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A 2 kg solid disk with a radius of 0.22 m has a tangential force of 300N applied to it.
a) What is the torque acting on the disk?
b) What is the moment of inertia of the disk?
c) What angular acceleration is produced by the torque?
d) If the disk starts from rest and the acceleration is constant for 3.0s, what is the angular velocity of the disk at the end of 3.0s?
e) Through what angle in radians has the disk rotated during this time?
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Answer :

ONYEBUCHINNAJIGO

(a) The torque acting on the disk is 66 Nm.

(b) The moment of inertia of the disk is 0.05 kgm².

(c) The angular acceleration is produced by the torque is 1,320 rad/s².

(d) The final angular velocity of the disk is  3,960 rad/s.

(e) The angle of rotation of the disk is 5,940 rad.

Torque acting on the disk

The torque acting on the disk is calculated as follows;

τ = Fr

τ = 300 x 0.22

τ = 66 Nm

Moment of inertia

The moment of inertia of a solid disk is calculated as follows;

I = ¹/₂MR²

I = ¹/₂ x 2 x (0.22)²

I = 0.05 kgm²

Angular acceleration of the disk

The angular acceleration of the disk is calculated as follows;

τ = Iα

[tex]\alpha = \frac{\tau }{I} \\\\\alpha = \frac{66}{0.05} \\\\\alpha = 1,320 \ rad/s^2[/tex]

Angular velocity of the disk after 3 s

ωf = ωi + αt

ωf = 0 + (1320 x 3)

ωf = 3,960 rad/s

Angle of rotation of the disk

ωf² = ωi²+ 2αθ

(3,960)² = 0 + 2(1320)θ

θ = (3,960²) / (2 x 1320)

θ = 5,940 rad

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