for 10 seconds. Calculate the change in its angular momentum and change in its kinetic energy.
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उत्तर
Given:
M = 8 kg, R = 10 cm = 0.1 m,
ω1 = 5 rad/s, τ = 0.01 Nm, t = 10 s
To find:
- Change in angular momentum (∆L)
- Change in K.E. (∆K.E.)
Formulae:
- I = `"MR"^2/2`
- τ = `I((ω_2 - ω_1)/t)`
- ∆L = I(ω2 - ω1)
- ∆K.E. = `1/2I(ω_2^2 - ω_1^2)`
Calculation:
From formula (i),
I = `(8 xx (0.1)^2)/2`
= 0.04 kgm2
From formula (ii),
ω2 = `((tau xx t))/I + ω_1`
= `(0.01 xx 10)/0.04 + 5`
= 7.5 rad/s
From formula (iii),
∆L = 0.04(7.5 – 5) = 0.1 kg m2/s
From formula (iv),
∆K.E. = `1/2 xx 0.04 xx (7.5^2 - 5^2)`
= 0.625 J
The change in its angular momentum and change in its kinetic energy is 0.1 kg m2/s and 0.625 J respectively.
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