Question

3. A disk 6.0cm in diameter and moment of inertia of 0.015kg-m’ initially at rest at t = 0, is spun up to 720-rpm over 6.0s about an axis through its center of mass. Assume the angular acceleration is constant. a) Find the angular velocity (rad/s) at 6.0s b) Find the angular acceleration (rad/s) c) Find the number of revolutions the disk spins through during that interval. d) What is the mass of the disk? e) What is the change in its kinetic energy (in J) from 0 to 6.0s?

Answer 1

Solution :

Here we have :

Diameter of disk : d = 6 cm = 0.06 m

So, radius of the disk : r = d/2 = 0.03 m

Moment of inertia of disk : I = 0.015 kg m²

Initial angular velocity : ω_i = 0 rad/s (rest)

Final angular velocity : ω_f = 720 rpm = 75.4 rad/s

Time taken : t = 6 sec

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Part (a) Solution :

Angular velocity at 6.0 sec : ω_f = 720 rpm = (720 x 2π rad) / (60 s)= 75.4 rad/s

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Part (b) Solution :

Since, Angular acceleration is given by :

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Part (c) Solution :

Here, Angular displacement of the disk is given by : θ = ω_i t + (1/2) α t²

∴ θ = (0 rad/s)(6 sec) + (1/2)(12.57 rad/s²)(6 sec)²

∴ θ = 226.19 rad = 36 revolutions

{ Since : 12 rev = 2π rad }

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Part (d) Solution :

Now, Moment of inertia of disk is given by : I = (1/2) m r²

∴ (0.015 kg m²) = (1/2) m (0.03 m)²

∴ m = 33.3 kg

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Part (e) Solution :

Since, Kinetic energy of the disk is given by : KE = (1/2) I ω

Thus initial kinetic energy will be given by : KE_i = (1/2) I (ω_i)² = (1/2)(0.015 kg m²)(0 rad/s)² = 0 J

And, Final kinetic energy will be given by : KE_f = (1/2) I (ω_f)² = (1/2)(0.015 kg m²)(75.4 rad/s)² = 42.64 J

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Therefore, Change in KE = KE_f - KE_i = (42.64 J) - (0 J) = 42.64 J

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